Rectilinear Kinematics

Rectilinear Kinematics
. Rectilinear Kinematics Hibbler Class Objectives: Students will be able to: Find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path

1. In dynamics, a particle is assumed to have _________.
READING QUIZ 1. In dynamics, a particle is assumed to have _________. A) both translation and rotational motions B) only a mass C) a mass but the size and shape cannot be neglected D) no mass or size or shape, it is just a point 2. The average speed is defined as __________. A) Dr/Dt B) Ds/Dt C) sT/Dt D) None of the above. 1. B 2. C

APPLICATIONS The motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles. Why? If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration?

APPLICATIONS (continued)
A sports car travels along a straight road. Can we treat the car as a particle? If the car accelerates at a constant rate, how can we determine its position and velocity at some instant?

An Overview of Mechanics
Mechanics: The study of how bodies react to forces acting on them. Statics: The study of bodies in equilibrium. Dynamics: 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics - concerned with the forces causing the motion

RECTILINEAR KINEMATICS: CONTINIOUS MOTION
(Section 12.2) A particle travels along a straight-line path defined by the coordinate axis s. The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft). The displacement of the particle is defined as its change in position. Vector form:  r = r’ - r Scalar form:  s = s’ - s The total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.

The instantaneous velocity is the time-derivative of position.
Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s. The average velocity of a particle during a time interval t is vavg = r / t The instantaneous velocity is the time-derivative of position. v = dr / dt Speed is the magnitude of velocity: v = ds / dt Average speed is the total distance traveled divided by elapsed time: (vsp)avg = sT / t

ACCELERATION Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2. The instantaneous acceleration is the time derivative of velocity. Vector form: a = dv / dt Scalar form: a = dv / dt = d2s / dt2 Acceleration can be positive (speed increasing) or negative (speed decreasing). As the book indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

SUMMARY OF KINEMATIC RELATIONS:
RECTILINEAR MOTION • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds • Integrate acceleration for velocity and position. Velocity: ò = t o v dt a dv s ds or ò = t o s dt v ds Position: • Note that so and vo represent the initial position and velocity of the particle at t = 0.

CONSTANT ACCELERATION
The three kinematic equations can be integrated for the special case when acceleration is constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2 downward. These equations are: t a v c o + = yields ò dt dv 2 c o s t (1/2) a v + = yields ò dt ds ) s - (s 2a (v v o c 2 + = yields ò ds a dv

EXAMPLE Given: A particle travels along a straight line to the right with a velocity of v = ( 4 t – 3 t2 ) m/s where t is in seconds. Also, s = 0 when t = 0. Find: The position and acceleration of the particle when t = 4 s. Plan: Establish the positive coordinate, s, in the direction the particle is traveling. Since the velocity is given as a function of time, take a derivative of it to calculate the acceleration. Conversely, integrate the velocity function to calculate the position.

ò EXAMPLE (continued) Solution:
1) Take a derivative of the velocity to determine the acceleration. a = dv / dt = d(4 t – 3 t2) / dt =4 – 6 t => a = – 20 m/s2 (or in the  direction) when t = 4 s 2) Calculate the distance traveled in 4s by integrating the velocity using so = 0: v = ds / dt => ds = v dt => => s – so = 2 t2 – t3 => s – 0 = 2(4)2 – (4)3 => s = – 32 m ( or ) ò = t o s (4 t – 3 t2) dt ds

CONCEPT QUIZ t = 2 s t = 7 s 3 m/s 5 m/s
1. A particle moves along a horizontal path with its velocity varying with time as shown. The average acceleration of the particle is _________. A) 0.4 m/s2 B) 0.4 m/s2 C) 1.6 m/s2 D) 1.6 m/s2 2. A particle has an initial velocity of 30 ft/s to the left. If it then passes through the same location 5 seconds later with a velocity of 50 ft/s to the right, the average velocity of the particle during the 5 s time interval is _______. A) 10 ft/s B) 40 ft/s C) 16 m/s D) 0 ft/s Answers: 1. D 2. D

GROUP PROBLEM SOLVING Given: Ball A is released from rest at a height of 40 ft at the same time that ball B is thrown upward, 5 ft from the ground. The balls pass one another at a height of 20 ft. Find: The speed at which ball B was thrown upward. Plan: Both balls experience a constant downward acceleration of 32.2 ft/s2 due to gravity. Apply the formulas for constant acceleration, with ac = ft/s2.

GROUP PROBLEM SOLVING (continued)
Solution: 1) First consider ball A. With the origin defined at the ground, ball A is released from rest ((vA)o = 0) at a height of 40 ft ((sA )o = 40 ft). Calculate the time required for ball A to drop to 20 ft (sA = 20 ft) using a position equation. sA = (sA )o + (vA)o t + (1/2) ac t2 So, 20 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t2) => t = s

Solution: 2) Now consider ball B. It is throw upward from a height of 5 ft ((sB)o = 5 ft). It must reach a height of 20 ft (sB = 20 ft) at the same time ball A reaches this height (t = s). Apply the position equation again to ball B using t = 1.115s. sB = (sB)o + (vB)ot + (1/2) ac t2 So, 20 ft = 5 + (vB)o(1.115) + (1/2)(-32.2)(1.115)2 => (vB)o = 31.4 ft/s

1. A particle has an initial velocity of 3 ft/s to the left at
ATTENTION QUIZ 1. A particle has an initial velocity of 3 ft/s to the left at s0 = 0 ft. Determine its position when t = 3 s if the acceleration is 2 ft/s2 to the right. A) 0.0 ft B) 6.0 ft C) 18.0 ft D) 9.0 ft 2. A particle is moving with an initial velocity of v = 12 ft/s and constant acceleration of 3.78 ft/s2 in the same direction as the velocity. Determine the distance the particle has traveled when the velocity reaches 30 ft/s. A) 50 ft B) 100 ft C) 150 ft D) 200 ft Answers: 1. A 2. B

RECTILINEAR KINEMATICS: ERRATIC MOTION
Today’s Objectives: Students will be able to: Determine position, velocity, and acceleration of a particle using graphs. In-Class Activities: Check Homework Reading Quiz Applications s-t, v-t, a-t, v-s, and a-s diagrams Concept Quiz Group Problem Solving Attention Quiz

READING QUIZ 1. The slope of a v-t graph at any instant represents instantaneous A) velocity. B) acceleration. C) position. D) jerk. 2. Displacement of a particle in a given time interval equals the area under the ___ graph during that time. A) a-t B) a-s C) v-t C) s-t Answers: 1. B 2. C 19

APPLICATION In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the tank truck, how can we determine its acceleration at position s = 1500 feet?

ERRATIC MOTION (Section 12.3)
Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics. The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve. 21

S-T GRAPH Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt). Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph. 22

V-T GRAPH Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv/dt). Therefore, the acceleration vs. time (or a-t) graph can be constructed by finding the slope at various points along the v-t graph. Also, the distance moved (displacement) of the particle is the area under the v-t graph during time t. 23

A-T GRAPH Given the acceleration vs. time or a-t curve, the change in velocity (v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle. 24

½ (v1² – vo²) = = area under the ò s2
A-S GRAPH A more complex case is presented by the acceleration versus position or a-s graph. The area under the a-s curve represents the change in velocity (recall ò a ds = ò v dv ). a-s graph ½ (v1² – vo²) = = area under the ò s2 s1 a ds This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance. 25

Thus, we can obtain an a-s plot from the v-s curve.
V-S GRAPH Another complex case is presented by the velocity vs. distance or v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. Recall the formula a = v (dv/ds). Thus, we can obtain an a-s plot from the v-s curve. 26

Given: The s-t graph for a sports car moving along a straight road.
EXAMPLE Given: The s-t graph for a sports car moving along a straight road. Find: The v-t graph and a-t graph over the time interval shown. What is your plan of attack for the problem? 27

when 0 < t < 5 s; v0-5 = ds/dt = d(3t2)/dt = 6 t m/s
EXAMPLE (continued) Solution: The v-t graph can be constructed by finding the slope of the s-t graph at key points. What are those? when 0 < t < 5 s; v0-5 = ds/dt = d(3t2)/dt = 6 t m/s when 5 < t < 10 s; v5-10 = ds/dt = d(30t75)/dt = 30 m/s v-t graph v(m/s) t(s) 30 5 10 28

when 0 < t < 5 s; a0-5 = dv/dt = d(6t)/dt = 6 m/s2
EXAMPLE (continued) Similarly, the a-t graph can be constructed by finding the slope at various points along the v-t graph. when 0 < t < 5 s; a0-5 = dv/dt = d(6t)/dt = 6 m/s2 when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2 a-t graph a(m/s2) t(s) 6 5 10 29

1. If a particle starts from rest and
CONCEPT QUIZ 1. If a particle starts from rest and accelerates according to the graph shown, the particle’s velocity at t = 20 s is A) 200 m/s B) 100 m/s C) D) 20 m/s Answers: 1. B 2. D 2. The particle in Problem 1 stops moving at t = _______. A) 10 s B) 20 s C) 30 s D) 40 s 30

Given: The v-t graph shown.
GROUP PROBLEM SOLVING Given: The v-t graph shown. Find: The a-t graph, average speed, and distance traveled for the s interval. Plan: Find slopes of the v-t curve and draw the a-t graph. Find the area under the curve. It is the distance traveled. Finally, calculate average speed (using basic definitions!). 31

Solution: Find the a–t graph: For 0 ≤ t ≤ a = dv/dt = 1.0 m/s² For 30 ≤ t ≤ a = dv/dt = -0.5 m/s² a-t graph -0.5 1 a(m/s²) 30 90 t(s) 32

Now find the distance traveled: Ds0-30 = ò v dt = (1/2) (30)2 = m Ds30-90 = ò v dt = (1/2) (-0.5)(90)2 + 45(90) – (1/2) (-0.5)(30)2 – 45(30) = 900 m s0-90 = = m vavg(0-90) = total distance / time = / 90 = 15 m/s 33

2. Select the correct a-t graph for the velocity curve shown.
ATTENTION QUIZ 1. If a car has the velocity curve shown, determine the time t necessary for the car to travel 100 meters. A) 8 s B) 4 s C) 10 s D) 6 s t v 6 s 75 2. Select the correct a-t graph for the velocity curve shown. A) B) C) D) a t t v Answers: 1. B 2. D 34

GENERAL & RECTANGULAR COMPONENTS
CURVILINEAR MOTION: GENERAL & RECTANGULAR COMPONENTS Today’s Objectives: Students will be able to: Describe the motion of a particle traveling along a curved path. Relate kinematic quantities in terms of the rectangular components of the vectors. In-Class Activities: • Check Homework • Reading Quiz • Applications • General Curvilinear Motion • Rectangular Components of Kinematic Vectors • Concept Quiz • Group Problem Solving • Attention Quiz

A) tangent to the hodograph. B) perpendicular to the hodograph.
READING QUIZ 1. In curvilinear motion, the direction of the instantaneous velocity is always A) tangent to the hodograph. B) perpendicular to the hodograph. C) tangent to the path. D) perpendicular to the path. 2. In curvilinear motion, the direction of the instantaneous acceleration is always A) tangent to the hodograph. B) perpendicular to the hodograph. C) tangent to the path. D) perpendicular to the path. Answers: 1. C 2. A 36

APPLICATIONS The path of motion of a plane can be tracked with radar and its x, y, and z coordinates (relative to a point on earth) recorded as a function of time. How can we determine the velocity or acceleration of the plane at any instant?

A roller coaster car travels down a fixed, helical path at a constant speed. How can we determine its position or acceleration at any instant? If you are designing the track, why is it important to be able to predict the acceleration of the car?

GENERAL CURVILINEAR MOTION
(Section 12.4) A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are used to describe the motion. A particle moves along a curve defined by the path function, s. The position of the particle at any instant is designated by the vector r = r(t). Both the magnitude and direction of r may vary with time. If the particle moves a distance Ds along the curve during time interval Dt, the displacement is determined by vector subtraction: D r = r’ - r

VELOCITY Velocity represents the rate of change in the position of a particle. The average velocity of the particle during the time increment Dt is vavg = Dr/Dt . The instantaneous velocity is the time-derivative of position v = dr/dt . The velocity vector, v, is always tangent to the path of motion. The magnitude of v is called the speed. Since the arc length Ds approaches the magnitude of Dr as t→0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector!

ACCELERATION Acceleration represents the rate of change in the velocity of a particle. If a particle’s velocity changes from v to v’ over a time increment Dt, the average acceleration during that increment is: aavg = Dv/Dt = (v - v’)/Dt The instantaneous acceleration is the time-derivative of velocity: a = dv/dt = d2r/dt2 A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.

CURVILINEAR MOTION: RECTANGULAR COMPONENTS
(Section 12.5) It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference. The position of the particle can be defined at any instant by the position vector r = x i + y j + z k . The x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t) . The magnitude of the position vector is: r = (x2 + y2 + z2)0.5 The direction of r is defined by the unit vector: ur = (1/r)r

The velocity vector is the time derivative of the position vector:
RECTANGULAR COMPONENTS: VELOCITY The velocity vector is the time derivative of the position vector: v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt Since the unit vectors i, j, k are constant in magnitude and direction, this equation reduces to v = vx i + vy j + vz k where vx = = dx/dt, vy = = dy/dt, vz = = dz/dt x y z • The magnitude of the velocity vector is v = [(vx)2 + (vy)2 + (vz)2]0.5 The direction of v is tangent to the path of motion.

vx x vy y vz z RECTANGULAR COMPONENTS: ACCELERATION
The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): a = dv/dt = d2r/dt2 = ax i + ay j + az k where ax = = = dvx /dt, ay = = = dvy /dt, az = = = dvz /dt vx x vy y vz z • •• The magnitude of the acceleration vector is a = [(ax)2 + (ay)2 + (az)2 ]0.5 The direction of a is usually not tangent to the path of the particle.

EXAMPLE Given: The motion of two particles (A and B) is described by the position vectors rA = [3t i + 9t(2 – t) j] m and rB = [3(t2 –2t +2) i + 3(t – 2) j] m. Find: The point at which the particles collide and their speeds just before the collision. Plan: 1) The particles will collide when their position vectors are equal, or rA = rB . 2) Their speeds can be determined by differentiating the position vectors.

EXAMPLE (continued) Solution: 1) The point of collision requires that rA = rB, so xA = xB and yA = yB . Set the x-components equal: 3t = 3(t2 – 2t + 2) Simplifying: t2 – 3t + 2 = 0 Solving: t = {3  [32 – 4(1)(2)]0.5}/2(1) => t = 2 or 1 s Set the y-components equal: 9t(2 – t) = 3(t – 2) Simplifying: 3t2 – 5t – 2 = 0 Solving: t = {5  [52 – 4(3)(–2)]0.5}/2(3) => t = 2 or – 1/3 s So, the particles collide when t = 2 s (only common time). Substituting this value into rA or rB yields xA = xB = 6 m and yA = yB = 0

2) Differentiate rA and rB to get the velocity vectors.
EXAMPLE (continued) 2) Differentiate rA and rB to get the velocity vectors. vA = drA/dt = = [ 3 i + (18 – 18t) j ] m/s At t = 2 s: vA = [ 3i – 18 j ] m/s j yA i xA . + • vB = drB/dt = xB i + yB j = [ (6t – 6) i + 3 j ] m/s At t = 2 s: vB = [ 6 i + 3 j ] m/s • • Speed is the magnitude of the velocity vector. vA = ( ) = 18.2 m/s vB = ( ) = 6.71 m/s

CHECK YOUR UNDERSTANDING QUIZ
1. If the position of a particle is defined by r = [(1.5t2 + 1) i + (4t – 1) j ] (m), its speed at t = 1 s is A) 2 m/s B) 3 m/s C) 5 m/s D) 7 m/s 2. The path of a particle is defined by y = 0.5x2. If the component of its velocity along the x-axis at x = 2 m is vx = 1 m/s, its velocity component along the y-axis at this position is A) 0.25 m/s B) 0.5 m/s C) 1 m/s D) 2 m/s Answers: 1. C 2. D 48

GROUP PROBLEM SOLVING Given: The velocity of the particle is v = [ 16 t2 i + 4 t3 j + (5 t + 2) k] m/s. When t = 0, x = y = z = 0. Find: The particle’s coordinate position and the magnitude of its acceleration when t = 2 s. Plan: Note that velocity vector is given as a function of time. 1) Determine the position and acceleration by integrating and differentiating v, respectively, using the initial conditions. 2) Determine the magnitude of the acceleration vector using t = 2 s.

Solution: 1) x-components: Velocity known as: vx = x = dx/dt = (16 t2 ) m/s Position: = (16 t2) dt  x = (16/3)t3 = 42.7 m at t = 2 s Acceleration: ax = x = vx = d/dt (16 t2) = 32 t = 64 m/s2 •• ò x dx t • 2) y-components: Velocity known as: vy = y = dy/dt = (4 t3 ) m/s Position: = (4 t3) dt  y = t4 = (16) m at t = 2 s Acceleration: ay = y = vy = d/dt (4 t3) = 12 t2 = 48 m/s2 •• ò y dy t •

3) z-components: Velocity is known as: vz = z = dz/dt = (5 t + 2) m/s Position: = (5 t + 2) dt  z = (5/2) t2 + 2t = 14 m at t=2s Acceleration: az = z = vz = d/dt (5 t + 2) = 5 m/s2 •• ò z dz t • 4) The position vector and magnitude of the acceleration vector are written using the component information found above. Position vector: r = [ 42.7 i + 16 j + 14 k] m. Acceleration vector: a = [ 64 i + 48 j + 5 k] m/s2 Magnitude: a = ( )0.5 = 80.2 m/s2

A) (4 i +8 j ) m/s2 B) (8 i -16 j ) m/s2 C) (8 i) m/s2 D) (8 j ) m/s2
ATTENTION QUIZ 1. If a particle has moved from A to B along the circular path in 4s, what is the average velocity of the particle ? A) 2.5 i m/s B) 2.5 i +1.25j m/s C)  i m/s D)  j m/s A B R=5m x y 2. The position of a particle is given as r = (4t2 i - 2x j) m. Determine the particle’s acceleration. A) (4 i +8 j ) m/s B) (8 i -16 j ) m/s2 C) (8 i) m/s D) (8 j ) m/s2 Answers: 1. A 2. B 52

MOTION OF A PROJECTILE Today’s Objectives: Students will be able to: Analyze the free-flight motion of a projectile. In-Class Activities: • Check Homework • Reading Quiz • Applications • Kinematic Equations for Projectile Motion • Concept Quiz • Group Problem Solving • Attention Quiz

The downward acceleration of an object in free-flight motion is
READING QUIZ The downward acceleration of an object in free-flight motion is A) zero. B) increasing with time. C) 9.81 m/s2. D) 9.81 ft/s2. 2. The horizontal component of velocity remains _________ during a free-flight motion. A) zero B) constant C) at 9.81 m/s2 D) at 32.2 ft/s2 Answers: C B 54

APPLICATIONS A good kicker instinctively knows at what angle, q, and initial velocity, vA, he must kick the ball to make a field goal. For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?

A basketball is shot at a certain angle. What parameters should the shooter consider in order for the basketball to pass through the basket? Distance, speed, the basket location, … anything else ?

A firefighter needs to know the maximum height on the wall she can project water from the hose. What parameters would you program into a wrist computer to find the angle, q, that she should use to hold the hose?

MOTION OF A PROJECTILE (Section 12.6)
Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., from gravity). For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant.

KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by: x = xo + (vox) t No air resistance is assumed! Why is ax equal to zero (assuming movement through the air)? 59

KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = – g. Application of the constant acceleration equations yields: vy = voy – g t y = yo + (voy) t – ½ g t2 vy2 = voy2 – 2 g (y – yo) For any given problem, only two of these three equations can be used. Why?

Find: The equation that defines y as a function of x.
EXAMPLE I Given: vo and θ Find: The equation that defines y as a function of x. Plan: Eliminate time from the kinematic equations. Solution: Using vx = vo cos θ and vy = vo sin θ We can write: x = (vo cos θ)t or y = (vo sin θ) t – ½ g (t)2 t = x vo cos θ y = (vo sin θ) { } { }2 x g x vo cos θ 2 vo cos θ – By substituting for t: 61

EXAMPLE I (continued) Simplifying the last equation, we get: y = (x tanq) – g x2 2vo2 (1 + tan2q) The above equation is called the “path equation” which describes the path of a particle in projectile motion. The equation shows that the path is parabolic.

EXAMPLE II Given: Projectile is fired with vA=150 m/s at point A. Find: The horizontal distance it travels (R) and the time in the air. Plan: Establish a fixed x, y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y-directions.

EXAMPLE II (continued)
Solution: 1) Place the coordinate system at point A. Then, write the equation for horizontal motion. + xB = xA + vAx tAB where xB = R, xA = 0, vAx = 150 (4/5) m/s Range, R will be R = 120 tAB 2) Now write a vertical motion equation. Use the distance equation. + yB = yA + vAy tAB – 0.5 g tAB2 where yB = – 150, yA = 0, and vAy = 150(3/5) m/s We get the following equation: –150 = 90 tAB (– 9.81) tAB2 Solving for tAB first, tAB = s. Then, R = 120 tAB = 120 (19.89) = 2387 m

A) (vo sin q)/g B) (2vo sin q)/g C) (vo cos q)/g D) (2vo cos q)/g
CONCEPT QUIZ 1. In a projectile motion problem, what is the maximum number of unknowns that can be solved? A) B) 2 C) D) 4 2. The time of flight of a projectile, fired over level ground, with initial velocity Vo at angle θ, is equal to? A) (vo sin q)/g B) (2vo sin q)/g C) (vo cos q)/g D) (2vo cos q)/g Answers: C B 65

GROUP PROBLEM SOLVING Establish a fixed x,y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x and y-directions. x y Given: A skier leaves the ski jump ramp at qA = 25o and hits the slope at B. Find: The skier’s initial speed vA. Plan:

Solution: Motion in x-direction: Using xB = xA + vox(tAB) => (4/5)100 = 0 + vA (cos 25) tAB = tAB= 80 vA (cos 25) 88.27 vA – 64 = 0 + vA(sin 25) { } 88.27 vA – ½ (9.81) { }2 Motion in y-direction: Using yB = yA + voy(tAB) – ½ g(tAB)2 vA = m/s

ATTENTION QUIZ 1. A projectile is given an initial velocity
vo at an angle f above the horizontal. The velocity of the projectile when it hits the slope is ____________ the initial velocity vo. A) less than B) equal to C) greater than D) None of the above. 2. A particle has an initial velocity vo at angle q with respect to the horizontal. The maximum height it can reach is when A) q = 30° B) q = 45° C) q = 60° D) q = 90° Answers: 1. A 2. D 68

NORMAL AND TANGENTIAL COMPONENTS
CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS Today’s Objectives: Students will be able to: Determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path. In-Class Activities: • Check Homework • Reading Quiz • Applications • Normal and Tangential Components of Velocity and Acceleration • Special Cases of Motion • Concept Quiz • Group Problem Solving • Attention Quiz 69

A) positive. B) negative. C) zero. D) constant.
READING QUIZ 1. If a particle moves along a curve with a constant speed, then its tangential component of acceleration is A) positive. B) negative. C) zero. D) constant. 2. The normal component of acceleration represents A) the time rate of change in the magnitude of the velocity. B) the time rate of change in the direction of the velocity. C) magnitude of the velocity. D) direction of the total acceleration. Answers: 1. C 2. B 70

APPLICATIONS Cars traveling along a clover-leaf interchange experience an acceleration due to a change in velocity as well as due to a change in direction of the velocity. If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration? Why would you care about the total acceleration of the car?

A roller coaster travels down a hill for which the path can be approximated by a function y = f(x). The roller coaster starts from rest and increases its speed at a constant rate. How can we determine its velocity and acceleration at the bottom? Why would we want to know these values?

NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)
When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle). The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion. The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

NORMAL AND TANGENTIAL COMPONENTS (continued)
The positive n and t directions are defined by the unit vectors un and ut, respectively. The center of curvature, O’, always lies on the concave side of the curve. The radius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point. The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.

VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always tangent to the path of motion (t-direction). The magnitude is determined by taking the time derivative of the path function, s(t). v = v ut where v = s = ds/dt . Here v defines the magnitude of the velocity (speed) and ut defines the direction of the velocity vector.

ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change of velocity: a = dv/dt = d(vut)/dt = vut + vut . Here v represents the change in the magnitude of velocity and ut represents the rate of change in the direction of ut. . . a = v ut + (v2/r) un = at ut + an un. After mathematical manipulation, the acceleration vector can be expressed as:

ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
So, there are two components to the acceleration vector: a = at ut + an un • The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity. at = v or at ds = v dv . • The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r • The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5

SPECIAL CASES OF MOTION
There are some special cases of motion to consider. 1) The particle moves along a straight line. r  => an = v2/r = => a = at = v The tangential component represents the time rate of change in the magnitude of the velocity. . 2) The particle moves along a curve at constant speed. at = v = => a = an = v2/r . The normal component represents the time rate of change in the direction of the velocity.

SPECIAL CASES OF MOTION (continued)
3) The tangential component of acceleration is constant, at = (at)c. In this case, s = so + vo t + (1/2) (at)c t2 v = vo + (at)c t v2 = (vo)2 + 2 (at)c (s – so) As before, so and vo are the initial position and velocity of the particle at t = 0. How are these equations related to projectile motion equations? Why? 4) The particle moves along a path expressed as y = f(x). The radius of curvature, r, at any point on the path can be calculated from r = ________________ ]3/2 (dy/dx)2 1 [ + 2 d2y/dx

THREE-DIMENSIONAL MOTION
If a particle moves along a space curve, the n and t axes are defined as before. At any point, the t-axis is tangent to the path and the n-axis points toward the center of curvature. The plane containing the n and t axes is called the osculating plane. A third axis can be defined, called the binomial axis, b. The binomial unit vector, ub, is directed perpendicular to the osculating plane, and its sense is defined by the cross product ub = ut × un. There is no motion, thus no velocity or acceleration, in the binomial direction.

EXAMPLE Given: A boat travels around a circular path, r = 40 m, at a speed that increases with time, v = ( t2) m/s. Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 10 s. Plan: The boat starts from rest (v = 0 when t = 0). 1) Calculate the velocity at t = 10 s using v(t). 2) Calculate the tangential and normal components of acceleration and then the magnitude of the acceleration vector.

EXAMPLE (continued) Solution: 1) The velocity vector is v = v ut , where the magnitude is given by v = (0.0625t2) m/s. At t = 10s: v = t2 = (10)2 = m/s 2) The acceleration vector is a = atut + anun = vut + (v2/r)un. . Tangential component: at = v = d(.0625 t2 )/dt = t m/s2 At t = 10s: at = 0.125t = 0.125(10) = 1.25 m/s2 . Normal component: an = v2/r m/s2 At t = 10s: an = (6.25)2 / (40) = m/s2 The magnitude of the acceleration is a = [(at)2 + (an)2]0.5 = [(1.25)2 + (0.9766)2]0.5 = 1.59 m/s2

1. A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing at a constant rate of 4 m/s2. What is the magnitude of its total acceleration at this instant? A) 3 m/s2 B) 4 m/s2 C) 5 m/s2 D) -5 m/s2 2. If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total acceleration at t = 1 s? A) 8 m/s B) 8.6 m/s C) 3.2 m/s D) 11.2 m/s Answers: 1. C 2. B 83

2. Calculate the radius of curvature of the path at B.
GROUP PROBLEM SOLVING Given: A roller coaster travels along a vertical parabolic path defined by the equation y = 0.01x2. At point B, it has a speed of 25 m/s, which is increasing at the rate of 3 m/s2. Find: The magnitude of the roller coaster’s acceleration when it is at point B. Plan: 1. The change in the speed of the car (3 m/s2) is the tangential component of the total acceleration. 2. Calculate the radius of curvature of the path at B. 3. Calculate the normal component of acceleration. 4. Determine the magnitude of the acceleration vector. 84

Solution: 1) The tangential component of acceleration is the rate of increase of the roller coaster’s speed, so at = v = 3 m/s2. . 2) Determine the radius of curvature at point B (x = 30 m): dy/dx = d(0.01x2)/dx = 0.02x, d2y/dx2 = d (0.02x)/dx = 0.02 At x =30 m, dy/dx = 0.02(30) = 0.6, d2y/dx2 = 0.02 => r = = [1 + (0.6)2]3/2/(0.02) = 79.3 m [1+(dy/dx)2]3/2 d2y/dx2 3) The normal component of acceleration is an = v2/r = (25)2/(79.3) = m/s2 4) The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5 = [(3)2 + (7.881)2]0.5 = 8.43 m/s2 85

1. The magnitude of the normal acceleration is
ATTENTION QUIZ 1. The magnitude of the normal acceleration is A) proportional to radius of curvature. B) inversely proportional to radius of curvature. C) sometimes negative. D) zero when velocity is constant. 2. The directions of the tangential acceleration and velocity are always A) perpendicular to each other. B) collinear. C) in the same direction. D) in opposite directions. Answers: 1. B 2. B 86

CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
Today’s Objectives: Students will be able to: Determine velocity and acceleration components using cylindrical coordinates. In-Class Activities: Check Homework Reading Quiz Applications Velocity Components Acceleration Components Concept Quiz Group Problem Solving Attention Quiz 87

A) transverse velocity. B) radial velocity.
READING QUIZ 1. In a polar coordinate system, the velocity vector can be written as v = vrur + vθuθ = rur + rquq. The term q is called A) transverse velocity. B) radial velocity. C) angular velocity. D) angular acceleration. . 2. The speed of a particle in a cylindrical coordinate system is A) r B) rq C) (rq)2 + (r)2 D) (rq)2 + (r)2 + (z)2 . Answers: 1. C 2. D 88

APPLICATIONS The cylindrical coordinate system is used in cases where the particle moves along a 3-D curve. In the figure shown, the box slides down the helical ramp. How would you find the box’s velocity components to know if the package will fly off the ramp?

CYLINDRICAL COMPONENTS
(Section 12.8) We can express the location of P in polar coordinates as r = r ur. Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, q, is measured counter-clockwise (CCW) from the horizontal. 90

VELOCITY in POLAR COORDINATES)
The instantaneous velocity is defined as: v = dr/dt = d(rur)/dt v = rur + r dur dt . Using the chain rule: dur/dt = (dur/dq)(dq/dt) We can prove that dur/dq = uθ so dur/dt = quθ Therefore: v = rur + rquθ . . Thus, the velocity vector has two components: r, called the radial component, and rq called the transverse component. The speed of the particle at any given instant is the sum of the squares of both components or v = (r q )2 + ( r )2 91

ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as: a = dv/dt = (d/dt)(rur + rquθ) . After manipulation, the acceleration can be expressed as a = (r – rq 2)ur + (rq + 2rq)uθ .. . . The magnitude of acceleration is a = (r – rq 2)2 + (rq + 2rq) 2 The term (r – rq 2) is the radial acceleration or ar . The term (rq + 2rq) is the transverse acceleration or aq . .. 92

CYLINDRICAL COORDINATES
If the particle P moves along a space curve, its position can be written as rP = rur + zuz Taking time derivatives and using the chain rule: Velocity: vP = rur + rquθ + zuz Acceleration: aP = (r – rq2)ur + (rq + 2rq)uθ + zuz .. . 93

Given: A car travels along acircular path.
EXAMPLE Given: A car travels along acircular path. r = 300 ft, q = 0.4 (rad/s), q = 0.2 (rad/s2) Find: Velocity and acceleration Plan: Use the polar coordinate system. . .. Solution: r = 300 ft, r = r = 0, and q = 0.4 (rad/s), q = 0.2 (rad/s2) .. . Substitute in the equation for velocity v = r ur + rq uθ = 0 ur (0.4) uθ v = (0)2 + (120)2 = 120 ft/s . 94

Substitute in the equation for acceleration:
EXAMPLE (continued) Substitute in the equation for acceleration: a = (r – rq 2)ur + (rq + 2rq)uθ a = [0 – 300(0.4)2] ur + [300(0.2) + 2(0)(0.4)] uθ a = – 48 ur + 60 uθ ft/s2 a = (– 48)2 + (60)2 = ft/s2 .. . 95

CONCEPT QUIZ .. . 1. If r is zero for a particle, the particle is
A) not moving. B) moving in a circular path. C) moving on a straight line. D) moving with constant velocity. . 2. If a particle moves in a circular path with constant velocity, its radial acceleration is A) zero B) r . C) − rq D) 2rq . .. . Answers: 1. B 2. C 96

GROUP PROBLEM SOLVING Given: The car’s speed is constant at 1.5 m/s. Find: The car’s acceleration (as a vector). Plan: f = tan-1( ) = ° Also, what is the relationship between f and q? 12 2p(10) Hint: The tangent to the ramp at any point is at an angle Plan: Use cylindrical coordinates. Since r is constant, all derivatives of r will be zero.

Solution: Since r is constant, the velocity only has 2 components: vq = rq = v cos f and vz = z = v sin f . Therefore: q = ( ) = rad/s q = 0 v cosf r . .. vz = z = v sinf = m/s z = 0 r = r = 0 . .. .. . a = (r – rq 2)ur + (rq + 2rq)uθ + zuz a = (-rq 2)ur = -10(0.147)2ur = ur m/s2

C) greater than its transverse component.
ATTENTION QUIZ 1. The radial component of velocity of a particle moving in a circular path is always A) zero. B) constant. C) greater than its transverse component. D) less than its transverse component. 2. The radial component of acceleration of a particle moving in a circular path is always A) negative. B) directed toward the center of the path. C) perpendicular to the transverse component of acceleration. D) All of the above. Answers: 1. A 2. D 99

ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES
Today’s Objectives: Students will be able to: Relate the positions, velocities, and accelerations of particles undergoing dependent motion. In-Class Activities: • Check Homework • Reading Quiz • Applications • Define Dependent Motion • Develop Position, Velocity, and Acceleration Relationships • Concept Quiz • Group Problem Solving • Attention Quiz

A) always independent. B) always dependent.
READING QUIZ 1. When particles are interconnected by a cable, the motions of the particles are ______ A) always independent. B) always dependent. C) not always dependent. D) None of the above. 2. If the motion of one particle is dependent on that of another particle, each coordinate axis system for the particles _______ A) should be directed along the path of motion. B) can be directed anywhere. C) should have the same origin. D) None of the above. Answers: 1. B 2. A 101

APPLICATIONS The cable and pulley system shown can be used to modify the speed of the mine car, A, relative to the speed of the motor, M. It is important to establish the relationships between the various motions in order to determine the power requirements for the motor and the tension in the cable. For instance, if the speed of the cable (P) is known because we know the motor characteristics, how can we determine the speed of the mine car? Will the slope of the track have any impact on the answer?

Rope and pulley arrangements are often used to assist in lifting heavy objects. The total lifting force required from the truck depends on both the weight and the acceleration of the cabinet. How can we determine the acceleration and velocity of the cabinet if the acceleration of the truck is known?

DEPENDENT MOTION (Section 12.9)
In many kinematics problems, the motion of one object will depend on the motion of another object. The blocks in this figure are connected by an inextensible cord wrapped around a pulley If block A moves downward along the inclined plane, block B will move up the other incline. The motion of each block can be related mathematically by defining position coordinates, sA and sB. Each coordinate axis is defined from a fixed point or datum line, measured positive along each plane in the direction of motion of each block.

DEPENDENT MOTION (continued)
In this example, position coordinates sA and sB can be defined from fixed datum lines extending from the center of the pulley along each incline to blocks A and B. If the cord has a fixed length, the position coordinates sA and sB are related mathematically by the equation sA + lCD + sB = lT Here lT is the total cord length and lCD is the length of cord passing over the arc CD on the pulley.

DEPENDENT MOTION (continued)
The velocities of blocks A and B can be related by differentiating the position equation. Note that lCD and lT remain constant, so dlCD/dt = dlT/dt = 0 dsA/dt + dsB/dt = 0 => vB = -vA The negative sign indicates that as A moves down the incline (positive sA direction), B moves up the incline (negative sB direction). Accelerations can be found by differentiating the velocity expression. Prove to yourself that aB = -aA .

DEPENDENT MOTION EXAMPLE
Consider a more complicated example. Position coordinates (sA and sB) are defined from fixed datum lines, measured along the direction of motion of each block. Note that sB is only defined to the center of the pulley above block B, since this block moves with the pulley. Also, h is a constant. The red colored segments of the cord remain constant in length during motion of the blocks.

DEPENDENT MOTION EXAMPLE (continued)
The position coordinates are related by the equation 2sB + h + sA = lT Where lT is the total cord length minus the lengths of the red segments. Since lT and h remain constant during the motion, the velocities and accelerations can be related by two successive time derivatives: 2vB = -vA and 2aB = -aA When block B moves downward (+sB), block A moves to the left (-sA). Remember to be consistent with your sign convention!

DEPENDENT MOTION EXAMPLE (continued)
This example can also be worked by defining the position coordinate for B (sB) from the bottom pulley instead of the top pulley. The position, velocity, and acceleration relations then become 2(h – sB) + h + sA = lT and 2vB = vA aB = aA Prove to yourself that the results are the same, even if the sign conventions are different than the previous formulation.

DEPENDENT MOTION: PROCEDURES
These procedures can be used to relate the dependent motion of particles moving along rectilinear paths (only the magnitudes of velocity and acceleration change, not their line of direction). 1. Define position coordinates from fixed datum lines, along the path of each particle. Different datum lines can be used for each particle. 2. Relate the position coordinates to the cord length. Segments of cord that do not change in length during the motion may be left out. 3. If a system contains more than one cord, relate the position of a point on one cord to a point on another cord. Separate equations are written for each cord. 4. Differentiate the position coordinate equation(s) to relate velocities and accelerations. Keep track of signs!

EXAMPLE Given: In the figure on the left, the cord at A is pulled down with a speed of 2 m/s. Find: The speed of block B. Plan: There are two cords involved in the motion in this example. There will be two position equations (one for each cord). Write these two equations, combine them, and then differentiate them.

EXAMPLE (continued) Solution:
1) Define the position coordinates from a fixed datum line. Three coordinates must be defined: one for point A (sA), one for block B (sB), and one for block C (sC). • Define the datum line through the top pulley (which has a fixed position). • sA can be defined to the point A. • sB can be defined to the center of the pulley above B. • sC is defined to the center of pulley C. • All coordinates are defined as positive down and along the direction of motion of each point/object. 112

vA + 4vB = 0 => vB = – 0.25vA = – 0.25(2) = – 0.5 m/s
EXAMPLE (continued) 2) Write position/length equations for each cord. Define l1 as the length of the first cord, minus any segments of constant length. Define l2 in a similar manner for the second cord: Cord 1: sA + 2sC = l1 Cord 2: sB + (sB – sC) = l2 3) Eliminating sC between the two equations, we get sA + 4sB = l1 + 2l2 4) Relate velocities by differentiating this expression. Note that l1 and l2 are constant lengths. vA + 4vB = => vB = – 0.25vA = – 0.25(2) = – 0.5 m/s The velocity of block B is 0.5 m/s up (negative sB direction).

1. Determine the speed of block B. A) 1 m/s B) 2 m/s
CONCEPT QUIZ 1. Determine the speed of block B. A) 1 m/s B) 2 m/s C) 4 m/s D) None of the above. 2. Two blocks are interconnected by a cable. Which of the following is correct ? A) vA= - vB B) (vx)A= - (vx)B C) (vy)A= - (vy)B D) All of the above. x y Answers: 1. A 2. D 114

GROUP PROBLEM SOLVING Given: In this pulley system, block A is moving downward with a speed of 4 ft/s while block C is moving up at 2 ft/s. Find: The speed of block B. Plan: All blocks are connected to a single cable, so only one position/length equation will be required. Define position coordinates for each block, write out the position relation, and then differentiate it to relate the velocities.

Solution: 1) A datum line can be drawn through the upper, fixed, pulleys and position coordinates defined from this line to each block (or the pulley above the block). 2) Defining sA, sB, and sC as shown, the position relation can be written: sA + 2sB + sC = l 3) Differentiate to relate velocities: vA + 2vB + vC = 0 4 + 2vB + (-2) =0 vB = -1 ft/s The velocity of block B is 1 ft/s up (negative sB direction).

A) (8i + 6j) m/s B) (4i + 3j) m/s C) (-8i - 6j) m/s D) (3i + 4j) m/s j
ATTENTION QUIZ 1. Determine the speed of block B when block A is moving down at 6 ft/s while block C is moving down at 18 ft/s . A) 24 ft/s B) 3 ft/s C) 12 ft/s D) 9 ft/s vA=6 ft/s vC=18 ft/s 2. Determine the velocity vector of block A when block B is moving downward with a speed of 10 m/s. A) (8i + 6j) m/s B) (4i + 3j) m/s C) (-8i - 6j) m/s D) (3i + 4j) m/s vB=10 m/s j i Answers: 1. C 2. B 117

RELATIVE-MOTION ANALYSIS OF TWO PARTICLES USING TRANSLATING AXES
Today’s Objectives: Students will be able to: Understand translating frames of reference. Use translating frames of reference to analyze relative motion. In-Class Activities: • Check Homework, • Reading Quiz • Applications • Relative Position, Velocity and Acceleration • Vector & Graphical Methods • Concept Quiz • Group Problem Solving • Attention Quiz

1. The velocity of B relative to A is defined as
READING QUIZ 1. The velocity of B relative to A is defined as A) vB – vA . B) vA – vB . C) vB + vA . D) vA + vB . 2. Since two dimensional vector addition forms a triangle, there can be at most _________ unknowns (either magnitudes and/or directions of the vectors). A) one B) two C) three D) four Answers: 1. A 2. B 119

APPLICATIONS When you try to hit a moving object, the position, velocity, and acceleration of the object all have to be accounted for by your mind. You are smarter than you thought! Here, the boy on the ground is at d = 10 ft when the girl in the window throws the ball to him. If the boy on the ground is running at a constant speed of 4 ft/s, how fast should the ball be thrown?

When fighter jets take off or land on an aircraft carrier, the velocity of the carrier becomes an issue. If the aircraft carrier is underway with a forward velocity of 50 km/hr and plane A takes off at a horizontal air speed of 200 km/hr (measured by someone on the water), how do we find the velocity of the plane relative to the carrier? How would you find the same thing for airplane B? How does the wind impact this sort of situation?

RELATIVE POSITION (Section 12.10)
The absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by rA and rB. The position of B relative to A is represented by rB/A = rB – rA Therefore, if rB = (10 i + 2 j ) m and rA = (4 i + 5 j ) m, then rB/A = (6 i – 3 j ) m.

RELATIVE VELOCITY To determine the relative velocity of B with respect to A, the time derivative of the relative position equation is taken. vB/A = vB – vA or vB = vA + vB/A In these equations, vB and vA are called absolute velocities and vB/A is the relative velocity of B with respect to A. Note that vB/A = - vA/B .

RELATIVE ACCELERATION
The time derivative of the relative velocity equation yields a similar vector relationship between the absolute and relative accelerations of particles A and B. These derivatives yield: aB/A = aB – aA or aB = aA + aB/A

SOLVING PROBLEMS Since the relative motion equations are vector equations, problems involving them may be solved in one of two ways. For instance, the velocity vectors in vB = vA + vB/A could be written as two dimensional (2-D) Cartesian vectors and the resulting 2-D scalar component equations solved for up to two unknowns. Alternatively, vector problems can be solved “graphically” by use of trigonometry. This approach usually makes use of the law of sines or the law of cosines. Could a CAD system be used to solve these types of problems?

LAWS OF SINES AND COSINES
b c C B A Since vector addition or subtraction forms a triangle, sine and cosine laws can be applied to solve for relative or absolute velocities and accelerations. As a review, their formulations are provided below. Law of Sines: C c B b A a sin = Law of Cosines: A bc c b a cos 2 - + = B ac C ab

EXAMPLE Given: vA = 650 km/h vB = 800 km/h Find: vB/A Plan:
a) Vector Method: Write vectors vA and vB in Cartesian form, then determine vB – vA b) Graphical Method: Draw vectors vA and vB from a common point. Apply the laws of sines and cosines to determine vB/A. 127

vB/A = vB – vA = (–1050 i – 692.8 j) km/h
EXAMPLE (continued) Solution: vA = (650 i ) km/h vB = –800 cos 60 i – 800 sin 60 j = ( –400 i – j) km/h a) Vector Method: vB/A = vB – vA = (–1050 i – j) km/h v A B / = (-1050)2 +(-692.8)2 = km/h  = tan-1( ) = 33.4 q 1050 692.8 128

(vB/A)2 = (800) 2 + (650) 2 − (800) (650) cos 120 vB/A = 1258 km/h
EXAMPLE (continued) b) Graphical Method: Note that the vector that measures the tip of B relative to A is vB/A. vB vA 120 vB/A Law of Cosines: (vB/A)2 = (800) 2 + (650) 2 − (800) (650) cos 120 vB/A = 1258 km/h vB/A vA sin(120 ) sin q = or  = 33.4 Law of Sines: 129

2. Determine the velocity of plane A with respect to plane B.
CONCEPT QUIZ 1. Two particles, A and B, are moving in the directions shown. What should be the angle q so that vB/A is minimum? A) 0° B) 180° C) 90° D) 270° A B q s ft v 3 = 4 2. Determine the velocity of plane A with respect to plane B. A) (400 i j ) km/hr B) (1220 i j ) km/hr C) (-181 i j ) km/hr D) (-1220 i j ) km/hr Answers 1. A 2. B 30 130

GROUP PROBLEM SOLVING Given: vA = 30 mi/h vB = 20 mi/h aB = 1200 mi/h2
aA = 0 mi/h2 Find: vB/A aB/A Plan: 131

vB = –20 sin(30) i + 20 cos(30) j = (–10 i + 17.32 j) mi/h
GROUP PROBLEM SOLVING Given: vA = 30 mi/h vB = 20 mi/h aB = 1200 mi/h2 aA = 0 mi/h2 Find: vB/A aB/A Plan: Write the velocity and acceleration vectors for A and B and determine vB/A and aB/A by using vector equations. Solution: The velocity of B is: vB = –20 sin(30) i + 20 cos(30) j = (–10 i j) mi/h 132

GROUP PROBLEM SOLVING (solution continued)
The velocity of A is: vA = –30 i (mi/h) The relative velocity of B with respect to A is (vB/A): vB/A = vB – vA = (–10i j) – (–30i) = (20 i j) mi/h or vB/A = (20)2 + (17.32)2 = mi/h q = tan-1( ) = 40.9° q 17.32 20 133

GROUP PROBLEM SOLVING (solution continued)
The acceleration of B is: aB = (at)B + (an)B = [– 1200 sin(30) i cos(30) j] + [ ( ) cos(30) i +( ) sin(30) j] aB = i j (mi/h2) 202 0.3 The acceleration of A is zero : aA = 0 The relative acceleration of B with respect to A is: aB/A = aB – aA = i j (mi/h2) aA/B = (554.7)2 + (1706)2 = 1790 mi/h2 b = tan-1(1706 / 554.7) = 72° b 134

ATTENTION QUIZ 1. Determine the relative velocity of particle B with respect to particle A. A) (48i + 30j) km/h B) (- 48i + 30j ) km/h C) (48i - 30j ) km/h D) (- 48i - 30j ) km/h B A vB=100 km/h vA=60 km/h 30 y x 2. If theta equals 90° and A and B start moving from the same point, what is the magnitude of rB/A at t = 5 s? A) 20 ft B) 15 ft C) 18 ft D) 25 ft Answers: 1. C 2. D A B q s ft v 3 = 4 135

NEWTON’S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES
Today’s Objectives: Students will be able to: Write the equation of motion for an accelerating body. Draw the free-body and kinetic diagrams for an accelerating body. In-Class Activities: • Check Homework • Reading Quiz • Applications • Newton’s Laws of Motion • Newton’s Law of Gravitational Attraction • Equation of Motion For A Particle or System of Particles • Concept Quiz • Group Problem Solving • Attention Quiz

A) external forces B) weight C) internal forces D) All of the above.
READING QUIZ 1. Newton’s second law can be written in mathematical form as F = ma. Within the summation of forces, F, ________ are(is) not included. A) external forces B) weight C) internal forces D) All of the above. 2. The equation of motion for a system of n-particles can be written as Fi =  miai = maG, where aG indicates _______. A) summation of each particle’s acceleration B) acceleration of the center of mass of the system C) acceleration of the largest particle D) None of the above. Answers: 1. C 2. B 137

APPLICATIONS The motion of an object depends on the forces acting on it. A parachutist relies on the atmospheric drag resistance force of her parachute to limit her velocity. Knowing the drag force, how can we determine the acceleration or velocity of the parachutist at any point in time? This has some importance when landing!

The baggage truck A tows a cart B, and a cart C. If we know the frictional force developed at the driving wheels of the truck, can we determine the acceleration of the truck? How? Can we also determine the horizontal force acting on the coupling between the truck and cart B? This is needed when designing the coupling (or understanding why it failed).

A freight elevator is lifted using a motor attached to a cable and pulley system as shown. How can we determine the tension force in the cable required to lift the elevator and load at a given acceleration? This is needed to decide what size cable should be used. Is the tension force in the cable greater than the weight of the elevator and its load?

NEWTON’S LAWS OF MOTION (Section 13.1)
The motion of a particle is governed by Newton’s three laws of motion. First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero. Second Law: If the resultant force on the particle is not zero, the particle experiences an acceleration in the same direction as the resultant force. This acceleration has a magnitude proportional to the resultant force. Third Law: Mutual forces of action and reaction between two particles are equal, opposite, and collinear.

NEWTON’S LAWS OF MOTION (continued)
The first and third laws were used in developing the concepts of statics. Newton’s second law forms the basis of the study of dynamics. Mathematically, Newton’s second law of motion can be written F = ma where F is the resultant unbalanced force acting on the particle, and a is the acceleration of the particle. The positive scalar m is called the mass of the particle. Newton’s second law cannot be used when the particle’s speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom).

NEWTON’S LAW OF GRAVITATIONAL ATTRACTION
Any two particles or bodies have a mutually attractive gravitational force acting between them. Newton postulated the law governing this gravitational force as F = G(m1m2/r2) where F = force of attraction between the two bodies, G = universal constant of gravitation , m1, m2 = mass of each body, and r = distance between centers of the two bodies. When near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the body. This force is called the weight of the body.

MASS AND WEIGHT It is important to understand the difference between the mass and weight of a body! Mass is an absolute property of a body. It is independent of the gravitational field in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, as defined by Newton’s second law of motion (m = F/a). The weight of a body is not absolute, since it depends on the gravitational field in which it is measured. Weight is defined as W = mg where g is the acceleration due to gravity.

UNITS: SI SYSTEM VS. FPS SYSTEM
SI system: In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in kilograms (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s2, then the weight is expressed in newtons (N). On the earth’s surface, g can be taken as g = 9.81 m/s2. W (N) = m (kg) g (m/s2) => N = kg·m/s2 FPS System: In the FPS system of units, weight is a base unit and mass is a derived unit. Weight is typically specified in pounds (lb), and mass is calculated from m = W/g. If g is specified in units of ft/s2, then the mass is expressed in slugs. On the earth’s surface, g is approximately 32.2 ft/s2. m (slugs) = W (lb)/g (ft/s2) => slug = lb·s2/ft

EQUATION OF MOTION (Section 13.2)
The motion of a particle is governed by Newton’s second law, relating the unbalanced forces on a particle to its acceleration. If more than one force acts on the particle, the equation of motion can be written F = FR = ma where FR is the resultant force, which is a vector summation of all the forces. To illustrate the equation, consider a particle acted on by two forces. First, draw the particle’s free-body diagram, showing all forces acting on the particle. Next, draw the kinetic diagram, showing the inertial force ma acting in the same direction as the resultant force FR.

INERTIAL FRAME OF REFERENCE
This equation of motion is only valid if the acceleration is measured in a Newtonian or inertial frame of reference. What does this mean? For problems concerned with motions at or near the earth’s surface, we typically assume our “inertial frame” to be fixed to the earth. We neglect any acceleration effects from the earth’s rotation. For problems involving satellites or rockets, the inertial frame of reference is often fixed to the stars.

EQUATION OF MOTION FOR A SYSTEM OF PARTICLES
(Section 13.3) The equation of motion can be extended to include systems of particles. This includes the motion of solids, liquids, or gas systems. As in statics, there are internal forces and external forces acting on the system. What is the difference between them? Using the definitions of m = mi as the total mass of all particles and aG as the acceleration of the center of mass G of the particles, then maG = miai . The text shows the details, but for a system of particles: F = maG where F is the sum of the external forces acting on the entire system.

KEY POINTS 1) Newton’s second law is a “law of nature”-- experimentally proven, not the result of an analytical proof. 2) Mass (property of an object) is a measure of the resistance to a change in velocity of the object. 3) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = m a, i.e., W = m g. 4) Unbalanced forces cause the acceleration of objects This condition is fundamental to all dynamics problems!

PROCEDURE FOR THE APPLICATION OF THE
EQUATION OF MOTION 1) Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used. 2) Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate components. 3) Draw the kinetic diagram, showing the particle’s inertial force, ma. Resolve this vector into its appropriate components. 4) Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5) It may be necessary to apply the proper kinematic relations to generate additional equations.

Find: Draw the free-body and kinetic diagrams of the block.
EXAMPLE 10 kg Given: A 10-kg block is subjected to the force F=500 N. A spring of stiffness k=500 N/m is mounted against the block. When s=0, the block is at rest and the spring is uncompressed. The contact surface is smooth. Find: Draw the free-body and kinetic diagrams of the block. Plan:

EXAMPLE Given: A 10-kg block is subjected to the force F=500 N. A spring of stiffness k=500 N/m is mounted against the block. When s=0, the block is at rest and the spring is uncompressed. The contact surface is smooth. Find: Draw the free-body and kinetic diagrams of the block. Plan: 1) Define an inertial coordinate system. 2) Draw the block’s free-body diagram, showing all external forces applied to the block in the proper directions. 3) Draw the block’s kinetic diagram, showing the inertial force vector ma in the proper direction.

EXAMPLE (continued) 10 a Solution:
1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of the block: The weight force (W) acts through the block’s center of mass. F is the applied load and Fs =500s (N) is the spring force, where s is the spring deformation. The normal force (N) is perpendicular to the surface. There is no friction force since the contact surface is smooth. y x W = 10 g Fs=500 s (N) N F=500 (N) 3 4 3) Draw the kinetic diagram of the block: The block will be moved to the right. The acceleration can be directed to the right if the block is speeding up or to the left if it is slowing down. 10 a 153

CONCEPT QUIZ 1. The block (mass = m) is moving upward with a speed v. Draw the FBD if the kinetic friction coefficient is k. A) B) C) D) None of the above. mg kmg N kN v Answers: 1. B 154

CONCEPT QUIZ (continued)
2. Packaging for oranges is tested using a machine that exerts ay = 20 m/s2 and ax = 3 m/s2, simultaneously. Select the correct FBD and kinetic diagram for this condition. A) B) C) D) = • may max W Ry Rx y x Answers: 2. A 155

GROUP PROBLEM SOLVING Given: The block and cylinder have a mass of m. The coefficient of kinetic friction at all surfaces of contact is m. Block A is moving to the right. Find: Draw the free-body and kinetic diagrams of each block. Plan: 1) Define an inertial coordinate system. 2) Draw the free-body diagrams for each block, showing all external forces. 3) Draw the kinetic diagrams for each block, showing the inertial forces.

Solution: 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of each block: y x 2T WB = mg Block B: y x Block A: T NA FfA = mNA WA = mg The friction force opposes the motion of block A relative to the surfaces on which it slides. Block A: maA Block B: maB 3) Draw the kinetic diagram of each block:

B) equal and opposite and do not affect the calculations.
ATTENTION QUIZ 1. Internal forces are not included in an equation of motion analysis because the internal forces are_____ A) equal to zero. B) equal and opposite and do not affect the calculations. C) negligibly small. D) not important. 2. A 10 lb block is initially moving down a ramp with a velocity of v. The force F is applied to bring the block to rest. Select the correct FBD. A) B) C) k10 10 N F kN v Answers 1. B 2. C 158

RECTANGULAR COORDINATES
EQUATIONS OF MOTION: RECTANGULAR COORDINATES Today’s Objectives: Students will be able to: Apply Newton’s second law to determine forces and accelerations for particles in rectilinear motion. In-Class Activities: • Check Homework • Reading Quiz • Applications • Equations of Motion Using Rectangular (Cartesian) Coordinates • Concept Quiz • Group Problem Solving • Attention Quiz

A) in the direction of its motion. B) a kinetic friction.
READING QUIZ 1. In dynamics, the friction force acting on a moving object is always ________ A) in the direction of its motion B) a kinetic friction. C) a static friction D) zero. 2. If a particle is connected to a spring, the elastic spring force is expressed by F = ks . The “s” in this equation is the A) spring constant. B) un-deformed length of the spring. C) difference between deformed length and un-deformed length. D) deformed length of the spring. Answers: 1. B 2. C 160

APPLICATIONS If a man is trying to move a 100 lb crate, how large a force F must he exert to start moving the crate? What factors influence how large this force must be to start moving the crate? If the crate starts moving, is there acceleration present? What would you have to know before you could find these answers?

Objects that move in air (or other fluid) have a drag force acting on them. This drag force is a function of velocity. If the dragster is traveling with a known velocity and the magnitude of the opposing drag force at any instant is given as a function of velocity, can we determine the time and distance required for dragster to come to a stop if its engine is shut off? How ?

RECTANGULAR COORDINATES
(Section 13.4) The equation of motion, F = m a, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities, or mass. Remember, unbalanced forces cause acceleration! Three scalar equations can be written from this vector equation. The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k) or, as scalar equations, Fx = max , Fy = may , and Fz = maz .

PROCEDURE FOR ANALYSIS
Free Body Diagram (always critical!!) Establish your coordinate system and draw the particle’s free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ‘ma’ vector (sometimes called the inertial force) on a separate diagram. Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic linear spring, a spring force equal to ‘k s’ should be included on the FBD.

PROCEDURE FOR ANALYSIS (continued)
Equations of Motion If the forces can be resolved directly from the free-body diagram (often the case in 2-D problems), use the scalar form of the equation of motion. In more complex cases (usually 3-D), a Cartesian vector is written for every force and a vector analysis is often best. A Cartesian vector formulation of the second law is F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2-D.

Kinematics The second law only provides solutions for forces and accelerations. If velocity or position have to be found, kinematics equations are used once the acceleration is found from the equation of motion. Any of the kinematics tools learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem!

Given: The 200 lb mine car is hoisted up the incline.
EXAMPLE aP/c= 4ft/s2 Given: The 200 lb mine car is hoisted up the incline. The motor M pulls in the cable with an acceleration of 4 ft/s2. Find: The acceleration of the mine car and the tension in the cable. Plan: Draw the free-body and kinetic diagrams of the car. Using a dependent motion equation, determine an acceleration relationship between cable and mine car. Apply the equation of motion to determine the cable tension. 167

= EXAMPLE (continued) Solution:
1) Draw the free-body and kinetic diagrams of the mine car: = 30° x y W = mg 3T N maC Since the motion is up the incline, rotate the x-y axes. Motion occurs only in the x-direction. We are also neglecting any friction in the wheel bearings, etc., on the cart.

The cable equation results in sp + 2 sc = lt
EXAMPLE (continued) The cable equation results in sp + 2 sc = lt Taking the derivative twice yields ap + 2 ac = 0 (eqn. 1) The relative acceleration equation is ap = ac + ap/c As the motor is mounted on the car, ap/c = 4 ft/s2 So, ap = ac + 4 ft/s2 (eqn. 2) Solving equations 1 and 2, yields aC=1.333 ft/s2 aP/c= 4ft/s2 sp sc

3) Apply the equation of motion in the x-direction:
EXAMPLE (continued) aP/c= 4ft/s2 3) Apply the equation of motion in the x-direction: +  Fx = max => 3T – mg(sin30°) = max => 3T – (200)(sin 30°) = (200/32.2) (1.333) => T = lb

1. If the cable has a tension of 3 N, determine the acceleration of block B. A) m/s2 B) m/s2 C) m/s2 D) m/s2 10 kg 4 kg k=0.4 2. Determine the acceleration of the block. A) 2.20 m/s2 B) 3.17 m/s2 C) 11.0 m/s2 D) 4.26 m/s2 5 kg 60 N • 30 Answers: 1. D 2. A 171

GROUP PROBLEM SOLVING Given: WA = 10 lb WB = 20 lb voA = 2 ft/s mk = 0.2 Find: vA when A has moved 4 feet to the right. Plan: This is not an easy problem, so think carefully about how to approach it!

GROUP PROBLEM SOLVING Given: WA = 10 lb WB = 20 lb voA = 2 ft/s mk = 0.2 Find: vA when A has moved 4 feet to the right. Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion to each. Using dependent motion equations, derive a relationship between aA and aB and use with the equation of motion formulas.

GROUP PROBLEM SOLVING (continued) Solution: Free-body and kinetic diagrams of B: 2T WB mBaB = Apply the equation of motion to B: ∑Fy= m ay WB – 2T = mB aB 20 – 2T = aB (1) 32.2 20 +

Free-body and kinetic diagrams of A:
GROUP PROBLEM SOLVING (continued) Free-body and kinetic diagrams of A: T N F = mkN WA mAaA = Apply the equations of motion to A: + ∑Fy = m ay = 0 N = WA = 10 lb F = kN = 2 lb ∑Fx= m ax F – T = mA aA 2 – T = aA 32.2 (2) 10 175

Now consider the kinematics. Constraint equation: sA + 2 sB = constant
GROUP PROBLEM SOLVING (continued) Now consider the kinematics. Constraint equation: sA + 2 sB = constant or vA + 2 vB = 0 Therefore aA + 2 aB = 0 aA = − 2 aB (3) (Notice aA is considered positive to the left and aB is positive downward.) sA sB Datums

GROUP PROBLEM SOLVING (continued) Now combine equations (1), (2), and (3). T = = 7.33 lb 22 3 aA = −17.16 ft/s2 = ft/s2 Now use the kinematic equation: vA = 11.9 ft/s (vA)2 = (v0A)2 +2 aA(sA − s0A) (vA)2 = (2)2 +2 (17.16)(4)

2. A 10 lb particle has forces of F1= (3i + 5j) lb and
ATTENTION QUIZ 1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s2 acceleration. A) N B) N C) N D) N T 60 a = 4 m/s2 2. A 10 lb particle has forces of F1= (3i + 5j) lb and F2= (-7i + 9j) lb acting on it. Determine the acceleration of the particle. A) (-0.4 i j) ft/s2 B) (-4 i + 14 j) ft/s2 C) (-12.9 i + 45 j) ft/s2 D) (13 i + 4 j) ft/s2 Answers: 1. C 2. C 178

NORMAL AND TANGENTIAL COORDINATES
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES Today’s Objectives: Students will be able to: Apply the equation of motion using normal and tangential coordinates. In-Class Activities: • Check Homework • Reading Quiz • Applications • Equation of Motion in n-t Coordinates • Concept Quiz • Group Problem Solving • Attention Quiz

A) impulse B) centripetal force C) tangential force D) inertia force
READING QUIZ 1. The “normal” component of the equation of motion is written as Fn=man, where Fn is referred to as the _______. A) impulse B) centripetal force C) tangential force D) inertia force 2. The positive n direction of the normal and tangential coordinates is ____________. A) normal to the tangential component B) always directed toward the center of curvature C) normal to the bi-normal component D) All of the above. Answers: 1. B 2. D 180

APPLICATIONS Race tracks are often banked in the turns to reduce the frictional forces required to keep the cars from sliding up to the outer rail at high speeds. If the car’s maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle (q) required to prevent the car from sliding up the track?

The picture shows a ride at the amusement park. The hydraulically-powered arms turn at a constant rate, which creates a centrifugal force on the riders. We need to determine the smallest angular velocity of the cars A and B so that the passengers do not loose contact with the seat. What parameters do we need for this calculation?

Satellites are held in orbit around the earth by using the earth’s gravitational pull as the centripetal force – the force acting to change the direction of the satellite’s velocity. Knowing the radius of orbit of the satellite, we need to determine the required speed of the satellite to maintain this orbit. What equation governs this situation?

NORMAL & TANGENTIAL COORDINATES (Section 13.5)
When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates. The normal direction (n) always points toward the path’s center of curvature. In a circle, the center of curvature is the center of the circle. The tangential direction (t) is tangent to the path, usually set as positive in the direction of motion of the particle.

EQUATIONS OF MOTION Since the equation of motion is a vector equation , F = ma, it may be written in terms of the n & t coordinates as Ftut + Fnun+ Fbub = mat+man Here Ft & Fn are the sums of the force components acting in the t & n directions, respectively. This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations: Ft = mat and Fn = man . Since there is no motion in the binormal (b) direction, we can also write Fb = 0.

NORMAL AND TANGENTIAL ACCERLERATIONS
The tangential acceleration, at = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of Ft, the particle’s speed will either be increasing or decreasing. The normal acceleration, an = v2/r, represents the time rate of change in the direction of the velocity vector. Remember, an always acts toward the path’s center of curvature. Thus, Fn will always be directed toward the center of the path. Recall, if the path of motion is defined as y = f(x), the radius of curvature at any point can be obtained from r = [1 + ( )2]3/2 dy dx d2y dx2

SOLVING PROBLEMS WITH n-t COORDINATES
• Use n-t coordinates when a particle is moving along a known, curved path. • Establish the n-t coordinate system on the particle. • Draw free-body and kinetic diagrams of the particle. The normal acceleration (an) always acts “inward” (the positive n-direction). The tangential acceleration (at) may act in either the positive or negative t direction. • Apply the equations of motion in scalar form and solve. • It may be necessary to employ the kinematic relations: at = dv/dt = v dv/ds an = v2/r

EXAMPLE Given: At the instant q = 45°, the boy with a mass of 75 kg, moves a speed of 6 m/s, which is increasing at 0.5 m/s2. Neglect his size and the mass of the seat and cords. The seat is pin connected to the frame BC. Find: Horizontal and vertical reactions of the seat on the boy. Plan: 1) Since the problem involves a curved path and requires finding the force perpendicular to the path, use n-t coordinates. Draw the boy’s free-body and kinetic diagrams. 2) Apply the equation of motion in the n-t directions.

1) The n-t coordinate system can be established on the boy at angle 45°. Approximating the boy and seat together as a particle, the free-body and kinetic diagrams can be drawn. W n t 45 Free-body diagram = Rx Ry n t man mat Kinetic diagram 189

EXAMPLE (continued) 2) Apply the equations of motion in the n-t directions. (a) Fn = man => – Rx cos 45° – Ry sin 45° +W sin 45° = man Using an = v2/r = 62/10, W = 75(9.81) N, and m = 75 kg, we get: – Rx cos 45° – Ry sin 45° = (75)(62/10) (1) (b) Ft = mat => – Rx sin 45° + Ry cos 45° – W cos 45° = mat we get: – Rx sin 45° + Ry cos 45° – 520.3= 75 (0.5) (2) Using equations (1) and (2), solve for Rx, Ry. Rx= –217 N, Ry=572 N

CONCEPT QUIZ 1. A 10 kg sack slides down a smooth surface. If the normal force on the surface at the flat spot, A, is 98.1 N () , the radius of curvature is ____. A) 0.2 m B) 0.4 m C) 1.0 m D) None of the above. A v=2m/s 2. A 20 lb block is moving along a smooth surface. If the normal force on the surface at A is 10 lb, the velocity is ________. A) 7.6 ft/s B) 9.6 ft/s C) 10.6 ft/s D) 12.6 ft/s =7 ft A Answers: 1. D 2. C 191

GROUP PROBLEM SOLVING Given: A 800 kg car is traveling over the hill having the shape of a parabola. When it is at point A, it is traveling at 9 m/s and increasing its speed at 3 m/s2. Find: The resultant normal force and resultant frictional force exerted on the road at point A. Plan: 1) Treat the car as a particle. Draw the free-body and kinetic diagrams. 2) Apply the equations of motion in the n-t directions. 3) Use calculus to determine the slope and radius of curvature of the path at point A.

Solution: 1) The n-t coordinate system can be established on the car at point A. Treat the car as a particle and draw the free-body and kinetic diagrams: t q n N F W = t n mat man W = mg = weight of car N = resultant normal force on road F = resultant friction force on road

2) Apply the equations of motion in the n-t directions:  Fn = man => W cos q – N = man Using W = mg and an = v2/r = (9)2/r => (800)(9.81) cos q – N = (800) (81/r) => N = 7848 cos q – 64800/r (1)  Ft = mat => W sin q – F = mat Using W = mg and at = 3 m/s2 (given) => (800)(9.81) sin q – F = (800) (3) => F = 7848 sin q – (2)

3) Determine r by differentiating y = f(x) at x = 80 m: y = 20(1 – x2/6400) => dy/dx = (–40) x / 6400 => d2y/dx2 = (–40) / 6400 r = = [1 + ( )2]3/2 dy dx d2y dx2 [1 + (–0.5)2]3/2 x = 80 m = m Determine q from the slope of the curve at A: tan q = dy/dx q = tan-1 (dy/dx) = tan-1 (-0.5) = 26.6° x = 80 m q dy dx 195

From Eq.(1): N = 7848 cos q – / r = 7848 cos (26.6°) – / = 6728 N From Eq.(2): F = 7848 sin q – 2400 = 7848 sin (26.6°) – 2400 = 1114 N

1. The tangential acceleration of an object
ATTENTION QUIZ 1. The tangential acceleration of an object A) represents the rate of change of the velocity vector’s direction. B) represents the rate of change in the magnitude of the velocity. C) is a function of the radius of curvature. D) Both B and C. 2. The block has a mass of 20 kg and a speed of v = 30 m/s at the instant it is at its lowest point. Determine the tension in the cord at this instant. A) 1596 N B) 1796 N C) 1996 N D) 2196 N  10 m v = 30m/s Answers: 1. B 2. C 197

EQUATIONS OF MOTION: CYLINDRICAL COORDINATES Today’s Objectives: Students will be able to: Analyze the kinetics of a particle using cylindrical coordinates. In-Class Activities: • Check Homework • Reading Quiz • Applications • Equations of Motion Using Cylindrical Coordinates • Angle Between Radial and Tangential Directions • Concept Quiz • Group Problem Solving • Attention Quiz

A) radial line. B) transverse direction.
READING QUIZ The normal force which the path exerts on a particle is always perpendicular to the _________ A) radial line. B) transverse direction. C) tangent to the path. D) None of the above. When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the __________ direction. A) radial B) tangential C) transverse D) None of the above. Answers: 1. C 2. B 199

APPLICATIONS The forces acting on the 100-lb boy can be analyzed using the cylindrical coordinate system. How would you write the equation describing the frictional force on the boy as he slides down this helical slide?

When an airplane executes the vertical loop shown above, the centrifugal force causes the normal force (apparent weight) on the pilot to be smaller than her actual weight. How would you calculate the velocity necessary for the pilot to experience weightlessness at A?

(Section 13.6) This approach to solving problems has some external similarity to the normal & tangential method just studied. However, the path may be more complex or the problem may have other attributes that make it desirable to use cylindrical coordinates. Equilibrium equations or “Equations of Motion” in cylindrical coordinates (using r, q , and z coordinates) may be expressed in scalar form as:  Fr = mar = m (r – r q 2 )  Fq = maq = m (r q – 2 r q )  Fz = maz = m z . ..

(continued) If the particle is constrained to move only in the r – q plane (i.e., the z coordinate is constant), then only the first two equations are used (as shown below). The coordinate system in such a case becomes a polar coordinate system. In this case, the path is only a function of q.  Fr = mar = m(r – rq 2 )  Fq = maq = m(rq – 2rq ) . .. Note that a fixed coordinate system is used, not a “body-centered” system as used in the n – t approach.

TANGENTIAL AND NORMAL FORCES
If a force P causes the particle to move along a path defined by r = f (q ), the normal force N exerted by the path on the particle is always perpendicular to the path’s tangent. The frictional force F always acts along the tangent in the opposite direction of motion. The directions of N and F can be specified relative to the radial coordinate by using angle y .

DETERMINATION OF ANGLE y
The angle y, defined as the angle between the extended radial line and the tangent to the curve, can be required to solve some problems. It can be determined from the following relationship. dr/d r dr r d = tan y If y is positive, it is measured counterclockwise from the radial line to the tangent. If it is negative, it is measured clockwise.

EXAMPLE Given: The ball (P) is guided along the vertical circular path. W = 0.5 lb, q = 0.4 rad/s, q = 0.8 rad/s2, rc = 0.4 ft Find: Force of the arm OA on the ball when q = 30. Plan: . ..

EXAMPLE Given: The ball (P) is guided along the vertical circular path. W = 0.5 lb, q = 0.4 rad/s, q = 0.8 rad/s2, rc = 0.4 ft Find: Force of the arm OA on the ball when q = 30. . .. Plan: Draw a FBD. Then develop the kinematic equations and finally solve the kinetics problem using cylindrical coordinates. Solution: Notice that r = 2rc cos q, therefore: r = -2rc sin q q r = -2rc cos q q 2 – 2rc sin q q . ..

= mar EXAMPLE (continued)
Free Body Diagram and Kinetic Diagram : Establish the r,  inertial coordinate system and draw the particle’s free body diagram. mg Ns NOA 2q ma mar = 208

EXAMPLE (continued) Kinematics: at q = 30 r = 2(0.4) cos(30) = ft r = - 2(0.4) sin(30)(0.4) = ft/s r = - 2(0.4) cos(30)(0.4)2 – 2(0.4) sin(30)(0.8) = ft/s2 .. . Acceleration components are ar = r – rq 2 = – (0.693)(0.4)2 = ft/s2 aq = rq + 2rq = (0.693)(0.8) + 2(-0.16)(0.4) = ft/s2 .. .

= mar EXAMPLE (continued) Equation of motion: r direction  Fr = mar
Ns cos(30) – 0.5 sin(30) = (-0.542) Ns = lb 0.5 32.2 = ma mar

= mar EXAMPLE (continued) Equation of motion:  direction  Fq = maq
NOA sin(30) – 0.5 cos(30) = (0.426) NOA = 0.3 lb 0.5 32.2 = ma mar

When a pilot flies an airplane in a
CONCEPT QUIZ When a pilot flies an airplane in a vertical loop of constant radius r at constant speed v, his apparent weight is maximum at A) Point A B) Point B (top of the loop) C) Point C D) Point D (bottom of the loop) r A B C D If needing to solve a problem involving the pilot’s weight at Point C, select the approach that would be best. A) Equations of Motion: Cylindrical Coordinates B) Equations of Motion: Normal & Tangential Coordinates C) Equations of Motion: Polar Coordinates D) No real difference – all are bad. E) Toss up between B and C. Answers: 1. D 2. C 212

GROUP PROBLEM SOLVING Given: The smooth particle is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path. The cord’s stiffness is k=30 N/m unstretched length = 0.25 m m=0.08 kg, r = (0.8 sin ) m Find: Forces of the guide on the particle when  = 60 and  = 5 rad/s, which is constant. Plan: Determine r and r by differentiating r. Draw Free Body Diagram & Kinetic Diagram. Solve for the accelerations, and apply the equation of motion to find the forces. . ..

Solution: Kinematics: r = 0.8 (sin  ) r = 0.8 (cos  )  r = -0.8 (sin  )  (cos  )  ·· When  = 60;  = 5 rad/s,  = 0 rad/s2. r = m r = 2 m/s r = m/s2 ·· Accelerations : ar = r − r q 2 = − (0.6928) 52 = m/s2 aq = r q + 2 r q = (0.6928) (2) 5 = 20 m/s2 ·· 214

Free Body Diagram & Kinetic Diagram mar ma = where the spring force F will be Fs = k s = 30 ( ) = N Kinetics:  Fr = mar => N cos 30 = 0.08 (-34.64) N = N

Kinetics:  F = ma => F - N sin 30 = 0.08 (20) F = N mar ma =

For the path defined by r = q 2 , the angle y at q = 0.5 rad is
ATTENTION QUIZ For the path defined by r = q 2 , the angle y at q = 0.5 rad is A) 10º B) 14º C) 26º D) 75º 2. If r = q 2 and q = 2t, find the magnitude of r and q when t = 2 seconds. A) 4 cm/sec, 2 rad/sec B) 4 cm/sec, 0 rad/sec2 C) 8 cm/sec, 16 rad/sec D) 16 cm/sec, 0 rad/sec2 ·· Answers: 1. B 2. D 217

THE WORK OF A FORCE, THE PRINCIPLE OF WORK AND ENERGY & SYSTEMS OF PARTICLES
Today’s Objectives: Students will be able to: Calculate the work of a force. Apply the principle of work and energy to a particle or system of particles. In-Class Activities: • Check Homework • Reading Quiz • Applications • Work of A Force • Principle of Work And Energy • Concept Quiz • Group Problem Solving • Attention Quiz

1. What is the work done by the force F? A) F s B) –F s
READING QUIZ 1. What is the work done by the force F? A) F s B) –F s C) Zero D) None of the above. s s1 s2 F 2. If a particle is moved from 1 to 2, the work done on the particle by the force, FR will be A) B) C) D) Anwers: 1. C 2. A 219

APPLICATIONS A roller coaster makes use of gravitational forces to assist the cars in reaching high speeds in the “valleys” of the track. How can we design the track (e.g., the height, h, and the radius of curvature, r) to control the forces experienced by the passengers?

Crash barrels are often used along roadways for crash protection. The barrels absorb the car’s kinetic energy by deforming. If we know the velocity of an oncoming car and the amount of energy that can be absorbed by each barrel, how can we design a crash cushion?

WORK AND ENERGY Another equation for working kinetics problems involving particles can be derived by integrating the equation of motion (F = ma) with respect to displacement. By substituting at = v (dv/ds) into Ft = mat, the result is integrated to yield an equation known as the principle of work and energy. This principle is useful for solving problems that involve force, velocity, and displacement. It can also be used to explore the concept of power. To use this principle, we must first understand how to calculate the work of a force.

WORK OF A FORCE (Section 14.1)
A force does work on a particle when the particle undergoes a displacement along the line of action of the force. Work is defined as the product of force and displacement components acting in the same direction. So, if the angle between the force and displacement vector is q, the increment of work dU done by the force is dU = F ds cos q By using the definition of the dot product and integrating, the total work can be written as ò r2 r1 U1-2 = F • dr

WORK OF A FORCE (continued)
If F is a function of position (a common case) this becomes ò = s2 s1 F cos q ds U1-2 If both F and q are constant (F = Fc), this equation further simplifies to U1-2 = Fc cos q (s2 - s1) Work is positive if the force and the movement are in the same direction. If they are opposing, then the work is negative. If the force and the displacement directions are perpendicular, the work is zero.

WORK OF A WEIGHT The work done by the gravitational force acting on a particle (or weight of an object) can be calculated by using - W (y2 − y1) = - W Dy - W dy = U1-2 = ò y2 y1 The work of a weight is the product of the magnitude of the particle’s weight and its vertical displacement. If Dy is upward, the work is negative since the weight force always acts downward.

The work of the spring force moving from position s1 to position s2 is
WORK OF A SPRING FORCE When stretched, a linear elastic spring develops a force of magnitude Fs = ks, where k is the spring stiffness and s is the displacement from the unstretched position. The work of the spring force moving from position s1 to position s2 is = 0.5 k (s2)2 – 0.5 k (s1)2 k s ds Fs ds U1-2 s2 s1 = ò If a particle is attached to the spring, the force Fs exerted on the particle is opposite to that exerted on the spring. Thus, the work done on the particle by the spring force will be negative or U1-2 = – [ 0.5 k (s2)2 – 0.5 k (s1)2 ] .

It is important to note the following about spring forces.
1. The equations above are for linear springs only! Recall that a linear spring develops a force according to F = ks (essentially the equation of a line). 2. The work of a spring is not just spring force times distance at some point, i.e., (ksi)(si). Beware, this is a trap that students often fall into! 3. Always double check the sign of the spring work after calculating it. It is positive work if the force put on the object by the spring and the movement are in the same direction. 227

PRINCIPLE OF WORK AND ENERGY
(Section 14.2 & Section 14.3) By integrating the equation of motion,  Ft = mat = mv(dv/ds), the principle of work and energy can be written as  U1-2 = 0.5 m (v2)2 – 0.5 m (v1)2 or T1 +  U1-2 = T2 U1-2 is the work done by all the forces acting on the particle as it moves from point 1 to point 2. Work can be either a positive or negative scalar. T1 and T2 are the kinetic energies of the particle at the initial and final position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)2. The kinetic energy is always a positive scalar (velocity is squared!). So, the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to final position is equal to the particle’s final kinetic energy.

PRINCIPLE OF WORK AND ENERGY (continued)
Note that the principle of work and energy (T1 +  U1-2 = T2) is not a vector equation! Each term results in a scalar value. Both kinetic energy and work have the same units, that of energy! In the SI system, the unit for energy is called a joule (J), where 1 J = 1 N·m. In the FPS system, units are ft·lb. The principle of work and energy cannot be used, in general, to determine forces directed normal to the path, since these forces do no work. The principle of work and energy can also be applied to a system of particles by summing the kinetic energies of all particles in the system and the work due to all forces acting on the system.

WORK OF FRICTION CAUSED BY SLIDING
The case of a body sliding over a rough surface merits special consideration. Consider a block which is moving over a rough surface. If the applied force P just balances the resultant frictional force k N, a constant velocity v would be maintained. The principle of work and energy would be applied as 0.5m (v)2 + P s – (k N) s = 0.5m (v)2 This equation is satisfied if P = k N. However, we know from experience that friction generates heat, a form of energy that does not seem to be accounted for in this equation. It can be shown that the work term (k N)s represents both the external work of the friction force and the internal work that is converted into heat.

Find: The distance s when the block stops. Plan:
EXAMPLE Given: When s = 0.6 m, the spring is not stretched or compressed, and the 10 kg block, which is subjected to a force of F= 100 N, has a speed of 5 m/s down the smooth plane. Find: The distance s when the block stops. Plan: Since this problem involves forces, velocity and displacement, apply the principle of work and energy to determine s. 231

Apply the principle of work and energy between position 1
EXAMPLE (continued) Solution: Apply the principle of work and energy between position 1 (s1 = 0.6 m) and position 2 (s2). Note that the normal force (N) does no work since it is always perpendicular to the displacement. T1 + U1-2 = T2 There is work done by three different forces; 1) work of a the force F =100 N; UF = 100 (s2− s1) = 100 (s2 − 0.6) 2) work of the block weight; UW = 10 (9.81) (s2− s1) sin 30 = (s2 − 0.6) 3) and, work of the spring force. US = (200) (s2−0.6)2 = -100 (s2 − 0.6)2 232

The work and energy equation will be T1 + U1-2 = T2
EXAMPLE (continued) The work and energy equation will be T1 + U1-2 = T2 0.5 (10) (s2 − 0.6) (s2 − 0.6) − 100(s2 − 0.6)2 = 0  (s2 − 0.6) − 100(s2 − 0.6)2 = 0 Solving for (s2 − 0.6), (s2 − 0.6) = { ± ( – 4×(-100)×125)0.5} / 2(-100) Selecting the positive root, indicating a positive spring deflection, (s2 − 0.6) = 2.09 m Therefore, s2 = 2.69 m 233

A) -[0.5 k(4 in)2 - 0.5 k(2 in)2] B) 0.5 k (2 in)2
CONCEPT QUIZ 1. A spring with an un-stretched length of 5 in expands from a length of 2 in to a length of 4 in. The work done on the spring is _________ in·lb . A) -[0.5 k(4 in) k(2 in)2] B) 0.5 k (2 in)2 C) -[0.5 k(3 in) k(1 in)2] D) 0.5 k(3 in) k(1 in)2 2. If a spring force is F = 5 s3 N/m and the spring is compressed by s = 0.5 m, the work done on a particle attached to the spring will be A) N · m B) – N · m C) N · m D) – N · m Answers: 1. C 2. D 234

GROUP PROBLEM SOLVING Given: Block A has a weight of 60 lb and block B has a weight of 40 lb. The coefficient of kinetic friction between the blocks and the incline is mk = Neglect the mass of the cord and pulleys. Find: The speed of block A after block B moves 2 ft up the plane, starting from rest. Plan: 1) Define the kinematic relationships between the blocks. 2) Draw the FBD of each block. 3) Apply the principle of work and energy to the system of blocks. Why choose this method?

Solution: 1) The kinematic relationships can be determined by defining position coordinates sA and sB, and then differentiating. sA sB Since the cable length is constant: 2sA + sB = l 2DsA + DsB = 0 When DsB = -2 ft => DsA = 1 ft and 2vA + vB = 0 => vB = -2vA Note that, by this definition of sA and sB, positive motion for each block is defined as downwards.

2) Draw the FBD of each block. y x NA mNA 2T WA A 60 30 B NB mNB T WB Sum forces in the y-direction for block A (note that there is no motion in y-direction): Fy = 0: NA – WA cos 60 = 0 NA = WA cos 60 Similarly, for block B: NB = WB cos 30

3) Apply the principle of work and energy to the system (the blocks start from rest). T1 + U1-2 = T2 [0.5mA(vA1)2 + .5mB(vB1)2] + [WA sin 60– 2T – mNA]DsA + [WB sin 30– T + mNB]DsB = [0.5mA(vA2) mB(vB2)2] where vA1 = vB1 = 0, DsA = 1ft, DsB = -2 ft, vB = -2vA, NA = WA cos 60, NB = WB cos 30 => [0 + 0] + [60 sin 60– 2T – 0.1(60 cos 60)] (1) + [40 sin 30 – T + 0.1(40 cos 30)] (-2) = [0.5(60/32.2)(vA2) (40/32.2)(-2vA2)2]

Again, the Work and Energy equation is: => [0 + 0] + [60 sin 60– 2T – 0.1(60 cos 60)] (1) + [40 sin 30 – T + 0.1(40 cos 30)] (-2) = [0.5(60/32.2)(vA2) (40/32.2)(-2vA2)2] Solving for the unknown velocity yeilds => vA2 = ft/s Note that the work due to the cable tension force on each block cancels out.

ATTENTION QUIZ 1. What is the work done by the normal force N if a 10 lb box is moved from A to B ? A) lb · ft B) lb · ft C) lb · ft D) lb · ft 2. Two blocks are initially at rest. How many equations would be needed to determine the velocity of block A after block B moves 4 m horizontally on the smooth surface? A) One B) Two C) Three D) Four Answers: 1. B 2. C (Work & Energy, length of cord equation, velocity of blocks equation) 2 kg 240

POWER AND EFFICIENCY Today’s Objectives: Students will be able to:
Determine the power generated by a machine, engine, or motor. Calculate the mechanical efficiency of a machine. In-Class Activities: • Check Homework • Reading Quiz • Applications • Define Power • Define Efficiency • Concept Quiz • Group Problem Solving • Attention Quiz

1. The formula definition of power is ___________. A) dU / dt B) F  v
READING QUIZ 1. The formula definition of power is ___________. A) dU / dt B) F  v C) F  dr/dt D) All of the above. 2. Kinetic energy results from _______. A) displacement B) velocity C) gravity D) friction Answers: 1. D 2. B 242

APPLICATIONS Engines and motors are often rated in terms of their power output. The power output of the motor lifting this elevator is related to the vertical force F acting on the elevator, causing it to move upwards. Given a desired lift velocity for the elevator (with a known maximum load), how can we determine the power requirement of the motor? 243

The speed at which a truck can climb a hill depends in part on the power output of the engine and the angle of inclination of the hill. For a given angle, how can we determine the speed of this truck, knowing the power transmitted by the engine to the wheels? Can we find the speed, if we know the power? If we know the engine power output and speed of the truck, can we determine the maximum angle of climb of this truck ? 244

P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
POWER AND EFFICIENCY (Section 14.4) Power is defined as the amount of work performed per unit of time. If a machine or engine performs a certain amount of work, dU, within a given time interval, dt, the power generated can be calculated as P = dU/dt Since the work can be expressed as dU = F • dr, the power can be written P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v Thus, power is a scalar defined as the product of the force and velocity components acting in the same direction. 245

POWER Using scalar notation, power can be written P = F • v = F v cos q where q is the angle between the force and velocity vectors. So if the velocity of a body acted on by a force F is known, the power can be determined by calculating the dot product or by multiplying force and velocity components. The unit of power in the SI system is the Watt (W) where 1 W = 1 J/s = 1 (N · m)/s . In the FPS system, power is usually expressed in units of horsepower (hp) where 1 hp = 550 (ft · lb)/s = 746 W .

e = (power output) / (power input)
EFFICIENCY The mechanical efficiency of a machine is the ratio of the useful power produced (output power) to the power supplied to the machine (input power) or e = (power output) / (power input) If energy input and removal occur at the same time, efficiency may also be expressed in terms of the ratio of output energy to input energy or e = (energy output) / (energy input) Machines will always have frictional forces. Since frictional forces dissipate energy, additional power will be required to overcome these forces. Consequently, the efficiency of a machine is always less than 1.

• Find the resultant external force acting on the body causing its motion. It may be necessary to draw a free-body diagram. • Determine the velocity of the point on the body at which the force is applied. Energy methods or the equation of motion and appropriate kinematic relations, may be necessary. • Multiply the force magnitude by the component of velocity acting in the direction of F to determine the power supplied to the body (P = F v cos q ). • In some cases, power may be found by calculating the work done per unit of time (P = dU/dt). • If the mechanical efficiency of a machine is known, either the power input or output can be determined.

EXAMPLE Given: A 50 kg block (A) is hoisted by the pulley system and motor M. The motor has an efficiency of At this instant, point P on the cable has a velocity of 12 m/s which is increasing at a rate of 6 m/s2. Neglect the mass of the pulleys and cable. Find: The power supplied to the motor at this instant. Plan: 1) Relate the cable and block velocities by defining position coordinates. Draw a FBD of the block. 2) Use the equation of motion to determine the cable tension. 3) Calculate the power supplied by the motor and then to the motor.

1) Define position coordinates to relate velocities.
EXAMPLE (continued) Solution: 1) Define position coordinates to relate velocities. Datum SA SP Here sP is defined to a point on the cable. Also sA is defined only to the lower pulley, since the block moves with the pulley. From kinematics, sP + 2 sA = l  aP + 2 aA = 0 aA = − aP / 2 = −3 m/s2 (↑) sB sm Draw the FBD and kinetic diagram of the block: 2T WA A mA aA = 250

EXAMPLE (continued) 2) The tension of the cable can be obtained by applying the equation of motion to the block. +↑ Fy = mA aA 2T − = 50 (3)  T = N 3) The power supplied by the motor is the product of the force applied to the cable and the velocity of the cable. Po = F • v = (320.3)(12) = 3844 W The power supplied to the motor is determined using the motor’s efficiency and the basic efficiency equation. Pi = Po/e = 3844/0.8 = 4804 W = 4.8 kW

A) (1000 lb)(260 ft/s) B) (2000 lb)(260 ft/s) cos 10
CONCEPT QUIZ 1. A motor pulls a 10 lb block up a smooth incline at a constant velocity of 4 ft/s. Find the power supplied by the motor. A) 8.4 ft·lb/s B) 20 ft·lb/s C) 34.6 ft·lb/s D) 40 ft·lb/s 30º 2. A twin engine jet aircraft is climbing at a 10 degree angle at 260 ft/s. The thrust developed by a jet engine is 1000 lb. The power developed by the aircraft is A) (1000 lb)(260 ft/s) B) (2000 lb)(260 ft/s) cos 10 C) (1000 lb)(260 ft/s) cos D) (2000 lb)(260 ft/s) Answers: B D 252

GROUP PROBLEM SOLVING Given: A sports car has a mass of 2000 kg and an engine efficiency of e = Moving forward, the wind creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. The car accelerates at 5 m/s2, starting from rest. Find: The engine’s input power when t = 4 s. Plan: 1) Draw a free body diagram of the car. 2) Apply the equation of motion and kinematic equations to find the car’s velocity at t = 4 s. 3) Determine the output power required for this motion. 4) Use the engine’s efficiency to determine input power.

GROUP PROBLEM SOLVING (continued) Solution: 1) Draw the FBD of the car. The drag force and weight are known forces. The normal force Nc and frictional force Fc represent the resultant forces of all four wheels. The frictional force between the wheels and road pushes the car forward. 2) The equation of motion can be applied in the x-direction, with ax = 5 m/s2: + Fx = max => Fc – 1.2v2 = (2000)(5) => Fc = (10, v2) N

GROUP PROBLEM SOLVING (continued) 3) The constant acceleration equations can be used to determine the car’s velocity. vx = vxo + axt = 0 + (5)(4) = 20 m/s 4) The power output of the car is calculated by multiplying the driving (frictional) force and the car’s velocity: Po = (Fc)(vx ) = [10,000 + (1.2)(20)2](20) = kW 5) The power developed by the engine (prior to its frictional losses) is obtained using the efficiency equation. Pi = Po/e = 209.6/0.65 = 322 kW

ATTENTION QUIZ 1. The power supplied by a machine will always be _________ the power supplied to the machine. A) less than B) equal to C) greater than D) A or B 2. A car is traveling a level road at 88 ft/s. The power being supplied to the wheels is 52,800 ft·lb/s. Find the combined friction force on the tires. A) lb B) 400 lb C) 600 lb D) 4.64 x 106 lb Answers: A C 256

CONSERVATIVE FORCES, POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Today’s Objectives: Students will be able to: Understand the concept of conservative forces and determine the potential energy of such forces. Apply the principle of conservation of energy. In-Class Activities: Check Homework Reading Quiz Applications Conservative Force Potential Energy Conservation of Energy Concept Quiz Group Problem Solving Attention Quiz

1. The potential energy of a spring is ________
READING QUIZ 1. The potential energy of a spring is ________ A) always negative. B) always positive. C) positive or negative. D) equal to ks. 2. When the potential energy of a conservative system increases, the kinetic energy _________ A) always decreases. B) always increases. C) could decrease or D) does not change. increase. Answers: 1. B 2. A (Conservation of Energy Principal) 258

APPLICATIONS The weight of the sacks resting on this platform causes potential energy to be stored in the supporting springs. As each sack is removed, the platform will rise slightly since some of the potential energy within the springs will be transformed into an increase in gravitational potential energy of the remaining sacks. If the sacks weigh 100 lb and the equivalent spring constant is k = 500 lb/ft, what is the energy stored in the springs?

The boy pulls the water balloon launcher back, stretching each of the four elastic cords. If we know the unstretched length and stiffness of each cord, can we estimate the maximum height and the maximum range of the water balloon when it is released from the current position ? 260

The roller coaster is released from rest at the top of the hill. As the coaster moves down the hill, potential energy is transformed into kinetic energy. What is the velocity of the coaster when it is at B and C? Also, how can we determine the minimum height of the hill so that the car travels around both inside loops without leaving the track?

CONSERVATIVE FORCE (Section 14.5)
A force F is said to be conservative if the work done is independent of the path followed by the force acting on a particle as it moves from A to B. This also means that the work done by the force F in a closed path (i.e., from A to B and then back to A) is zero. Thus, we say the work is conserved. ò r F = d x y z A B F The work done by a conservative force depends only on the positions of the particle, and is independent of its velocity or acceleration.

CONSERVATIVE FORCE (continued)
A more rigorous definition of a conservative force makes use of a potential function (V) and partial differential calculus, as explained in the text. However, even without the use of the these mathematical relationships, much can be understood and accomplished. The “conservative” potential energy of a particle/system is typically written using the potential function V. There are two major components to V commonly encountered in mechanical systems, the potential energy from gravity and the potential energy from springs or other elastic elements. Vtotal = Vgravity + Vsprings

POTENTIAL ENERGY Potential energy is a measure of the amount of work a conservative force will do when a body changes position. In general, for any conservative force system, we can define the potential function (V) as a function of position. The work done by conservative forces as the particle moves equals the change in the value of the potential function (e.g., the sum of Vgravity and Vsprings). It is important to become familiar with the two types of potential energy and how to calculate their magnitudes.

POTENTIAL ENERGY DUE TO GRAVITY
The potential function (formula) for a gravitational force, e.g., weight (W = mg), is the force multiplied by its elevation from a datum. The datum can be defined at any convenient location. Vg = ± W y Vg is positive if y is above the datum and negative if y is below the datum. Remember, YOU get to set the datum. 265

ELASTIC POTENTIAL ENERGY
Recall that the force of an elastic spring is F = ks. It is important to realize that the potential energy of a spring, while it looks similar, is a different formula. k s2 2 1 V e = Ve (where ‘e’ denotes an elastic spring) has the distance “s” raised to a power (the result of an integration) or Notice that the potential function Ve always yields positive energy.

CONSERVATION OF ENERGY (Section 14.6)
When a particle is acted upon by a system of conservative forces, the work done by these forces is conserved and the sum of kinetic energy and potential energy remains constant. In other words, as the particle moves, kinetic energy is converted to potential energy and vice versa. This principle is called the principle of conservation of energy and is expressed as 2 1 V T + = = Constant T1 stands for the kinetic energy at state 1 and V1 is the potential energy function for state 1. T2 and V2 represent these energy states at state 2. Recall, the kinetic energy is defined as T = ½ mv2.

EXAMPLE Given: The 2 kg collar is moving down with the velocity of 4 m/s at A. The spring constant is 30 N/m. The unstretched length of the spring is 1 m. Find: The velocity of the collar when s = 1 m. Plan: Apply the conservation of energy equation between A and C. Set the gravitational potential energy datum at point A or point C (in this example, choose point A—why?).

EXAMPLE (continued) Solution: Note that the potential energy at C has two parts. VC = (VC)e + (VC)g VC = 0.5 (30) (√5 – 1)2 – 2 (9.81) 1 The kinetic energy at C is TC = 0.5 (2) v2 Similarly, the potential and kinetic energies at A will be VA = 0.5 (30) (2 – 1)2, TA = 0.5 (2) 42 The energy conservation equation becomes TA + VA = TC + VC . [ 0.5(30) (√5 – 1)2 – 2(9.81)1 ] (2) v2 = [0.5 (30) (2 – 1)2 ]+ 0.5 (2) 42  v = 5.26 m/s

C) +10 ft·lb. D) None of the above.
CONCEPT QUIZ 1. If the work done by a conservative force on a particle as it moves between two positions is –10 ft·lb, the change in its potential energy is _______ A) 0 ft·lb. B) -10 ft·lb. C) +10 ft·lb. D) None of the above. 2. Recall that the work of a spring is U1-2 = -½ k(s22 – s12) and can be either positive or negative. The potential energy of a spring is V = ½ ks2. Its value is __________ A) always negative B) either positive or negative. C) always positive D) an imaginary number! Answers: 1. C 2. C 270

GROUP PROBLEM SOLVING Given: The 800 kg roller coaster starts from A with a speed of 3 m/s. Find: The minimum height, h, of the hill so that the car travels around inside loop at B without leaving the track. Also find the normal reaction on the car when the car is at C for this height of A. Plan: Note that only kinetic energy and potential energy due to gravity are involved. Determine the velocity at B using the equation of equilibrium and then apply the conservation of energy equation to find minimum height h .

Solution: 1) Placing the datum at A: TA + VA = TB + VB  0.5 (800) = 0.5 (800) (vB)2 − 800(9.81) (h − 20) (1) 2) Find the required velocity of the coaster at B so it doesn’t leave the track. Equation of motion applied at B: r 2 v m ma F n = å 10 800 (9.81) = 800 (vB)2  vB = m/s man mg NB  0 = 272

Now using the energy conservation, eq. (1), the minimum h can be determined. 0.5 (800) = 0.5 (800) (9.905)2 − 800(9.81) (h − 20)  h= 24.5 m 3) To find the normal reaction at C, we need vc. TA + VA = TC + VC  0.5 (800) = 0.5 (800) (vC)2 − 800(9.81) (24.5 − 14)  VC = m/s Equation of motion applied at B: r 2 v m F n = å 7 NC+800 (9.81) = 800 14.662  man mg NC =  NC = 16.8 kN 273

C) the same amount of work D) It is a mystery!
ATTENTION QUIZ 1. The principle of conservation of energy is usually ______ to apply than the principle of work & energy. A) harder B) easier C) the same amount of work D) It is a mystery! 2. If the pendulum is released from the horizontal position, the velocity of its bob in the vertical position is _____ A) 3.8 m/s. B) 6.9 m/s. C) m/s. D) 21 m/s. Answers: 1. B 2. A 274

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM
Today’s Objectives: Students will be able to: Calculate the linear momentum of a particle and linear impulse of a force. Apply the principle of linear impulse and momentum. In-Class Activities: • Check Homework • Reading Quiz • Applications • Linear Momentum and Impulse • Principle of Linear Impulse and Momentum • Concept Quiz • Group Problem Solving • Attention Quiz

A) friction force B) equation of motion
READING QUIZ 1. The linear impulse and momentum equation is obtained by integrating the ______ with respect to time. A) friction force B) equation of motion C) kinetic energy D) potential energy 2. Which parameter is not involved in the linear impulse and momentum equation? A) Velocity B) Displacement C) Time D) Force Answers: 1. B 2. B 276

APPLICATIONS A dent in an automotive fender can be removed using an impulse tool, which delivers a force over a very short time interval. To do so the weight is gripped and jerked upwards, striking the stop ring. How can we determine the magnitude of the linear impulse applied to the fender? Could you analyze a carpenter’s hammer striking a nail in the same fashion? Sure!

When a stake is struck by a sledgehammer, a large impulse force is delivered to the stake and drives it into the ground. If we know the initial speed of the sledgehammer and the duration of impact, how can we determine the magnitude of the impulsive force delivered to the stake?

(Section 15.1) The next method we will consider for solving particle kinetics problems is obtained by integrating the equation of motion with respect to time. The result is referred to as the principle of impulse and momentum. It can be applied to problems involving both linear and angular motion. This principle is useful for solving problems that involve force, velocity, and time. It can also be used to analyze the mechanics of impact (taken up in a later section).

(continued) The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. The equation of motion can be written F = m a = m (dv/dt) Separating variables and integrating between the limits v = v1 at t = t1 and v = v2 at t = t2 results in mv2 – mv1 dv m F dt v2 v1 t2 t1 = ò å This equation represents the principle of linear impulse and momentum. It relates the particle’s final velocity (v2) and initial velocity (v1) and the forces acting on the particle as a function of time.

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (continued)
Linear momentum: The vector mv is called the linear momentum, denoted as L. This vector has the same direction as v. The linear momentum vector has units of (kg·m)/s or (slug·ft)/s. Linear impulse: The integral F dt is the linear impulse, denoted I. It is a vector quantity measuring the effect of a force during its time interval of action. I acts in the same direction as F and has units of N·s or lb·s. The impulse may be determined by direct integration. Graphically, it can be represented by the area under the force versus time curve. If F is constant, then I = F (t2 – t1) .

(continued) The particle’s initial momentum plus the sum of all the impulses applied from t1 to t2 is equal to the particle’s final momentum. The principle of linear impulse and momentum in vector form is written as F dt t2 t1 å ò mv1 + = mv2 The two momentum diagrams indicate direction and magnitude of the particle’s initial and final momentum, mv1 and mv2. The impulse diagram is similar to a free body diagram, but includes the time duration of the forces acting on the particle.

ò IMPULSE AND MOMENTUM: SCALAR EQUATIONS
Since the principle of linear impulse and momentum is a vector equation, it can be resolved into its x, y, z component scalar equations: m(vx)1 +  Fx dt = m(vx)2 m(vy)1 +  Fy dt = m(vy)2 m(vz)1 +  Fz dt = m(vz)2 t2 t1 ò The scalar equations provide a convenient means for applying the principle of linear impulse and momentum once the velocity and force vectors have been resolved into x, y, z components.

PROBLEM SOLVING • Establish the x, y, z coordinate system. • Draw the particle’s free body diagram and establish the direction of the particle’s initial and final velocities, drawing the impulse and momentum diagrams for the particle. Show the linear momenta and force impulse vectors. • Resolve the force and velocity (or impulse and momentum) vectors into their x, y, z components, and apply the principle of linear impulse and momentum using its scalar form. • Forces as functions of time must be integrated to obtain impulses. If a force is constant, its impulse is the product of the force’s magnitude and time interval over which it acts.

EXAMPLE Given: A 0.5 kg ball strikes the rough ground and rebounds with the velocities shown. Neglect the ball’s weight during the time it impacts the ground. Find: The magnitude of impulsive force exerted on the ball. Plan: 1) Draw the momentum and impulse diagrams of the ball as it hits the surface. 2) Apply the principle of impulse and momentum to determine the impulsive force.

+ = EXAMPLE (continued) Solution:
1) The impulse and momentum diagrams can be drawn as: + =  W dt  0  N dt  0  F dt mv2 30° mv1 45° The impulse caused by the ball’s weight and the normal force N can be neglected because their magnitudes are very small as compared to the impulse from the ground.

ò ò ò  EXAMPLE (continued)
2) The principle of impulse and momentum can be applied along the direction of motion: mv1 +  F dt = mv2 t2 t1 ò 0.5 (25 cos 45° i − 25 sin 45° j) +  F dt = 0.5 (10 cos 30° i + 10 sin 30° j) t2 t1 ò  The impulsive force vector is I =  F dt = (4.509 i j ) Ns Magnitude: I = √ = 12.2 Ns ò t1 t2

1. Calculate the impulse due to the force. A) 20 kg·m/s B) 10 kg·m/s C) 5 N·s D) 15 N·s F 10 N t 2 s Force curve 2. A constant force F is applied for 2 s to change the particle’s velocity from v1 to v2. Determine the force F if the particle’s mass is 2 kg. A) (17.3 j) N B) (–10 i j) N C) (20 i j) N D) ( 10 i j) N v2=20 m/s v1=10 m/s 60 Answers: 1. B 2. C 288

GROUP PROBLEM SOLVING Given: The 20 kg crate is resting on the floor. The motor M pulls on the cable with a force of F, which has a magnitude that varies as shown on the graph. Find: The speed of the crate when t = 6 s. Plan: 1) Determine the force needed to begin lifting the crate, and then the time needed for the motor to generate this force. 2) After the crate starts moving, apply the principle of impulse and momentum to determine the speed of the crate at t = 6 s. 289

Solution: 1) The crate begins moving when the cable force F exceeds the crate weight. Solve for the force, then the time. F = mg = (20) (9.81) = N F = N = 50 t t = s 2) Apply the principle of impulse and momentum from the time the crate starts lifting at t1 = s to t2 = 6 s. Note that there are two external forces (cable force and weight) we need to consider. A. The impulse due to cable force: F dt = [0.5(250) 5 + (250) 1] – 0.5(196.2)3.924= Ns ò 3.924 6 +↑ 290

B. The impulse due to weight: +↑ (− mg) dt = − (6 − 3.924) = − Ns ò 3.924 6 Now, apply the principle of impulse and momentum mv1 +  F dt = mv2 where v1 = 0 t2 t1 ò − = (20) v2 => v2 = 4.14 m/s +↑ 291

ATTENTION QUIZ 1. Jet engines on the 100 Mg VTOL aircraft exert a constant vertical force of 981 kN as it hovers. Determine the net impulse on the aircraft over t = 10 s. A) -981 kN·s B) 0 kN·s C) 981 kN·s D) kN·s 30 2. A 100 lb cabinet is placed on a smooth surface. If a force of a 100 lb is applied for 2 s, determine the net impulse on the cabinet during this time interval. A) 0 lb·s B) 100 lb·s C) 200 lb·s D) 300 lb·s Answers: 1. B (impulse of thrust – impulse of weight) 2. B (impulse of F – impulse of W sin30) 292

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM AND CONSERVATION OF LINEAR MOMENTUM FOR SYSTEMS OF PARTICLES Today’s Objectives: Students will be able to: Apply the principle of linear impulse and momentum to a system of particles. Understand the conditions for conservation of momentum. In-Class Activities: • Check Homework • Reading Quiz • Applications • Linear Impulse and Momentum for a System of Particles • Conservation of Linear Momentum • Concept Quiz • Group Problem Solving • Attention Quiz 293

A) equal the external impulses. B) sum to zero.
READING QUIZ The internal impulses acting on a system of particles always __________ A) equal the external impulses. B) sum to zero. C) equal the impulse of weight. D) None of the above. 2. If an impulse-momentum analysis is considered during the very short time of interaction, as shown in the picture, weight is a/an __________ A) impulsive force. B) explosive force. C) non-impulsive force. D) internal force. Answers: 1. B 2. C (it is non-impulsive since the weight will have a negligible effect on the change in momentum) 294

APPLICATIONS As the wheels of this pitching machine rotate, they apply frictional impulses to the ball, thereby giving it linear momentum in the direction of Fdt and F ’dt. The weight impulse, Wt is very small since the time the ball is in contact with the wheels is very small. Does the release velocity of the ball depend on the mass of the ball?

This large crane-mounted hammer is used to drive piles into the ground. Conservation of momentum can be used to find the velocity of the pile just after impact, assuming the hammer does not rebound off the pile. If the hammer rebounds, does the pile velocity change from the case when the hammer doesn’t rebound ? Why ? In the impulse-momentum analysis, do we have to consider the impulses of the weights of the hammer and pile and the resistance force ? Why or why not ?

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES
(Section 15.2) For the system of particles shown, the internal forces fi between particles always occur in pairs with equal magnitude and opposite directions. Thus the internal impulses sum to zero. The linear impulse and momentum equation for this system only includes the impulse of external forces. mi(vi)2 dt Fi mi(vi)1 t2 t1 å = + ò

MOTION OF THE CENTER OF MASS
For a system of particles, we can define a “fictitious” center of mass of an aggregate particle of mass mtot, where mtot is the sum (å mi) of all the particles. This system of particles then has an aggregate velocity of vG = (å mivi) / mtot. The motion of this fictitious mass is based on motion of the center of mass for the system. The position vector rG = (å miri) / mtot describes the motion of the center of mass.

CONSERVATION OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES (Section 15
When the sum of external impulses acting on a system of objects is zero, the linear impulse-momentum equation simplifies to å mi(vi)1 = å mi(vi)2 This equation is referred to as the conservation of linear momentum. Conservation of linear momentum is often applied when particles collide or interact. When particles impact, only impulsive forces cause a change of linear momentum. The sledgehammer applies an impulsive force to the stake. The weight of the stake is considered negligible, or non-impulsive, as compared to the force of the sledgehammer. Also, provided the stake is driven into soft ground with little resistance, the impulse of the ground acting on the stake is considered non-impulsive.

= EXAMPLE I vA vB y A vi vC B M C x z
Given: M = 100 kg, vi = 20j (m/s) mA = 20 kg, vA = 50i + 50j (m/s) mB = 30 kg, vB = -30i – 50k (m/s) An explosion has broken the mass m into 3 smaller particles, a, b and c. z x y M vi vC C vB B vA A = Find: The velocity of fragment C after the explosion. Plan: Since the internal forces of the explosion cancel out, we can apply the conservation of linear momentum to the SYSTEM.

100(20j) = 20(50i + 50j) + 30(-30i-50k) + 50(vcx i + vcy j + vcz k)
EXAMPLE I (continued) Solution: mvi = mAvA + mBvB + mCvC 100(20j) = 20(50i + 50j) + 30(-30i-50k) + 50(vcx i + vcy j + vcz k) Equating the components on the left and right side yields: 0 = 1000 – (vcx) vcx = -2 m/s 2000 = (vcy) vcy = 20 m/s 0 = (vcz) vcz = 30 m/s So vc = (-2i + 20j + 30k) m/s immediately after the explosion.

EXAMPLE II Given: Two rail cars with masses of mA = 20 Mg and mB = 15 Mg and velocities as shown. Find: The speed of the car A after collision if the cars collide and rebound such that B moves to the right with a speed of 2 m/s. Also find the average impulsive force between the cars if the collision place in 0.5 s. Plan: Use conservation of linear momentum to find the velocity of the car A after collision (all internal impulses cancel). Then use the principle of impulse and momentum to find the impulsive force by looking at only one car.

Solution: Conservation of linear momentum (x-dir): mA(vA1) + mB(vB1) = mA(vA2)+ mB(vB2) 20,000 (3) + 15,000 (-1.5) = (20,000) vA2 + 15,000 (2) vA2 = m/s Impulse and momentum on car A (x-dir): mA (vA1)+ ∫ F dt = mA (vA2) 20,000 (3) - ∫ F dt = 20,000 (0.375) ∫ F dt = 52,500 N·s The average force is ∫ F dt = 52,500 N·s = Favg(0.5 sec); Favg = 105 kN

C) not subject to Newton’s second law. D) Both A and C.
CONCEPT QUIZ 1) Over the short time span of a tennis ball hitting the racket during a player’s serve, the ball’s weight can be considered _____________ A) nonimpulsive. B) impulsive. C) not subject to Newton’s second law. D) Both A and C. 2) A drill rod is used with a air hammer for making holes in hard rock so explosives can be placed in them. How many impulsive forces act on the drill rod during the drilling? A) None B) One C) Two D) Three Answers: A. A B. C (The impact on the top of the rod and the reaction at the end in the hard rock. This assumes the weight of the rod itself is small in comparision to the other two forces.) 304

GROUP PROBLEM SOLVING Given: The free-rolling ramp has a weight of 120 lb. The 80 lb crate slides from rest at A, 15 ft down the ramp to B. Assume that the ramp is smooth, and neglect the mass of the wheels. Find: The ramp’s speed when the crate reaches B. Plan: Use the energy conservation equation as well as conservation of linear momentum and the relative velocity equation (you thought you could safely forget it?) to find the velocity of the ramp.

Solution: Energy conservation equation: (3/5) (15) = 0.5 (80/32.2)(vB) (120/32.2)(vr)2 + To find the relations between vB and vr, use conservation of linear momentum: → 0 = (120/32.2) vr − (80/32.2) vBx  vBx =1.5 vr (1) Since vB = vr + vB/r  -vBx i + vBy j = vr i + vB/r (−4/5 i −3/5 j)  -vBx = vr − (4/5) vB/r (2) vBy = − (3/5) vB/r (3) Eliminating vB/r from Eqs. (2) and (3) and Substituting Eq. (1) results in vBy =1.875 vr 306

Then, energy conservation equation can be rewritten ; (3/5) (15) = 0.5 (80/32.2)(vB) (120/32.2)(vr)2 (3/5) (15) = 0.5 (80/32.2) [(1.5 vr)2 +(1.875 vr)2] + 0.5 (120/32.2) (vr)2 720 = (vr)2 vr = 8.93 ft/s

ATTENTION QUIZ 1. The 20 g bullet is fired horizontally at 1200 m/s into the g block resting on a smooth surface. If the bullet becomes embedded in the block, what is the velocity of the block immediately after impact. A) 1125 m/s B) 80 m/s C) 1200 m/s D) 75 m/s 1200 m/s 2. The 200-g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900-g, moving at 47 m/s. During the impact with the bat, how many impulses of importance are used to find the final velocity of the ball? A) Zero B) One C) Two D) Three Answers: A. D B. C BAT vball vbat 308

IMPACT Today’s Objectives: Students will be able to:
Understand and analyze the mechanics of impact. Analyze the motion of bodies undergoing a collision, in both central and oblique cases of impact. In-Class Activities: • Check Homework • Reading Quiz • Applications • Central Impact • Coefficient of Restitution • Oblique Impact • Concept Quiz • Group Problem Solving • Attention Quiz 309

A) central impact. B) oblique impact.
READING QUIZ 1. When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be ________ A) central impact. B) oblique impact. C) major impact. D) None of the above. 2. The ratio of the restitution impulse to the deformation impulse is called _________ A) impulse ratio. B) restitution coefficient. C) energy ratio. D) mechanical efficiency. Answers: 1. B 2. B 310

APPLICATIONS The quality of a tennis ball is measured by the height of its bounce. This can be quantified by the coefficient of restitution of the ball. If the height from which the ball is dropped and the height of its resulting bounce are known, how can we determine the coefficient of restitution of the ball? 311

In the game of billiards, it is important to be able to predict the trajectory and speed of a ball after it is struck by another ball. If we know the velocity of ball A before the impact, how can we determine the magnitude and direction of the velocity of ball B after the impact? What parameters do we need to know for this? 312

IMPACT (Section 15.4) Impact occurs when two bodies collide during a very short time period, causing large impulsive forces to be exerted between the bodies. Common examples of impact are a hammer striking a nail or a bat striking a ball. The line of impact is a line through the mass centers of the colliding particles. In general, there are two types of impact: Central impact occurs when the directions of motion of the two colliding particles are along the line of impact. Oblique impact occurs when the direction of motion of one or both of the particles is at an angle to the line of impact. 313

CENTRAL IMPACT Central impact happens when the velocities of the two objects are along the line of impact (recall that the line of impact is a line through the particles’ mass centers). vA vB Line of impact Once the particles contact, they may deform if they are non-rigid. In any case, energy is transferred between the two particles. There are two primary equations used when solving impact problems. The textbook provides extensive detail on their derivation. 314

CENTRAL IMPACT (continued)
In most problems, the initial velocities of the particles, (vA)1 and (vB)1, are known, and it is necessary to determine the final velocities, (vA)2 and (vB)2. So the first equation used is the conservation of linear momentum, applied along the line of impact. (mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2 This provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So another equation is needed. The principle of impulse and momentum is used to develop this equation, which involves the coefficient of restitution, or e. 315

CENTRAL IMPACT (continued)
The coefficient of restitution, e, is the ratio of the particles’ relative separation velocity after impact, (vB)2 – (vA)2, to the particles’ relative approach velocity before impact, (vA)1 – (vB)1. The coefficient of restitution is also an indicator of the energy lost during the impact. The equation defining the coefficient of restitution, e, is (vA)1 - (vB)1 (vB)2 – (vA)2 e = If a value for e is specified, this relation provides the second equation necessary to solve for (vA)2 and (vB)2. 316

COEFFICIENT OF RESTITUTION
In general, e has a value between zero and one. The two limiting conditions can be considered: • Elastic impact (e = 1): In a perfectly elastic collision, no energy is lost and the relative separation velocity equals the relative approach velocity of the particles. In practical situations, this condition cannot be achieved. • Plastic impact (e = 0): In a plastic impact, the relative separation velocity is zero. The particles stick together and move with a common velocity after the impact. Some typical values of e are: Steel on steel: 0.5 – 0.8 Wood on wood: 0.4 – 0.6 Lead on lead: 0.12 – 0.18 Glass on glass: 0.93 – 0.95 317

 U1-2 =  T2 −  T1 where Ti = 0.5mi (vi)2
IMPACT: ENERGY LOSSES Once the particles’ velocities before and after the collision have been determined, the energy loss during the collision can be calculated on the basis of the difference in the particles’ kinetic energy. The energy loss is  U1-2 =  T2 −  T1 where Ti = 0.5mi (vi)2 During a collision, some of the particles’ initial kinetic energy will be lost in the form of heat, sound, or due to localized deformation. In a plastic collision (e = 0), the energy lost is a maximum, although it does not necessarily go to zero. Why? 318

OBLIQUE IMPACT In an oblique impact, one or both of the particles’ motion is at an angle to the line of impact. Typically, there will be four unknowns: the magnitudes and directions of the final velocities. The four equations required to solve for the unknowns are: Conservation of momentum and the coefficient of restitution equation are applied along the line of impact (x-axis): mA(vAx)1 + mB(vBx)1 = mA(vAx)2 + mB(vBx)2 e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1] Momentum of each particle is conserved in the direction perpendicular to the line of impact (y-axis): mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2 319

• In most impact problems, the initial velocities of the particles and the coefficient of restitution, e, are known, with the final velocities to be determined. • Define the x-y axes. Typically, the x-axis is defined along the line of impact and the y-axis is in the plane of contact perpendicular to the x-axis. • For both central and oblique impact problems, the following equations apply along the line of impact (x-dir.):  m(vx)1 =  m(vx)2 and e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1] • For oblique impact problems, the following equations are also required, applied perpendicular to the line of impact (y-dir.): mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2 320

EXAMPLE Given: The ball strikes the smooth wall
with a velocity (vb)1 = 20 m/s. The coefficient of restitution between the ball and the wall is e = 0.75. Find: The velocity of the ball just after the impact. Plan: The collision is an oblique impact, with the line of impact perpendicular to the plane (through the relative centers of mass). Thus, the coefficient of restitution applies perpendicular to the wall and the momentum of the ball is conserved along the wall. 321

The momentum of the ball is conserved in the y-dir:
EXAMPLE (continued) Solution: Solve the impact problem by using x-y axes defined along and perpendicular to the line of impact, respectively: The momentum of the ball is conserved in the y-dir: m(vb)1 sin 30° = m(vb)2 sin  (vb)2 sin  = 10 m/s (1) The coefficient of restitution applies in the x-dir: e = [ 0 – (vbx)2 ] / [ (vbx)1 – 0 ]  = [ 0 – (-vb)2 cos  ] / [ 20 cos 30° – 0]  (vb)2 cos  = m/s (2) Using Eqs. (1) and (2) and solving for the velocity and  yields: (vb)2 = ( )0.5 = 16.4 m/s  = tan-1(10/12.99)=37.6° 322

C) Impossible to tell D) Don’t pick this one!
CONCEPT QUIZ 1. Two balls impact with a coefficient of restitution of Can one of the balls leave the impact with a kinetic energy greater than before the impact? A) Yes B) No C) Impossible to tell D) Don’t pick this one! 2. Under what condition is the energy lost during a collision maximum? A) e = B) e = 0.0 C) e = D) Collision is non-elastic. Answers: 1. A (Nothing was said about the masses of the balls!) 2. B ( D merely means that e is less than 1.0) 323

GROUP PROBLEM SOLVING Plan:
Given: A 2 kg crate B is released from rest, falls a distance h = 0.5 m, and strikes plate P (3 kg mass). The coefficient of restitution between B and P is e = 0.6, and the spring stiffness is k = 30 N/m. Find: The velocity of crate B just after the collision. Plan: 1) Determine the speed of the crate just before the collision using projectile motion or an energy method. 2) Analyze the collision as a central impact problem. 324

Solution: 1) Determine the speed of block B just before impact by using conservation of energy (why?). Define the gravitational datum at the initial position of the block (h1 = 0) and note the block is released from rest (v1 = 0): T1 + V1 = T2 + V2 0.5m(v1)2 + mgh1 = 0.5m(v2)2 + mgh2 0 + 0 = 0.5(2)(v2)2 + (2)(9.81)(-0.5) v2 = m/s This is the speed of the block just before the collision. Plate (P) is at rest, velocity of zero, before the collision. 325

2) Analyze the collision as a central impact problem. (vB)2 (vP)2 (vP)1 = 0 (vB)1 = m/s B P Apply conservation of momentum to the system in the vertical direction: + mB(vB)1 + mP(vP)1 = mB(vB)2 + mP(vP)2 (2)(-3.132) + 0 = (2)(vB)2 + (3)(vP)2 Using the coefficient of restitution: + e = [(vP)2 – (vB)2]/[(vB)1 – (vP)1] => = [(vP)2 – (vB)2]/[ – 0] => = (vP)2 – (vB)2 Solving the two equations simultaneously yields (vB)2 = m/s and (vP)2 = m/s Both the block and plate will travel down after the collision. 326

C) less than (vx) 1 D) greater than (vx) 1 v x y
ATTENTION QUIZ 1. Block B (1 kg) is moving on the smooth surface at 10 m/s when it squarely strikes block A (3 kg), which is at rest. If the velocity of block A after the collision is 4 m/s to the right, (vB)2 is A) 2 m/s B) 7 m/s C) 7 m/s D) 2 m/s B A vB=10 m/s 2. A particle strikes the smooth surface with a velocity of 30 m/s. If e = 0.8, (vx) 2 is _____ after the collision. A) zero B) equal to (vx) 1 C) less than (vx) 1 D) greater than (vx) 1  v 30 m/s x y 30 Answers: 1. D 2. C 327

ANGULAR MOMENTUM, MOMENT OF A FORCE AND PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM
Today’s Objectives: Students will be able to: Determine the angular momentum of a particle and apply the principle of angular impulse & momentum. Use conservation of angular momentum to solve problems. In-Class Activities: • Check Homework • Reading Quiz • Applications • Angular Momentum • Angular Impulse and Momentum Principle • Conservation of Angular Momentum • Concept Quiz • Group Problem Solving • Attention Quiz 328

A) angular momentum of the particle.
READING QUIZ 1. Select the correct expression for the angular momentum of a particle about a point. A) r × v B) r × (m v) C) v × r D) (m v) × r 2. The sum of the moments of all external forces acting on a particle is equal to A) angular momentum of the particle. B) linear momentum of the particle. C) time rate of change of angular momentum. D) time rate of change of linear momentum. Answers: A. B B. A 329

APPLICATIONS Planets and most satellites move in elliptical orbits. This motion is caused by gravitational attraction forces. Since these forces act in pairs, the sum of the moments of the forces acting on the system will be zero. This means that angular momentum is conserved. If the angular momentum is constant, does it mean the linear momentum is also constant? Why or why not? 330

The passengers on the amusement-park ride experience conservation of angular momentum about the axis of rotation (the z-axis). As shown on the free body diagram, the line of action of the normal force, N, passes through the z-axis and the weight’s line of action is parallel to it. Therefore, the sum of moments of these two forces about the z-axis is zero. If the passenger moves away from the z-axis, will his speed increase or decrease? Why? 331

ANGULAR MOMENTUM (Section 15.5)
The angular momentum of a particle about point O is defined as the “moment” of the particle’s linear momentum about O. i j k Ho = r × mv = rx ry rz mvx mvy mvz The magnitude of Ho is (Ho)z = mv d 332

RELATIONSHIP BETWEEN MOMENT OF A FORCE AND ANGULAR MOMENTUM
(Section 15.6) The resultant force acting on the particle is equal to the time rate of change of the particle’s linear momentum. Showing the time derivative using the familiar “dot” notation results in the equation F = L = mv We can prove that the resultant moment acting on the particle about point O is equal to the time rate of change of the particle’s angular momentum about point O or Mo = r × F = Ho 333

PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM (Section 15.7)
Considering the relationship between moment and time rate of change of angular momentum Mo = Ho = dHo/dt By integrating between the time interval t1 to t2 å ò - = 2 1 ) ( t Ho dt Mo + or This equation is referred to as the principle of angular impulse and momentum. The second term on the left side,  Mo dt, is called the angular impulse. In cases of 2D motion, it can be applied as a scalar equation using components about the z-axis. 334

CONSERVATION OF ANGULAR MOMENTUM
When the sum of angular impulses acting on a particle or a system of particles is zero during the time t1 to t2, the angular momentum is conserved. Thus, (HO)1 = (HO)2 An example of this condition occurs when a particle is subjected only to a central force. In the figure, the force F is always directed toward point O. Thus, the angular impulse of F about O is always zero, and angular momentum of the particle about O is conserved. 335

Given: A satellite has an elliptical orbit about earth.
EXAMPLE Given: A satellite has an elliptical orbit about earth. msatellite = 700 kg mearth = × 1024 kg vA = 10 km/s rA = 15 × 106 m fA = 70° Find: The speed, vB, of the satellite at its closest distance, rB, from the center of the earth. Plan: Apply the principles of conservation of energy and conservation of angular momentum to the system. 336

EXAMPLE (continued) Solution: 1 G ms me 1 G ms me 2 rA 2 rB 1024) ×
Conservation of energy: TA + VA = TB + VB becomes ms vA2 – = ms vB2 – where G = 66.73×10-12 m3/(kg·s2). Dividing through by ms and substituting values yields: G ms me G ms me 2 rA rB rB 1024) × -12(5.976 10 73 . 66 v 5 106 x 15 2 ) 000 , ( B - = 23.4 × 106 = 0.5 (vB)2 – (3.99 × 1014)/rB or 337

EXAMPLE (continued) Solution:
Now use Conservation of Angular Momentum. (rA ms vA) sin fA = rB ms vB (15 × 106)(10,000) sin 70° = rB vB or rB = ( × 109)/vB Solving the two equations for rB and vB yields rB = 13.8 × 106 m vB = 10.2 km/s 338

A) x direction. B) y direction. C) z direction. D) x - y direction.
CONCEPT QUIZ 1. If a particle moves in the x - y plane, its angular momentum vector is in the A) x direction. B) y direction. C) z direction. D) x - y direction. If there are no external impulses acting on a particle A) only linear momentum is conserved. B) only angular momentum is conserved. C) both linear momentum and angular momentum are conserved. D) neither linear momentum nor angular momentum are conserved. Answers: A. C B. C 339

A moment acts on the shaft as shown, M = 0.5t + 0.8 lb·ft).
GROUP PROBLEM SOLVING Given: The four 5 lb spheres are rigidly attached to the crossbar frame, which has a negligible weight. A moment acts on the shaft as shown, M = 0.5t lb·ft). Find: The velocity of the spheres after 4 seconds, starting from rest. Plan: Apply the principle of angular impulse and momentum about the axis of rotation (z-axis). 340

Solution: Angular momentum: HZ = r × mv reduces to a scalar equation. (HZ)1 = 0 and (HZ)2 = 4×{(5/32.2) (0.6) v2} = v2 Angular impulse: M dt = (0.5t + 0.8) dt = [(0.5/2) t t] = 7.2 lb·ft·s ò 2 1 t 4 Apply the principle of angular impulse and momentum. = v  v2 = 19.4 ft/s 341

A) Conservation of energy B) Conservation of angular momentum
ATTENTION QUIZ 1. A ball is traveling on a smooth surface in a 3 ft radius circle with a speed of 6 ft/s. If the attached cord is pulled down with a constant speed of 2 ft/s, which of the following principles can be applied to solve for the velocity of the ball when r = 2 ft? A) Conservation of energy B) Conservation of angular momentum C) Conservation of linear momentum D) Conservation of mass Answers: 1. A 2. A 2. If a particle moves in the z - y plane, its angular momentum vector is in the A) x direction. B) y direction. C) z direction. D) z - y direction. 342

PLANAR RIGID BODY MOTION: TRANSLATION & ROTATION
Today’s Objectives : Students will be able to: Analyze the kinematics of a rigid body undergoing planar translation or rotation about a fixed axis. In-Class Activities : • Check Homework • Reading Quiz • Applications • Types of Rigid-Body Motion • Planar Translation • Rotation About a Fixed Axis • Concept Quiz • Group Problem Solving • Attention Quiz

A) are usually different B) are always the same
READING QUIZ 1. If a rigid body is in translation only, the velocity at points A and B on the rigid body _______ . A) are usually different B) are always the same C) depend on their position D) depend on their relative position 2. If a rigid body is rotating with a constant angular velocity about a fixed axis, the velocity vector at point P is _______. A)   rp B) rp  C) drp/dt D) All of the above. Answers: 1. B A (should be w x r, not r x w) 344

APPLICATIONS Passengers on this amusement ride are subjected to curvilinear translation since the vehicle moves in a circular path but they always remains upright. If the angular motion of the rotating arms is known, how can we determine the velocity and acceleration experienced by the passengers? Why would we want to know these values? Does each passenger feel the same acceleration?

Gears, pulleys and cams, which rotate about fixed axes, are often used in machinery to generate motion and transmit forces. The angular motion of these components must be understood to properly design the system. To do this design, we need to relate the angular motions of contacting bodies that rotate about different fixed axes. How is this different than the analyses we did in earlier chapters?

RIGID BODY MOTION (Section 16.1)
There are cases where an object cannot be treated as a particle. In these cases the size or shape of the body must be considered. Rotation of the body about its center of mass requires a different approach. For example, in the design of gears, cams, and links in machinery or mechanisms, rotation of the body is an important aspect in the analysis of motion. We will now start to study rigid body motion. The analysis will be limited to planar motion. A body is said to undergo planar motion when all parts of the body move along paths equidistant from a fixed plane.

PLANAR RIGID BODY MOTION
There are three types of planar rigid body motion.

PLANAR RIGID BODY MOTION (continued)
Translation: Translation occurs if every line segment on the body remains parallel to its original direction during the motion. When all points move along straight lines, the motion is called rectilinear translation. When the paths of motion are curved lines, the motion is called curvilinear translation.

Rotation about a fixed axis: In this case, all the particles of the body, except those on the axis of rotation, move along circular paths in planes perpendicular to the axis of rotation. General plane motion: In this case, the body undergoes both translation and rotation. Translation occurs within a plane and rotation occurs about an axis perpendicular to this plane.

An example of bodies undergoing the three types of motion is shown in this mechanism. The wheel and crank undergo rotation about a fixed axis. In this case, both axes of rotation are at the location of the pins and perpendicular to the plane of the figure. The piston undergoes rectilinear translation since it is constrained to slide in a straight line. The connecting rod undergoes curvilinear translation, since it will remain horizontal as it moves along a circular path. The connecting rod undergoes general plane motion, as it will both translate and rotate.

RIGID-BODY MOTION: TRANSLATION
(Section 16.2) The positions of two points A and B on a translating body can be related by rB = rA + rB/A where rA & rB are the absolute position vectors defined from the fixed x-y coordinate system, and rB/A is the relative-position vector between B and A. The velocity at B is vB = vA+ drB/A/dt . Now drB/A/dt = 0 since rB/A is constant. So, vB = vA, and by following similar logic, aB = aA. Note, all points in a rigid body subjected to translation move with the same velocity and acceleration.

RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS (Section 16.3)
When a body rotates about a fixed axis, any point P in the body travels along a circular path. The angular position of P is defined by q. The change in angular position, d, is called the angular displacement, with units of either radians or revolutions. They are related by 1 revolution = (2) radians Angular velocity, , is obtained by taking the time derivative of angular displacement:  = d/dt (rad/s) + Similarly, angular acceleration is  = d2/dt2 = d/dt or  = (d/d) rad/s2

RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS (continued)
If the angular acceleration of the body is constant, a = aC, the equations for angular velocity and acceleration can be integrated to yield the set of algebraic equations below. w = w0 + aC t q = q0 + w0 t aC t2 w2 = (w0)2 + 2aC (q – q0) q0 and w0 are the initial values of the body’s angular position and angular velocity. Note these equations are very similar to the constant acceleration relations developed for the rectilinear motion of a particle.

RIGID-BODY ROTATION: VELOCITY OF POINT P
The magnitude of the velocity of P is equal to wr (the text provides the derivation). The velocity’s direction is tangent to the circular path of P. In the vector formulation, the magnitude and direction of v can be determined from the cross product of w and rp . Here rp is a vector from any point on the axis of rotation to P. v = w × rp = w × r The direction of v is determined by the right-hand rule.

RIGID-BODY ROTATION: ACCELERATION OF POINT P
The acceleration of P is expressed in terms of its normal (an) and tangential (at) components. In scalar form, these are at = a r and an = w2 r. The tangential component, at, represents the time rate of change in the velocity's magnitude. It is directed tangent to the path of motion. The normal component, an, represents the time rate of change in the velocity’s direction. It is directed toward the center of the circular path.

RIGID-BODY ROTATION: ACCELERATION OF POINT P (continued)
Using the vector formulation, the acceleration of P can also be defined by differentiating the velocity. a = dv/dt = dw/dt × rP + w × drP/dt = a × rP + w × (w × rP) It can be shown that this equation reduces to a = a × r – w2r = at + an The magnitude of the acceleration vector is a = (at)2 + (an)2

a = at + an = a × rP + w × (w × rP) = a × r – w2r
ROTATION ABOUT A FIXED AXIS: PROCEDURE • Establish a sign convention along the axis of rotation. • If a relationship is known between any two of the variables (a, w, q, or t), the other variables can be determined from the equations: w = dq/dt a = dw/dt a dq = w dw • If a is constant, use the equations for constant angular acceleration. • To determine the motion of a point, the scalar equations v = w r, at = a r, an = w2r , and a = (at)2 + (an)2 can be used. • Alternatively, the vector form of the equations can be used (with i, j, k components). v = w × rP = w × r a = at + an = a × rP + w × (w × rP) = a × r – w2r

EXAMPLE Given: The motor gives the blade an angular acceleration  = 20 e-0.6t rad/s2, where t is in seconds. The initial conditions are that when t = 0, the blade is at rest. Find: The velocity and acceleration of the tip P of one of the blades when t =3 s. How many revolutions has the blade turned in 3 s ? Plan: 1) Determine the angular velocity and displacement of the blade using kinematics of angular motion. 2) The magnitudes of the velocity and acceleration of point P can be determined from the scalar equations of motion for a point on a rotating body. Why scalar?

EXAMPLE (continued) Solution: 1) Since the angular acceleration is given as a function of time,  = 20 e-0.6t rad/s2 , the angular velocity and displacement can be found by integration. =   dt = 20  e-0.6t dt w = e-0.6t 20 (-0.6) when t = 3 s, w = rad/s  =  w dt  =  e-0.6t dt = e-0.6t 20 (-0.6) (-0.6)2 Angular displacement when t = 3 s,  = rad = 1.46 rev. Also , when t = 3 s,  = 20 e-0.6(3) b= rad/s2

The magnitude of the acceleration of P is determined by
EXAMPLE (continued) 2) The velocity of point P on the the fan, at a radius of 1.75 ft, is determined as vP = w r = (5.510)(1.75) = 9.64 ft/s The normal and tangential components of acceleration of point P are calculated as an = (w)2 r = (5.510)2 (1.75) = ft/s2 at = a r = (3.306)(1.75) = ft/s2 The magnitude of the acceleration of P is determined by aP = (an)2 + (at)2 = (53.13)2 + (5.786)2 = 53.4 ft/s2

A) (4 i + 32 j) ft/s2 B) (4 i - 32 j) ft/s2
CONCEPT QUIZ 1. A disk is rotating at 4 rad/s. If it is subjected to a constant angular acceleration of 2 rad/s2, determine the acceleration at B. A) (4 i + 32 j) ft/s B) (4 i - 32 j) ft/s2 C) (- 4 i + 32 j) ft/s D) (- 4 i -32 j) ft/s2 O A B x y 2 ft 2 rad/s2 2. A Frisbee is thrown and curves to the right. It is experiencing A) rectilinear translation. B) curvilinear translation. C) pure rotation D) general plane motion. Answer: 1. C 2. D 362

GROUP PROBLEM SOLVING Given: Starting from rest when gear A is given a constant angular acceleration, aA = 4.5 rad/s2. The cord is wrapped around pulley D which is rigidly attached to gear B. Find: The velocity of cylinder C and the distance it travels in 3 seconds. Plan: 1) The angular acceleration of gear B (and pulley D) can be related to aA. 2) The acceleration of cylinder C can be determined by using the equations for motion of a point on a rotating body since (at)D at point P is the same as ac. 3) The velocity and distance of C can be found by using the constant acceleration equations.

Solution: 1) Gear A and B will have the same speed and tangential component of acceleration at the point where they mesh. Thus, at = aArA = aBrB  (4.5)(75) = aB(225)  aB = 1.5 rad/s2 Since gear B and pulley D turn together, aD = aB = 1.5 rad/s2 2) Assuming the cord attached to pulley D is inextensible and does not slip, the velocity and acceleration of cylinder C will be the same as the velocity and tangential component of acceleration along the pulley D: aC = (at)D = aD rD = (1.5)(0.125) = m/s2

3) Since aA is constant, aD and aC will be constant. The constant acceleration equation for rectilinear motion can be used to determine the velocity and displacement of cylinder C when t = 3 s (s0= v0 = 0): vc = v0 + aC t = (3) = m/s sc = s0 + v0 t + (0.5) aC t2 = (0.5) (3)2 = m

ATTENTION QUIZ 1. The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular velocity of 4 rad/s, determine the speed of point P when the blades have turned 2 revolutions (when w = 8.14 rad/s). A) 14.2 ft/s B) ft/s C) 23.1 ft/s D) ft/s Answers: 1. A (using v=rw and the angular velocity provided) 2. D (Answer B is tangential acc. & C is the normal acc.) 2. Determine the magnitude of the acceleration at P when the blades have turned the 2 revolutions. A) 0 ft/s B) 3.5 ft/s2 C) ft/s D) 116 ft/s2 366

ABSOLUTE MOTION ANALYSIS
Today’s Objective: Students will be able to: Determine the velocity and acceleration of a rigid body undergoing general plane motion using an absolute motion analysis. In-Class Activities: • Check Homework • Reading Quiz • Applications • General Plane Motion • Concept Quiz • Group Problem Solving • Attention Quiz 367

1. A body subjected to general plane motion undergoes a/an
READING QUIZ 1. A body subjected to general plane motion undergoes a/an A) translation. B) rotation. C) simultaneous translation and rotation. D) out-of-plane movement. 2. In general plane motion, if the rigid body is represented by a slab, the slab rotates A) about an axis perpendicular to the plane. B) about an axis parallel to the plane. C) about an axis lying in the plane. D) None of the above. Answers: 1. C 2. A 368

APPLICATIONS The dumping bin on the truck rotates about a fixed axis passing through the pin at A. It is operated by the extension of the hydraulic cylinder BC. The angular position of the bin can be specified using the angular position coordinate  and the position of point C on the bin is specified using the coordinate s. As a part of the design process for the truck, an engineer had to relate the velocity at which the hydraulic cylinder extends and the resulting angular velocity of the bin. 369

The large window is opened using a hydraulic cylinder AB. The position B of the hydraulic cylinder rod is related to the angular position, q, of the window. A designer has to relate the translational velocity at B of the hydraulic cylinder and the angular velocity and acceleration of the window? How would you go about the task? 370

The position of the piston, x, can be defined as a function of the angular position of the crank, q. By differentiating x with respect to time, the velocity of the piston can be related to the angular velocity, w, of the crank. This is necessary when designing an engine. The stroke of the piston is defined as the total distance moved by the piston as the crank angle varies from 0 to 180°. How does the length of crank AB affect the stroke? 371

ABSOLUTE MOTION ANALYSIS
(Section 16.4) The figure below shows the window using a hydraulic cylinder AB. The absolute motion analysis method relates the position of a point, B, on a rigid body undergoing rectilinear motion to the angular position, q , of a line contained in the body. Once a relationship in the form of sB = f (q) is established, the velocity and acceleration of point B are obtained in terms of the angular velocity and angular acceleration of the rigid body by taking the first and second time derivatives of the position function. Usually the chain rule must be used when taking the derivatives of the position coordinate equation. 372

EXAMPLE I Given: The platform is constrained to move vertically by the smooth vertical guides. The cam C rotates with a constant angular velocity, . Find: The velocity and acceleration of platform P as a function of the angle q of cam C. Use the fixed reference point O and define the position of the platform, P, in terms of the parameter q. Take successive time derivatives of the position equation to find the velocity and acceleration. Plan: 373

By differentiating with respect to time, vP = r cos q (q) = rw cos q 
EXAMPLE I (continued) Solution: By geometry, y = r + r sin q By differentiating with respect to time, vP = r cos q (q) = rw cos q  O Note that the cam rotates with a constant angular velocity. Differentiating vP to find the acceleration, aP = d(rw cos q) / dt = rw(-sin q) (w) = − rw2 sin q 374

Given: Crank AB rotates at a constant velocity of w = 150 rad/s .
EXAMPLE II Given: Crank AB rotates at a constant velocity of w = 150 rad/s . Find: The velocity of point P when q = 30°. Plan: Define x as a function of q and differentiate with respect to time. 375

xP = 0.2 cos q + (0.75)2 – (0.2 sin q)2 Solution: vP = -0.2w sin q + (0.5)[(0.75)2 – (0.2sin q)2]-0.5(-2)(0.2sin q)(0.2cos q ) w vP = -0.2w sin q – [0.5(0.2)2 sin2q w] / (0.75)2 – (0.2 sin q)2 At q = 30°, w = 150 rad/s and vP = ft/s = 18.5 ft/s 376

1. The position, s, is given as a function of angular position, q, as s = 10 sin 2q . The velocity, v, is A) 20 cos 2q B) 20 sin 2q C) 20 w cos 2q D) 20 w sin 2q 2. If s = 10 sin 2q, the acceleration, a, is A) 20 a sin 2q B) 20 a cos 2q − 40 w2 sin 2q C) 20 a cos 2q D) -40 a sin2 q Answers: C B 377

Find: The velocity and acceleration of the plate when q =30.
GROUP PROBLEM SOLVING Given: The circular cam is rotating about the fixed point O with w=4 rad/s, a = 2 rad/s2 with the dimensions shown. Find: The velocity and acceleration of the plate when q =30. Set the coordinate x to be the distance between O and the plate. Relate x to the angular position, q . Then take time derivatives of the position to find the velocity and acceleration. Plan: 378

Solution: Relate x, the distance between O and the plate, to q. x = 120 sin q (mm) 2) Take time derivatives of the position to find the velocity and acceleration. vC= dx/dt = 120 cos q (q) = 120  cos q (mm/s) aC= d vC/dt = 120  cos q (-sin q)() (mm/s2)  When q = 30, w = 4 rad/s, and a = 2 rad/s2 . Substituting, vC = 120 (4) cos30 = 416 mm/s aC = 120 (2) cos30 + 120(42 )(-sin 30)= -752 mm/s2 379

ATTENTION QUIZ 1. The sliders shown below are confined to move in the horizontal and vertical slots. If vA=10 m/s, determine the connecting bar’s angular velocity when  = 30. A) 10 rad/s B) 10 rad/s C) 8.7 rad/s D) 8.7 rad/s 2. If vA=10 m/s and aA=10 m/s, determine the angular acceleration, a, when  = 30. A) 0 rad/s2 B) rad/s2 C) rad/s D) -173 rad/s2 Answers: 1. A 2. D (See example 1 for equations!) 380

RELATIVE MOTION ANALYSIS: VELOCITY
Today’s Objectives: Students will be able to: Describe the velocity of a rigid body in terms of translation and rotation components. Perform a relative-motion velocity analysis of a point on the body. In-Class Activities: • Check Homework • Reading Quiz • Applications • Translation and Rotation Components of Velocity • Relative Velocity Analysis • Concept Quiz • Group Problem Solving • Attention Quiz

A) be attached to the selected point for analysis.
READING QUIZ 1. When a relative-motion analysis involving two sets of coordinate axes is used, the x’ - y’ coordinate system will A) be attached to the selected point for analysis. B) rotate with the body. C) not be allowed to translate with respect to the fixed frame. D) None of the above. 2. In the relative velocity equation, vB/A is A) the relative velocity of B with respect to A. B) due to the rotational motion. C)  × rB/A . D) All of the above. Answers: A. A B. D 382

Which link is undergoing general plane motion? Link AB or link BC?
APPLICATIONS As the slider block A moves horizontally to the left with vA, it causes the link CB to rotate counterclockwise. Thus vB is directed tangent to its circular path. Which link is undergoing general plane motion? Link AB or link BC? How can the angular velocity,  of link AB be found? 383

Planetary gear systems are used in many automobile automatic transmissions. By locking or releasing different gears, this system can operate the car at different speeds. How can we relate the angular velocities of the various gears in the system?

RELATIVE MOTION ANALYSIS (Section 16.5)
When a body is subjected to general plane motion, it undergoes a combination of translation and rotation. Point A is called the base point in this analysis. It generally has a known motion. The x’- y’ frame translates with the body, but does not rotate. The displacement of point B can be written: drB = drA drB/A Disp. due to translation and rotation Disp. due to translation Disp. due to rotation

RELATIVE MOTION ANALYSIS: VELOCITY
The velocity at B is given as : (drB/dt) = (drA/dt) + (drB/A/dt) or vB = vA + vB/A Since the body is taken as rotating about A, vB/A = drB/A/dt = w × rB/A Here w will only have a k component since the axis of rotation is perpendicular to the plane of translation.

RELATIVE MOTION ANALYSIS: VELOCITY (continued)
vB = vA + w × rB/A When using the relative velocity equation, points A and B should generally be points on the body with a known motion. Often these points are pin connections in linkages. For example, point A on link AB must move along a horizontal path, whereas point B moves on a circular path. The directions of vA and vB are known since they are always tangent to their paths of motion.

RELATIVE MOTION ANALYSIS: VELOCITY (continued)
vB = vA + w × rB/A When a wheel rolls without slipping, point A is often selected to be at the point of contact with the ground. Since there is no slipping, point A has zero velocity. Furthermore, point B at the center of the wheel moves along a horizontal path. Thus, vB has a known direction, e.g., parallel to the surface.

The relative velocity equation can be applied using either a Cartesian vector analysis or by writing scalar x and y component equations directly. Scalar Analysis: 1. Establish the fixed x-y coordinate directions and draw a kinematic diagram for the body. Then establish the magnitude and direction of the relative velocity vector vB/A. 2. Write the equation vB = vA + vB/A. In the kinematic diagram, represent the vectors graphically by showing their magnitudes and directions underneath each term. 3. Write the scalar equations from the x and y components of these graphical representations of the vectors. Solve for the unknowns.

Vector Analysis: 1. Establish the fixed x - y coordinate directions and draw the kinematic diagram of the body, showing the vectors vA, vB, rB/A and w. If the magnitudes are unknown, the sense of direction may be assumed. 2. Express the vectors in Cartesian vector form (CVN) and substitute them into vB = vA + w × rB/A. Evaluate the cross product and equate respective i and j components to obtain two scalar equations. 3. If the solution yields a negative answer, the sense of direction of the vector is opposite to that assumed.

EXAMPLE I Given: Roller A is moving to the right at 3 m/s. Find: The velocity of B at the instant  = 30. Plan: 1. Establish the fixed x - y directions and draw a kinematic diagram of the bar and rollers. 2. Express each of the velocity vectors for A and B in terms of their i, j, k components and solve vB = vA + w × rB/A.

Express the velocity vectors in CVN vB = vA + w × rB/A
EXAMPLE I (continued) Solution: Kinematic diagram: y y Express the velocity vectors in CVN vB = vA + w × rB/A -vB j = 3 i + [ wk × (-1.5cos30i +1.5sin 30j )] -vB j = 3 i – w j – 0.75 w i Equating the i and j components gives: 0 = 3 – 0.75 w -vB = – w Solving: w = 4 rad/s or w = 4 rad/s k vB = 5.2 m/s or vB = -5.2 m/s j

EXAMPLE II Given: Crank rotates OA with an angular velocity of 12 rad/s. Find: The velocity of piston B and the angular velocity of rod AB. Plan: Notice that point A moves on a circular path. The directions of vA is tangent to its path of motion. Draw a kinematic diagram of rod AB and use vB = vA + wAB × rB/A.

Solution: Kinematic diagram of AB: Since crack OA rotates with an angular velocity of 12 rad/s, the velocity at A will be: vA = -0.3(12) i = -3.6 i m/s Rod AB. Write the relative-velocity equation: vB = vA + wAB × rB/A vB j = -3.6 i + wAB k × (0.6cos30 i − 0.6sin30 j ) vB j = -3.6 i wAB j wAB i By comparing the i, j components: i: 0 = wAB  wAB = 12 rad/s j: vB = wAB  vB = 6.24 m/s 394

1. If the disk is moving with a velocity at point O of 15 ft/s and  = 2 rad/s, determine the velocity at A. A) 0 ft/s B) 4 ft/s C) 15 ft/s D) 11 ft/s 2 ft V=15 ft/s  A O 2. If the velocity at A is zero, then determine the angular velocity, . A) 30 rad/s B) 0 rad/s C) 7.5 rad/s D) 15 rad/s Answers: A. D B. C 395

GROUP PROBLEM SOLVING Given: The ring gear R is rotating at wR = 3 rad/s, and the sun gear S is held fixed, wS = 0. Find: The angular velocity of the each of the planet gears P and of shaft A. Plan: Draw the kinematic diagram of gears. Then, apply the relative velocity equations to the gears and solve for unknowns.

Solution: Kinematic diagram of gears. y x A C B vB= 0 160 mm vA 3 rad/s R S P Since the ring gear R is rotating at wR = 3 rad/s, the velocity at point A will be ; vA = -3(160) i = -480 i mm/s Also note that vB = 0 since the gear R is held fixed wS = 0. Applying the relative velocity equation to points A and B; vB = vA + wP × rB/A 0 = i + (P k)×(- 80 j)  0 = i + 80 P i P = 6 rad/s 397

Solution: Apply the relative velocity equation at point B and C to Gear P in order to find the velocity at B. vC = vB + wP × rC/B = 0 + (6 k)×(40 j) = i mm/s y x A C B vB= 0 160 mm vA 3 rad/s R S P Note that the shaft A has a circular motion with the radius of 120 mm. The angular velocity of the shaft is wA = vC / r = -240 / 120 = -2 rad/s. The shaft A is rotating in counter-clockwise direction ! 398

1. Which equation could be used to find
ATTENTION QUIZ 1. Which equation could be used to find the velocity of the center of the gear, C, if the velocity vA is known? A) vB = vA + wgear × rB/A B) vA = vC + wgear × rA/C C) vB = vC + wgear × rC/B D) vA = vC + wgear × rC/A vA 2. If the bar’s velocity at A is 3 m/s, what “base” point (first term on the RHS of the velocity equation) would be best used to simplify finding the bar’s angular velocity when  = 60º? A) A B) B C) C D) No difference. A 4 m B  C Answers: 1. B 2. A 399

INSTANTANEOUS CENTER OF ZERO VELOCITY
Today’s Objectives: Students will be able to: Locate the instantaneous center of zero velocity. Use the instantaneous center to determine the velocity of any point on a rigid body in general plane motion. In-Class Activities: • Check Homework • Reading Quiz • Applications • Location of the Instantaneous Center • Velocity Analysis • Concept Quiz • Group Problem Solving • Attention Quiz 400

A) velocity B) acceleration C) velocity and acceleration D) force
READING QUIZ 1. If applicable, the method of instantaneous center can be used to determine the __________ of any point on a rigid body. A) velocity B) acceleration C) velocity and acceleration D) force 2. The velocity of any point on a rigid body is __________ to the relative position vector extending from the IC to the point. A) always parallel B) always perpendicular C) in the opposite direction D) in the same direction Answers: 1. A 2. B 401

Which point on the wheel has the maximum velocity?
APPLICATIONS The instantaneous center (IC) of zero velocity for this bicycle wheel is at the point in contact with ground. The velocity direction at any point on the rim is perpendicular to the line connecting the point to the IC. Which point on the wheel has the maximum velocity? Does a larger wheel mean the bike will go faster for the same rider effort in pedaling than a smaller wheel? 402

As the board slides down the wall (to the left), it is subjected to general plane motion (both translation and rotation). Since the directions of the velocities of ends A and B are known, the IC is located as shown. How can this result help you analyze other situations? What is the direction of the velocity of the center of gravity of the board? 403

INSTANTANEOUS CENTER OF ZERO VELOCITY
(Section 16-6) For any body undergoing planar motion, there always exists a point in the plane of motion at which the velocity is instantaneously zero (if it is rigidly connected to the body). This point is called the instantaneous center (IC) of zero velocity. It may or may not lie on the body! If the location of this point can be determined, the velocity analysis can be simplified because the body appears to rotate about this point at that instant. 404

LOCATION OF THE INSTANTANEOUS CENTER
To locate the IC, we can use the fact that the velocity of a point on a body is always perpendicular to the relative position vector from the IC to the point. Several possibilities exist. First, consider the case when velocity vA of a point A on the body and the angular velocity w of the body are known. In this case, the IC is located along the line drawn perpendicular to vA at A, a distance rA/IC = vA/w from A. Note that the IC lies up and to the right of A since vA must cause a clockwise angular velocity w about the IC. 405

(continued) A second case is when the lines of action of two non-parallel velocities, vA and vB, are known. First, construct line segments from A and B perpendicular to vA and vB. The point of intersection of these two line segments locates the IC of the body. 406

(continued) A third case is when the magnitude and direction of two parallel velocities at A and B are known. Here the location of the IC is determined by proportional triangles. As a special case, note that if the body is translating only (vA = vB), then the IC would be located at infinity. Then w equals zero, as expected. 407

VELOCITY ANALYSIS The velocity of any point on a body undergoing general plane motion can be determined easily once the instantaneous center of zero velocity of the body is located. Since the body seems to rotate about the IC at any instant, as shown in this kinematic diagram, the magnitude of velocity of any arbitrary point is v = w r, where r is the radial distance from the IC to the point. The velocity’s line of action is perpendicular to its associated radial line. 408

Find: The angular velocities of links AB and BD.
EXAMPLE I Given: A linkage undergoing motion as shown. The velocity of the block, vD, is 3 m/s. Find: The angular velocities of links AB and BD. Plan: Locate the instantaneous center of zero velocity of link BD and then solve for the angular velocities. 409

EXAMPLE I Solution: Since D moves to the right, it causes link AB to rotate clockwise about point A. The instantaneous center of velocity for BD is located at the intersection of the line segments drawn perpendicular to vB and vD. Note that vB is perpendicular to link AB. Therefore we can see that the IC is located along the extension of link AB. 410

wBD = vD/rD/IC = 3/0.566 = 5.3 rad/s
EXAMPLE I (continued) Using these facts, rB/IC = 0.4 tan 45° = 0.4 m rD/IC = 0.4/cos 45° = m Since the magnitude of vD is known, the angular velocity of link BD can be found from vD = wBD rD/IC . wBD = vD/rD/IC = 3/0.566 = 5.3 rad/s Link AB is subjected to rotation about A. wAB = vB/rB/A = (rB/IC)wBD/rB/A = 0.4(5.3)/0.4 = 5.3 rad/s 411

Find: The velocities of points A and B at the instant shown.
EXAMPLE II Given: The wheel rolls on its hub without slipping on the horizontal surface with vC = 2 ft/s. Find: The velocities of points A and B at the instant shown. Plan: Locate the IC of the wheel. Then calculate the velocities at A and B. 412

Solution: Note that the wheel rolls without slipping. Thus the IC is at the contact point with the surface. The angular velocity of the wheel can be found from w = vC/rC/IC = 2/3 =0.667 rad/s Or, w = k (rad/s) The velocity at A and B will be vA = w × rA/IC = ( ) k × (3 i + 3j) = (2 i - 2 j) in/s vB = w × rB/IC = ( ) k × (11 j) = 7.34 i in/s vA = √( ) = 2.83 in/s, vB = 7.34 in/s 413

C) the midpoint of the line connecting the two points.
CONCEPT QUIZ 1. When the velocities of two points on a body are equal in magnitude and parallel but in opposite directions, the IC is located at A) infinity. B) one of the two points. C) the midpoint of the line connecting the two points. D) None of the above. 2. When the direction of velocities of two points on a body are perpendicular to each other, the IC is located at A) infinity. B) one of the two points. C) the midpoint of the line connecting the two points. D) None of the above. Answers: C D 414

GROUP PROBLEM SOLVING Given: The four bar linkage is moving with wCD equal to 6 rad/s CCW. Find: The velocity of point E on link BC and angular velocity of link AB. Plan: This is an example of the second case in the lecture notes. Since the direction of Point B’s velocity must be perpendicular to AB, and Point C’s velocity must be perpendicular to CD, the location of the instantaneous center, I, for link BC can be found.

Link CD: vC 0.6 m wCD = 6 rad/s vC = 0.6(6) = 3.6 m/s C D Link AB: A B 30° 1.2 m wAB vB wBC B C I vB vC = 3.6 m/s vE 60° 30° 0.6 m Link BC: E From triangle CBI IC = m IB = 0.6/sin 60° = m vC = (IC)wBC wBC = vC/IC = 3.6/0.346 wBC = rad/s

vB = (IB)wBC = 0.693(10.39) = 7.2 m/s From link AB, vB is also equal to 1.2 wAB. Therefore 7.2 = 1.2 wAB => wAB = 6 rad/s vE = (IE)wBC where distance IE = = m vE = 0.458(10.39) = 4.76 m/s where q = tan-1(0.3/0.346) = 40.9° q

ATTENTION QUIZ 1. The wheel shown has a radius of 15 in and rotates clockwise at a rate of w = 3 rad/s. What is vB? A) 5 in/s B) 15 in/s C) 0 in/s D) 45 in/s 2. Point A on the rod has a velocity of 8 m/s to the right. Where is the IC for the rod? A) Point A. B) Point B. C) Point C. D) Point D. • C D • Answers: 1. D 2. C 418

RELATIVE MOTION ANALYSIS: ACCELERATION
Today’s Objectives: Students will be able to: Resolve the acceleration of a point on a body into components of translation and rotation. Determine the acceleration of a point on a body by using a relative acceleration analysis. In-Class Activities: • Check Homework • Reading Quiz • Applications • Translation and Rotation Components of Acceleration • Relative Acceleration Analysis • Roll-Without-Slip Motion • Concept Quiz • Group Problem Solving • Attention Quiz 419

READING QUIZ 1. If two bodies contact one another without slipping, and the points in contact move along different paths, the tangential components of acceleration will be ______ and the normal components of acceleration will be _________. A) the same, the same B) the same, different C) different, the same D) different, different 2. When considering a point on a rigid body in general plane motion, A) It’s total acceleration consists of both absolute acceleration and relative acceleration components. B) It’s total acceleration consists of only absolute acceleration components. C) It’s relative acceleration component is always normal to the path. D) None of the above. Answers: B A 420

APPLICATIONS In the mechanism for a window, link AC rotates about a fixed axis through C, and AB undergoes general plane motion. Since point A moves along a curved path, it has two components of acceleration while point B, sliding in a straight track, has only one. The components of acceleration of these points can be inferred since their motions are known. How can we determine the accelerations of the links in the mechanism?

In an automotive engine, the forces delivered to the crankshaft, and the angular acceleration of the crankshaft, depend on the speed and acceleration of the piston. How can we relate the accelerations of the piston, connection rod, and crankshaft to each other?

RELATIVE MOTION ANALYSIS: ACCELERATION (Section 16-7)
The equation relating the accelerations of two points on the body is determined by differentiating the velocity equation with respect to time. / + dt dv A B = These are absolute accelerations of points A and B. They are measured from a set of fixed x,y axes. This term is the acceleration of B with respect to A and includes both tangential and normal components. The result is aB = aA + (aB/A)t + (aB/A)n

RELATIVE MOTION ANALYSIS: ACCELERATION (continued)
aB = aA (aB/A)t + (aB/A)n Graphically: The relative tangential acceleration component (aB/A)t is ( ×rB/A) and perpendicular to rB/A. The relative normal acceleration component (aB/A)n is (−2 rB/A) and the direction is always from B towards A.

RELATIVE MOTION ANALYSIS: ACCELERATION (continued)
Since the relative acceleration components can be expressed as (aB/A)t =   rB/A and (aB/A)n = - 2 rB/A, the relative acceleration equation becomes aB = aA +   rB/A − 2 rB/A Note that the last term in the relative acceleration equation is not a cross product. It is the product of a scalar (square of the magnitude of angular velocity, w2) and the relative position vector, rB/A.

APPLICATION OF THE RELATIVE ACCELERATION EQUATION
In applying the relative acceleration equation, the two points used in the analysis (A and B) should generally be selected as points which have a known motion, such as pin connections with other bodies. In this mechanism, point B is known to travel along a circular path, so aB can be expressed in terms of its normal and tangential components. Note that point B on link BC will have the same acceleration as point B on link AB. Point C, connecting link BC and the piston, moves along a straight-line path. Hence, aC is directed horizontally.

1. Establish a fixed coordinate system. 2. Draw the kinematic diagram of the body. 3. Indicate on it aA, aB, , , and rB/A. If the points A and B move along curved paths, then their accelerations should be indicated in terms of their tangential and normal components, i.e., aA = (aA)t + (aA)n and aB = (aB)t + (aB)n. 4. Apply the relative acceleration equation: aB = aA +   rB/A − 2 rB/A 5. If the solution yields a negative answer for an unknown magnitude, this indicates that the sense of direction of the vector is opposite to that shown on the diagram.

w = vA/rA/IC = vA / (3) = 2 rad/s
EXAMPLE I Given: Point A on rod AB has an acceleration of 5 m/s2 and a velocity of 6 m/s at the instant shown. Solution: First, we need to find the angular velocity of the rod at this instant. Locating the instant center (IC) for rod AB, we can determine : w = vA/rA/IC = vA / (3) = 2 rad/s IC Find: The angular acceleration of the rod and the acceleration at B at this instant. Plan: Follow the problem solving procedure!

EXAMPLE I (continued) Since points A and B both move along straight-line paths, aA = -5 j m/s2 aB = aB i m/s2 Applying the relative acceleration equation aB = aA + a × rB/A – w2rB/A aB i = - 5 j + a k × (3 i – 4 j) – 22 (3 i – 4 j) aB i = - 5 j + 4 a i + 3 a j – (12 i – 16 j)

So with aB i = - 5 j + 4 a i + 3 a j – (12 i – 16 j) , we can solve.
EXAMPLE I (continued) So with aB i = - 5 j + 4 a i + 3 a j – (12 i – 16 j) , we can solve. By comparing the i, j components; aB = 4 a – 12 0 = a Solving: aB = m/s2  = rad/s2 430

(aA)t = (aA’)t (which implies aB rB = aC rC ).
BODIES IN CONTACT Consider two bodies in contact with one another without slipping, where the points in contact move along different paths. In this case, the tangential components of acceleration will be the same, i. e., (aA)t = (aA’)t (which implies aB rB = aC rC ). The normal components of acceleration will not be the same. (aA)n  (aA’)n so aA  aA’

ROLLING MOTION Another common type of problem encountered in dynamics involves rolling motion without slip; e.g., a ball, cylinder, or disk rolling without slipping. This situation can be analyzed using relative velocity and acceleration equations. As the cylinder rolls, point G (center) moves along a straight line. If w and a are known, the relative velocity and acceleration equations can be applied to A, at the instant A is in contact with the ground. The point A is the instantaneous center of zero velocity, however it is not a point of zero acceleration.

ROLLING MOTION (continued)
• Velocity: Since no slip occurs, vA = 0 when A is in contact with ground. From the kinematic diagram: vG = vA + w  rG/A vG i = 0 + (-wk)  (r j) vG = wr or vG = wr i • Acceleration: Since G moves along a straight-line path, aG is horizontal. Just before A touches ground, its velocity is directed downward, and just after contact, its velocity is directed upward. Thus, point A accelerates upward as it leaves the ground. aG = aA + a  rG/A – w2rG/A => aG i = aA j + (-a k)  (r j) – w2(r j) Evaluating and equating i and j components: aG = ar and aA = w2r or aG = ar i and aA = w2r j

EXAMPLE II Given: The gear rolls on the fixed rack. Find: The accelerations of point A at this instant. Plan: Follow the solution procedure! Solution: Since the gear rolls on the fixed rack without slip, aO is directed to the right with a magnitude of: aO = r = (6 rad/s2)(0.3 m)=1.8 m/s2

So now with aO = 1.8 m/s2, we can apply the relative acceleration equation between points O and A. aA = aO + a × rA/O – w2 rA/O aA = 1.8i + (-6k)×(0.3j) –122 (0.3j) = (3.6 i – 43.2j) m/s2 y x 1.8 m/s2

A) - 2r i −r j B) - r i + 2r j C) 2r i −r j D) Zero.
CONCEPT QUIZ 1. If a ball rolls without slipping, select the tangential and normal components of the relative acceleration of point A with respect to G. A) r i + 2r j B) - r i + 2r j C) 2r i − r j D) Zero. 2. What are the tangential and normal components of the relative acceleration of point B with respect to G. A) - 2r i −r j B) - r i + 2r j C) 2r i −r j D) Zero. Answers: B A 436

GROUP PROBLEM SOLVING Given: The smember AB is rotating with AB=3 rad/s, AB=2 rad/s2 at this instant. Find: The velocity and acceleration of the slider block C. Plan: Follow the solution procedure! Note that Point B is rotating. So what components of acceleration will it be experiencing?

Solution: Since Point B is rotating, its velocity and acceleration will be: vB = (wAB) rB/A = (3) 7 = 21 in/s aBn = (wAB)2 rB/A= (3)2 7 = 63 in/s2 aBt = (aAB) rB/A = (2) 7 = 14 in/s2 vB = (-21 i ) in/s aB = (-14 i −63 j ) in/s2 438

Now apply the relative velocity equation between points B and C to find the angular velocity of link BC. vC = vB + wBC× rC/B (-0.8 vC i −0.6 vC j) = (-21 i ) + wBC k × (-5 i −12 j) = ( wBC) i −5 wBC j By comparing the i, j components; -0.8 vC = wBC -0.6 vC = - 5 wBC Solving: wBC = rad/s vC = in/s 439

Now, apply the relative acceleration equation between points B and C. aC = aB + aBC × rC/B – w2BC rC/B (-0.8 aC i − 0.6 aC j) = (-14 i − 63 j) + aBC k × (-5 i −12 j) – (1.125)2 (-5 i −12 j) (- 0.8 aC i − 0.6 aC j) = ( aBC ) i + (- 63 – 5 aBC ) j By comparing the i, j components; - 0.8 aC = aBC - 0.6 aC = –5 aBC

Solving these two i, j component equations - 0.8 aC = aBC - 0.6 aC = –5 aBC Yields aBC = -3.0 rad/s2 aC = in/s2

C) 200 ft/sec2 D) None of above.
ATTENTION QUIZ Two bodies contact one another without slipping. If the tangential component of the acceleration of point A on gear B is 100 ft/sec2, determine the tangential component of the acceleration of point A’ on gear C. A) 50 ft/sec B) 100 ft/sec2 C) 200 ft/sec D) None of above. Answers: B A 2. If the tangential component of the acceleration of point A on gear B is 100 ft/sec2, determine the angular acceleration of gear B. A) 50 rad/sec B) 100 rad/sec2 C) 200 rad/sec D) None of above. 442

MOMENT OF INERTIA Today’s Objectives: Students will be able to:
Determine the mass moment of inertia of a rigid body or a system of rigid bodies. In-Class Activities: • Check Homework • Reading Quiz • Applications • Mass Moment of Inertia • Parallel-Axis Theorem • Composite Bodies • Concept Quiz • Group Problem Solving • Attention Quiz 443

A) translational motion B) deformation
READING QUIZ 1. Mass moment of inertia is a measure of the resistance of a body to _____________. A) translational motion B) deformation C) angular acceleration D) impulsive motion 2. Mass moment of inertia is always ____________. A) a negative quantity B) a positive quantity C) an integer value D) zero about an axis perpendicular to the plane of motion Answers: C B 444

APPLICATIONS The large flywheel in the picture is connected to a large metal cutter. The flywheel mass is used to help provide a uniform motion to the cutting blade. What property of the flywheel is most important for this use? How can we determine a value for this property? Why is most of the mass of the flywheel located near the flywheel’s circumference? Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 10.9 445

The crank on the oil-pump rig undergoes rotation about a fixed axis that is not at its mass center. The crank develops a kinetic energy directly related to its mass moment of inertia. As the crank rotates, its kinetic energy is converted to potential energy and vice versa. Is the mass moment of inertia of the crank about its axis of rotation smaller or larger than its moment of inertia about its center of mass? 446

Consider a rigid body with a center of
MASS MOMENT OF INERTIA Consider a rigid body with a center of mass at G. It is free to rotate about the z axis, which passes through G. Now, if we apply a torque T about the z axis to the body, the body begins to rotate with an angular acceleration of . T and  are related by the equation T = I  . In this equation, I is the mass moment of inertia (MMI) about the z axis. The MMI of a body is a property that measures the resistance of the body to angular acceleration. The MMI is often used when analyzing rotational motion. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 10.9 447

MASS MOMENT OF INERTIA (continued)
Consider a rigid body and the arbitrary axis P shown in the figure. The MMI about the P axis is defined as I = m r2 dm, where r, the “moment arm,” is the perpendicular distance from the axis to the arbitrary element dm. The mass moment of inertia is always a positive quantity and has a unit of kg ·m2 or slug · ft2. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 10.9 448

MASS MOMENT OF INERTIA (continued)
The figures below show the mass moment of inertia formulations for two flat plate shapes commonly used when working with three dimensional bodies. The shapes are often used as the differential element being integrated over the entire body. 449

When using direct integration, only symmetric bodies having surfaces generated by revolving a curve about an axis will be considered. Shell element • If a shell element having a height z, radius r = y, and thickness dy is chosen for integration, then the volume element is dV = (2py)(z)dy. • This element may be used to find the moment of inertia Iz since the entire element, due to its thinness, lies at the same perpendicular distance y from the z-axis. Disk element • If a disk element having a radius y and a thickness dz is chosen for integration, then the volume dV = (py2)dz. • Using the moment of inertia of the disk element, we can integrate to determine the moment of inertia of the entire body. 450

Given: The volume shown with r = 5 slug/ft3.
EXAMPLE I Given: The volume shown with r = 5 slug/ft3. Find: The mass moment of inertia of this body about the y-axis. Plan: Find the mass moment of inertia of a disk element about the y-axis, dIy, and integrate. 451

ò EXAMPLE I (continued) Solution:
The moment of inertia of a disk about an axis perpendicular to its plane is I = 0.5 m r2. Thus, for the disk element, we have dIy = 0.5 (dm) x2 where the differential mass dm = r dV = rpx2 dy. slug•ft2 873 . 18 p(5) dy 8 2 rp rpx4 Iy 1 = ò y 452

PARALLEL-AXIS THEOREM
If the mass moment of inertia of a body about an axis passing through the body’s mass center is known, then the moment of inertia about any other parallel axis may be determined by using the parallel axis theorem, I = IG + md2 where IG = mass moment of inertia about the body’s mass center m = mass of the body d = perpendicular distance between the parallel axes 453

Radius of Gyration and CompoSite Bodies
The mass moment of inertia of a body about a specific axis can be defined using the radius of gyration (k). The radius of gyration has units of length and is a measure of the distribution of the body’s mass about the axis at which the moment of inertia is defined. I = m k2 or k = (I/m) Composite Bodies If a body is constructed of a number of simple shapes, such as disks, spheres, or rods, the mass moment of inertia of the body about any axis can be determined by algebraically adding together all the mass moments of inertia, found about the same axis, of the different shapes. 454

EXAMPLE II Given: The pendulum consists of a slender rod with a mass 10 kg and sphere with a mass of 15 kg. Find: The pendulum’s MMI about an axis perpendicular to the screen and passing through point O. Plan: q r p Follow steps similar to finding the MoI for a composite area (as done in statics). The pendulum’s can be divided into a slender rod (r) and sphere (s) Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 10.9 455

Solution: 1. The center of mass for rod is at point Gr, m from Point O. The center of mass for sphere is at Gs, 0.55 m from point O. 2. The MMI data for a slender rod and sphere are given on the inside back cover of the textbook. Using those data and the parallel-axis theorem, calculate the following. IO = IG + (m) (d) 2 IOr = (1/12) (10)(0.45)2 +10 (0.225)2 = kg·m2 IOs = (2/5) (15) (0.1) (0.55)2 = kg·m2 3. Now add the two MMIs about point O. IO = IOr + IOs = kg·m2 Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 10.9 456

1. The mass moment of inertia of a rod of mass m and length L about a transverse axis located at its end is _________ . A) (1/12) m L2 B) (1/6) m L2 C) (1/3) m L2 D) m L2 2. The mass moment of inertia of a thin ring of mass m and radius R about the z axis is ___________ . A) (1/2) m R2 B) m R2 C) (1/4) m R2 D) 2 m R2 Answers: C B 457

GROUP PROBLEM SOLVING Given: The pendulum consists of a 5 kg plate and a 3 kg slender rod. Find: The radius of gyration of the pendulum about an axis perpendicular to the screen and passing through point G. Plan: Determine the MMI of the pendulum using the method for composite bodies. Then determine the radius of gyration using the MMI and mass values.

Solution: 1. Separate the pendulum into a square plate (P) and a slender rod (R). 2. The center of mass of the plate and rod are 2.25 m and 1 m from point O, respectively. y = ( y m) / ( m ) = {(1) 3 + (2.25) 5} / (3+5) = m 

3. The MMI data on plates and slender rods are given on the inside cover of the textbook. Using those data and the parallel-axis theorem, IP = (1/12) 5 ( ) + 5 (2.251.781)2 = kg·m2 IR = (1/12) 3 (2)2 + 3 (1.781  1)2 = kg·m2 4. IO = IP + IR = = 4.45 kg·m2 5. Total mass (m) equals 8 kg Radius of gyration k = IO / m = m 460

C) zero D) None of the above
ATTENTION QUIZ 1. The mass moment of inertia of any body about its center of mass is always ____________. A) maximum B) minimum C) zero D) None of the above 2. If the mass of body A and B are equal but kA = 2kB, then ____________. A) IA = 2IB B) IA = (1/2)IB C) IA = 4IB D) IA = (1/4)IB Answers: 1. B 2. C 461

ROTATION ABOUT A FIXED AXIS
EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS Today’s Objectives: Students will be able to: Analyze the planar kinetics of a rigid body undergoing rotational motion. In-Class Activities: • Check Homework • Reading Quiz • Applications • Rotation About an Axis • Equations of Motion • Concept Quiz • Group Problem Solving • Attention Quiz 462

B) tangent to the path of motion of G
READING QUIZ 1. In rotational motion, the normal component of acceleration at the body’s center of gravity (G) is always __________. A) zero B) tangent to the path of motion of G C) directed from G toward the center of rotation D) directed from the center of rotation toward G 2. If a rigid body rotates about point O, the sum of the moments of the external forces acting on the body about point O equals? A) IGa B) IOa C) m aG D) m aO Answers: 1. C 2. B 463

APPLICATIONS Pin at the center of rotation. The crank on the oil-pump rig undergoes rotation about a fixed axis, caused by the driving torque, M, from a motor. As the crank turns, a dynamic reaction is produced at the pin. This reaction is a function of angular velocity, angular acceleration, and the orientation of the crank. If the motor exerts a constant torque M on the crank, does the crank turn at a constant angular velocity? Is this desirable for such a machine?

The pendulum of the Charpy impact machine is released from rest when  = 0°. Its angular velocity () begins to increase. Can we determine the angular velocity when it is in vertical position ? On which property (P) of the pendulum does the angular acceleration () depend ? What is the relationship between P and ? Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 10.9 465

EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS (Section 17.4)
When a rigid body rotates about a fixed axis perpendicular to the plane of the body at point O, the body’s center of gravity G moves in a circular path of radius rG. Thus, the acceleration of point G can be represented by a tangential component (aG)t = rG a and a normal component (aG)n = rG w2. Since the body experiences an angular acceleration, its inertia creates a moment of magnitude, Iga, equal to the moment of the external forces about point G. Thus, the scalar equations of motion can be stated as:  Fn = m (aG)n = m rG w2  Ft = m (aG)t = m rG a  MG = IG a

EQUATIONS OF MOTION (continued)
Note that the MG moment equation may be replaced by a moment summation about any arbitrary point. Summing the moment about the center of rotation O yields MO = IGa + rG m (aG) t = [IG + m(rG)2] a From the parallel axis theorem, IO = IG + m(rG)2, therefore the term in parentheses represents IO. Consequently, we can write the three equations of motion for the body as: Fn = m (aG)n = m rG w2 Ft = m (aG)t = m rG a MO = IO a

Problems involving the kinetics of a rigid body rotating about a fixed axis can be solved using the following process. 1. Establish an inertial coordinate system and specify the sign and direction of (aG)n and (aG)t. 2. Draw a free body diagram accounting for all external forces and couples. Show the resulting inertia forces and couple (typically on a separate kinetic diagram). 3. Compute the mass moment of inertia IG or IO. 4. Write the three equations of motion and identify the unknowns. Solve for the unknowns. 5. Use kinematics if there are more than three unknowns (since the equations of motion allow for only three unknowns).

Given: The uniform slender rod has a mass of 15 kg.
EXAMPLE Given: The uniform slender rod has a mass of 15 kg. Find: The reactions at the pin O and the angular acceleration of the rod just after the cord is cut. Since the mass center, G, moves in a circle of radius 0.15 m, it’s acceleration has a normal component toward O and a tangential component acting downward and perpendicular to rG. Apply the problem solving procedure. G Plan:

= EXAMPLE (continued) Solution: FBD & Kinetic Diagram
rG Equations of motion: + Fn = man = mrGw  Ox = 0 N + Ft = mat = mrGa  -Oy + 15(9.81) = 15(0.15)a (1) + MO = IG a + m rG a (rG)  (0.15) 15(9.81)= IG a + m(rG)2 a Using IG = (ml2)/12 and rG = (0.15), we can write: IG a + m(rG)2 a = [(15×0.92)/ (0.15)2] a = a

= EXAMPLE (continued) After substituting:
rG After substituting: 22.07 = 1.35 a  a = 16.4 rad/s2 From Eq (1) : -Oy + 15(9.81) = 15(0.15)a  Oy = 15(9.81) − 15(0.15)16.4 = 110 N

C) m (l/2) w2 D) None of the above
CONCEPT QUIZ 1. If a rigid bar of length l (above) is released from rest in the horizontal position (q = 0), the magnitude of its angular acceleration is at maximum when A) q = B) q = 90 C) q = 180 D) q = 0 and 180 2. In the above problem, when q = 90°, the horizontal component of the reaction at pin O is __________. A) zero B) m g C) m (l/2) w D) None of the above Answers: 1. D ( angular accel = [mg (l/2) cos (theta)] over [I_G] ) 2. A 472

GROUP PROBLEM SOLVING Given: Wsphere = 30 lb, Wrod = 10 lb Find: The reaction at the pin O just after the cord AB is cut. Plan: Draw the free body diagram and kinetic diagram of the rod and sphere as one unit. Then apply the equations of motion.

Solution: FBD and kinetic diagram; 30 lb 10 lb Oy Ox msphere(3a) mrod(1.0a) (IG)spherea (IG)roda msphere(3)(0)2 mrod(1.0)(0)2 = Equations of motion: Fn = m(aG)n: Ox = (30/32.2)(3)(0)2 + (10/32.2)(1.0)(0)2  Ox = 0 lb

30 lb 10 lb Oy Ox msphere(3a) mrod(1.0a) (IG)spherea (IG)roda msphere(3)(0)2 mrod(1.0)(0)2 = Ft = m(aG)t: - Oy = (30/32.2) (3a) + (10/32.2) (1.0a)  Oy = 40 – a MO = Ioa: 30(3.0) + 10(1.0) = [ 0.4 (30/32.2) (1)2 + (30/32.2) (3)2 ]sphere a + [ (1/12) (10/32.2) (2)2 + (10/32.2) (1)2 ]rod a  100 = a Therefore, a = 10.9 rad/s2, Oy = 6.14 lb 475

A) aa > ab B) aa < ab C) aa = ab D) None of the above
m T 10 lb ATTENTION QUIZ 1. A drum of mass m is set into motion in two ways: (a) by a constant 40 N force, and, (b) by a block of weight 40 N. If aa and ab represent the angular acceleration of the drum in each case, select the true statement. A) aa > ab B) aa < ab C) aa = ab D) None of the above Answers: 1. A 2. B 2. In case (b) above, what is the tension T in the cable? A) T = 40 N B) T < 40 N C) T > 40 N D) None of the above 476

EQUATIONS OF MOTION: GENERAL PLANE MOTION
Today’s Objectives: Students will be able to: Analyze the planar kinetics of a rigid body undergoing general plane motion. In-Class Activities: • Check Homework • Reading Quiz • Applications • Equations of Motion • Frictional Rolling Problems • Concept Quiz • Group Problem Solving • Attention Quiz

A) not be equal to a r; less than sN B) be equal to a r; equal to kN
READING QUIZ 1. If a disk rolls on a rough surface without slipping, the acceleration af the center of gravity (G) will _________ and the friction force will be __________. A) not be equal to a r; less than sN B) be equal to a r; equal to kN C) be equal to a r; less than sN D) None of the above 2. If a rigid body experiences general plane motion, the sum of the moments of external forces acting on the body about any point P is equal to __________. A) IP a B) IP a + maP C) m aG D) IG a + rGP × maP Answers: C A 478

APPLICATIONS As the soil compactor accelerates forward, the front roller experiences general plane motion (both translation and rotation). What are the loads experienced by the roller shaft or bearings? The forces shown on the roller’s FBD cause the accelerations shown on the kinetic diagram Is the point A the IC? =

The lawn roller is pushed forward with a force of 200 N when the handle is at 45°. How can we determine its translation acceleration and angular acceleration? Does the acceleration depend on the coefficient’s of static and kinetic friction?

During an impact, the center of gravity of this crash dummy will decelerate with the vehicle, but also experience another acceleration due to its rotation about point A. Why? How can engineers use this information to determine the forces exerted by the seat belt on a passenger during a crash?

EQUATIONS OF MOTION: GENERAL PLANE MOTION (Section 17.5)
When a rigid body is subjected to external forces and couple-moments, it can undergo both translational motion and rotational motion. This combination is called general plane motion.  Fx = m (aG)x  Fy = m (aG)y  MG = IG a Using an x-y inertial coordinate system, the equations of motions about the center of mass, G, may be written as:

EQUATIONS OF MOTION: GENERAL PLANE MOTION (continued)
Sometimes, it may be convenient to write the moment equation about a point P other than G. Then the equations of motion are written as follows:  Fx = m (aG)x  Fy = m (aG)y  MP =  (Mk )P In this case,  (Mk )P represents the sum of the moments of IGa and maG about point P.

FRICTIONAL ROLLING PROBLEMS
When analyzing the rolling motion of wheels, cylinders, or disks, it may not be known if the body rolls without slipping or if it slides as it rolls. For example, consider a disk with mass m and radius r, subjected to a known force P. The equations of motion will be:  Fx = m(aG)x => P  F = m aG  Fy = m(aG)y => N  mg = 0  MG = IGa => F r = IG a There are 4 unknowns (F, N, a, and aG) in these three equations. 484

FRICTIONAL ROLLING PROBLEMS (continued)
Hence, we have to make an assumption to provide another equation. Then, we can solve for the unknowns. The 4th equation can be obtained from the slip or non-slip condition of the disk. Case 1: Assume no slipping and use aG = a r as the 4th equation and DO NOT use Ff = sN. After solving, you will need to verify that the assumption was correct by checking if Ff  sN. Case 2: Assume slipping and use Ff = kN as the 4th equation In this case, aG  ar.

Problems involving the kinetics of a rigid body undergoing general plane motion can be solved using the following procedure. 1. Establish the x-y inertial coordinate system. Draw both the free body diagram and kinetic diagram for the body. 2. Specify the direction and sense of the acceleration of the mass center, aG, and the angular acceleration a of the body. If necessary, compute the body’s mass moment of inertia IG. 3. If the moment equation Mp= (Mk)p is used, use the kinetic diagram to help visualize the moments developed by the components m(aG)x, m(aG)y, and IGa. 4. Apply the three equations of motion.

5. Identify the unknowns. If necessary (i.e., there are four unknowns), make your slip-no slip assumption (typically no slipping, or the use of aG = a r, is assumed first). 6. Use kinematic equations as necessary to complete the solution. 7. If a slip-no slip assumption was made, check its validity!!! Key points to consider: 1. Be consistent in using the assumed directions The direction of aG must be consistent with a. 2. If Ff = kN is used, Ff must oppose the motion. As a test, assume no friction and observe the resulting motion This may help visualize the correct direction of Ff.

EXAMPLE Given: A spool has a mass of 200 kg and a radius of gyration (kG) of 0.3 m. The coefficient of kinetic friction between the spool and the ground is k = 0.1. Find: The angular acceleration (a) of the spool and the tension in the cable. Plan: Focus on the spool. Follow the solution procedure (draw a FBD, etc.) and identify the unknowns.

The free body diagram and kinetic diagram for the body are: IG a maG = 1962 N Equations of motion: Fy = m (aG)y : NB − 1962 = 0  NB = 1962 N

EXAMPLE (continued) Note that aG = (0.4) a. Why ? Fx = m (aG)x: T – 0.1 NB = 200 aG = 200 (0.4) a  T – = 80 a MG = IG a : – T(0.4) – 0.1 NB (0.6) = 20 (0.3)2 a  450 – T(0.4) – (0.6) = 1.8 a Solving these two equations, we get a = 7.50 rad/s2, T = 797 N

A) right, left B) left, right C) right, right D) left, left 
CONCEPT QUIZ An 80 kg spool (kG = 0.3 m) is on a rough surface and a cable exerts a 30 N load to the right. The friction force at A acts to the __________ and the aG should be directed to the __________ . A) right, left B) left, right C) right, right D) left, left 30N A 0.2m 0.75m  G • 2. For the situation above, the moment equation about G is? A) 0.75 (FfA) - 0.2(30) = - (80)(0.32) B) -0.2(30) = - (80)(0.32) C) 0.75 (FfA) - 0.2(30) = - (80)(0.32) + 80aG D) None of the above Answers: C A (Positive has been set to be CCW) 491

GROUP PROBLEM SOLVING Given: A 80 kg lawn roller has a radius of gyration of kG = m. It is pushed forward with a force of 200 N. Find: The angular acceleration if s = 0.12 and k = 0.1. Plan: Follow the problem solving procedure. Solution: The moment of inertia of the roller about G is IG = m(kG)2 = (80)(0.175)2 = 2.45 kg·m2

FBD: Equations of motion:  Fx = m(aG)x FA – 200 cos 45 = 80 aG  Fy = m(aG)y NA – – 200 sin 45 = 0  MG = IG a – 0.2 FA = 2.45 a We have 4 unknowns: NA, FA, aG and a. Another equation is needed to allow solving for the unknowns.

The three equations we have now are: FA – 200 cos 45° = 80 aG NA – – 200 sin 45° = 0 – 0.2 FA= 2.45 a First, assume the wheel is not slipping. Thus, we can write aG = r a = 0.2 a Now solving the four equations yields: NA = N, FA = 61.4 N, a = rad/s2 , aG = -1.0 m/s2 The no-slip assumption must be checked. Is FA = 61.4 N  s NA= N ? Yes, therefore, the wheel rolls without slip.

2. Select the equation that best represents the “no-slip” assumption.
ATTENTION QUIZ 1. A slender 100 kg beam is suspended by a cable. The moment equation about point A is? A) 3(10) = 1/12(100)(42)  B) 3(10) = 1/3(100)(42)  C) 3(10) = 1/12(100)(42)  + (100 aGx)(2) D) None of the above 4m 10 N 3m A 2. Select the equation that best represents the “no-slip” assumption. A) Ff = s N B) Ff = k N C) aG = r a D) None of the above Answers: 1. C 2. C 495

KINETIC ENERGY, WORK, PRINCIPLE OF WORK AND ENERGY
Today’s Objectives: Students will be able to: Define the various ways a force and couple do work. Apply the principle of work and energy to a rigid body. In-Class Activities: • Check Homework • Reading Quiz • Applications • Kinetic Energy • Work of a Force or Couple • Principle of Work and Energy • Concept Quiz • Group Problem Solving • Attention Quiz 496

1. Kinetic energy due to rotation of the body is defined as
READING QUIZ 1. Kinetic energy due to rotation of the body is defined as A) (1/2) m (vG) B) (1/2) m (vG)2 + (1/2) IG w2. C) (1/2) IG w D) IG w2. 2. When calculating work done by forces, the work of an internal force does not have to be considered because ____________. A) internal forces do not exist B) the forces act in equal but opposite collinear pairs C) the body is at rest initially D) the body can deform Answers: 1. C 2. B 497

APPLICATIONS The work of the torque (or moment) developed by the driving gears on the two motors on the concrete mixer is transformed into the rotational kinetic energy of the mixing drum. If the motor gear characteristics are known, how would you find the rotational velocity of the mixing drum? 498

The work done by the soil compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of the roller and wheels (excluding the internal kinetic energy developed by the moving parts of the engine and drive train). Are the kinetic energies of the frame and the roller related to each other? If so, how? 499

KINETIC ENERGY (Section 18.1)
The kinetic energy of a rigid body can be expressed as the sum of its translational and rotational kinetic energies In equation form, a body in general plane motion has kinetic energy given by: T = 1/2 m (vG)2 + 1/2 IG w2 Several simplifications can occur. 1. Pure Translation: When a rigid body is subjected to only curvilinear or rectilinear translation, the rotational kinetic energy is zero (w = 0). Therefore, T = 1/2 m (vG)2 500

KINETIC ENERGY (continued)
2. Pure Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy. Thus, T = 0.5 m (vG) IG w2 Since vG = rGw, we can express the kinetic energy of the body as: T = 0.5 [ IG + m(rG)2 ] w2 = 0.5 IO w2 If the rotation occurs about the mass center, G, then what is the value of vG? In this case, the velocity of the mass center is equal to zero So the kinetic energy equation reduces to: T = 0.5 IG w2 501

THE WORK OF A FORCE (Section 18.2)
Recall that the work done by a force can be written as: UF =  F•dr =  (F cos q ) ds. When the force is constant, this equation reduces to UFc = (Fc cos q )s where Fccosq represents the component of the force acting in the direction of the displacement, s. s Work of a weight: As before, the work can be expressed as Uw = -WDy. Remember, if the force and movement are in the same direction, the work is positive. Work of a spring force: For a linear spring, the work is: Us = -0.5k[(s2)2 – (s1)2] 502

There are some external forces that do no work.
For instance, reactions at fixed supports do no work because the displacement at their point of application is zero. Normal forces and friction forces acting on bodies as they roll without slipping over a rough surface also do no work since there is no instantaneous displacement of the point in contact with ground (it is an instant center, IC). Internal forces do no work because they always act in equal and opposite pairs. Thus, the sum of their work is zero. 503

THE WORK OF A COUPLE (Section 18.3)
When a body subjected to a couple experiences general plane motion, the two couple forces do work only when the body undergoes rotation. If the body rotates through an angular displacement dq, the work of the couple moment, M, is: ò = q2 q1 M dq UM If the couple moment, M, is constant, then UM = M (q2 – q1) Here the work is positive, provided M and (q2 – q1) are in the same direction. 504

PRINCIPLE OF WORK AND ENERGY (Section 18.4)
Recall the statement of the principle of work and energy used earlier: T1 + SU1-2 = T2 In the case of general plane motion, this equation states that the sum of the initial kinetic energy (both translational and rotational) and the work done by all external forces and couple moments equals the body’s final kinetic energy (translational and rotational). This equation is a scalar equation. It can be applied to a system of rigid bodies by summing contributions from all bodies. 505

EXAMPLE Given: The disk weighs 40 lb and has a radius of gyration (kG) of 0.6 ft. A 15 ft·lb moment is applied and the spring has a spring constant of 10 lb/ft. Find: The angular velocity of the wheel when point G moves 0.5 ft. The wheel starts from rest and rolls without slipping. The spring is initially un-stretched. Plan: Use the principle of work and energy to solve the problem since distance is the primary parameter. Draw a free body diagram of the disk and calculate the work of the external forces. 506

Free body diagram of the disk:
EXAMPLE (continued) Solution: Free body diagram of the disk: Since the disk rolls without slipping on a horizontal surface, only the spring force and couple moment M do work. Why don’t forces FB and NB do any work? Since the spring is attached to the top of the wheel, it will stretch twice the amount of displacement of G, or 1 ft. 507

Work: U1-2 = -0.5k[(s2)2 – (s1)2] + M(q2 – q1)
EXAMPLE (continued) Work: U1-2 = -0.5k[(s2)2 – (s1)2] + M(q2 – q1) U1-2 = -0.5(10)(12 – 0) + 15(0.5/0.8) = ft·lb Kinematic relation: vG = r w = 0.8w Kinetic energy: T1 = 0 T2 = 0.5m (vG) IG w2 T2 = 0.5(40/32.2)(0.8w) (40/32.2)(0.6)2w2 T2 = w2 Work and energy: T1 + U1-2 = T2 = w2 w = 2.65 rad/s 508

C) equal to its rotational kinetic energy D) Cannot be determined
CONCEPT QUIZ 1. If a rigid body rotates about its center of gravity, its translational kinetic energy is ___________ at all times. A) constant B) zero C) equal to its rotational kinetic energy D) Cannot be determined 2. A rigid bar of mass m and length L is released from rest in the horizontal position. What is the rod’s angular velocity when it has rotated through 90°? A) g/3L B) 3g/L C) 12g/L D) g/L Answer: B m L • 509

Find: The angular velocity of the pendulum when q = 90°.
GROUP PROBLEM SOLVING Given: The 50 kg pendulum of the Charpy impact machine is released from rest when q = 0. The radius of gyration kA = 1.75 m. Find: The angular velocity of the pendulum when q = 90°. Plan: Since the problem involves distance, the principle of work and energy is an efficient solution method. The only force involved doing work is the weight, so only its work need be determined. 510

Solution: Calculate the vertical distance the mass center moves. Dy = 1.25 sin q Then, determine the work due to the weight. Uw = -WDy U1-2 = W (1.25 sin q ) = 50(9.81) (1.25 sin 90) = N·m The mass moment of inertia about A is: IA = m (kA)2 = 50(1.75)2 = kg·m2 511

Kinetic energy: T1 = 0 T2 = 0.5m(vG) IG w2 = 0.5 IA w2 = 0.5 (153.1) w2 Now apply the principle of work and energy equation: T1 + U1-2 = T2 = w2 w = 2.83 rad/s 512

C) The two are equal. D) Cannot be determined.
ATTENTION QUIZ 1. A disk and a sphere, each of mass m and radius r, are released from rest. After 2 full turns, which body has a larger angular velocity? Assume roll without slip. A) Sphere B) Disk C) The two are equal D) Cannot be determined. q r 2. A slender bar of mass m and length L is released from rest in a horizontal position. The work done by its weight when it has rotated through 90° is? A) m g (p/2) B) m g L C) m g (L/2) D) -m g (L/2) Answer: A (body with smaller MMI will have a larger angular velocity) C m L • 513

LINEAR AND ANGULAR MOMENTUM, PRINCIPLE OF IMPULSE AND MOMENTUM
Today’s Objectives: Students will be able to: Develop formulations for the linear and angular momentum of a body. Apply the principle of linear and angular impulse and momentum. In-Class Activities: • Check Homework • Reading Quiz • Applications • Linear and Angular Momentum • Principle of Impulse and Momentum • Concept Quiz • Group Problem Solving • Attention Quiz 514

READING QUIZ 1. The angular momentum of a rotating two-dimensional rigid body about its center of mass G is ___________. A) m vG B) IG vG C) m w D) IG w 2. If a rigid body rotates about a fixed axis passing through its center of mass, the body’s linear momentum is __________. A) a constant B) zero C) m vG D) IG w Answers: 1. D 2. B 515

APPLICATIONS The swing bridge opens and closes by turning using a motor located under the center of the deck at A that applies a torque M to the bridge. If the bridge was supported at its end B, would the same torque open the bridge in the same time, or would it open slower or faster? What are the benefits of making the bridge with the variable depth (thickness) substructure as shown? 516

As the pendulum of the Charpy tester swings downward, its angular momentum and linear momentum both increase. By calculating its momenta in the vertical position, we can calculate the impulse the pendulum exerts when it hits the test specimen. As the pendulum rotates about point O, what is its angular momentum about point O? 517

The space shuttle has several engines that exert thrust on the shuttle when they are fired. By firing different engines, the pilot can control the motion and direction of the shuttle. If only engine A is fired, about which axis does the shuttle tend to rotate? 518

LINEAR AND ANGULAR MOMENTUM (Section 19.1)
The linear momentum of a rigid body is defined as L = m vG This equation states that the linear momentum vector L has a magnitude equal to (mvG) and a direction defined by vG. The angular momentum of a rigid body is defined as HG = IG w Remember that the direction of HG is perpendicular to the plane of rotation. 519

LINEAR AND ANGULAR MOMENTUM (continued)
Translation. When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because w = 0. Therefore, L = m vG and HG = 0 520

Rotation about a fixed axis. When a rigid body is rotating about a fixed axis passing through point O, the body’s linear momentum and angular momentum about G are: L = mvG HG = IG w It is sometimes convenient to compute the angular momentum of the body about the center of rotation O. HO = ( rG × mvG) + IG w = IO w 521

General plane motion. When a rigid body is subjected to general plane motion, both the linear momentum and the angular momentum computed about G are required. L = m vG HG = IGw The angular momentum about point A is HA = IGw + mvG (d) 522

PRINCIPLE OF IMPULSE AND MOMENTUM
(Section 19.2) As in the case of particle motion, the principle of impulse and momentum for a rigid body is developed by combining the equation of motion with kinematics. The resulting equations allow a direct solution to problems involving force, velocity, and time. Linear impulse-linear momentum equation: L F dt = L2 or (mvG) F dt = (mvG)2 t2 t1 å ò Angular impulse-angular momentum equation: (HG) MG dt = (HG)2 or IGw MG dt = IGw2 t2 t1 å ò 523

PRINCIPLE OF IMPULSE AND MOMENTUM (continued)
The previous relations can be represented graphically by drawing the impulse-momentum diagram. To summarize, if motion is occurring in the x-y plane, the linear impulse-linear momentum relation can be applied to the x and y directions and the angular momentum-angular impulse relation is applied about a z-axis passing through any point (i.e., G). Therefore, the principle yields three scalar equations describing the planar motion of the body. 524

• Establish the x, y, z inertial frame of reference. • Draw the impulse-momentum diagrams for the body. • Compute IG, as necessary. • Apply the equations of impulse and momentum (one vector and one scalar or the three scalar equations). • If more than three unknowns are involved, kinematic equations relating the velocity of the mass center G and the angular velocity w should be used to furnish additional equations. 525

The wheel is subjected to a couple moment of 300 Nm.
EXAMPLE Given: The 300 kg wheel has a radius of gyration about its mass center O of kO = 0.4 m. The wheel is subjected to a couple moment of 300 Nm. A Find: The angular velocity after 6 seconds if it starts from rest and no slipping occurs. Plan: Time as a parameter should make you think Impulse and Momentum! Since the body rolls without slipping, point A is the center of rotation. Therefore, applying the angular impulse and momentum relationships along with kinematics should solve the problem. 526

+ = ò EXAMPLE (continued) Solution: (m vG)1 (m vG)2 IGw1 IG w2 M t F t
W t r + = A y x Impulse-momentum diagram: Kinematics: (vG)2 = r w2 Impulse & Momentum: (HA) MA dt = (HA)2 0 + M t = m(vG)2 r + IG w2 = m r2 w2 + m(kO)2 w2 = m{r2 + (kO)2}w2 t2 t1 å ò 300 ( ) 300 (6) m {r2 + (kO)2} M t w2 = 11.5 rad/s = 527

A) larger than B) less than C) the same as D) None of the above
CONCEPT QUIZ A) larger than B) less than C) the same as D) None of the above 1. If a slab is rotating about its center of mass G, its angular momentum about any arbitrary point P is __________ its angular momentum computed about G (i.e., IG w). Answers: C B 2. The linear momentum of the slab in question 1 is __________. A) constant B) zero C) increasing linearly D) decreasing linearly with time with time 528

GROUP PROBLEM SOLVING Given: A gear set with: mA = 10 kg mB = 50 lb kA = 0.08 m kB = 0.15 m M = 10 N·m Find: The angular velocity of gear B after 5 seconds if the gears start turning from rest. Plan: Time is a parameter, thus Impulse and Momentum is recommended. First, relate the angular velocities of the two gears using kinematics. Then apply angular impulse and momentum to both gears.

Solution: Impulse-momentum diagrams: Note that the initial momentum is zero for both gears. Gear A: = Ft Ay t Ax t WAt M t IA wA rA y x Gear B: = F t WBt Bxt Byt rB IB wB 530

Kinematics: rAwA = rBwB Angular impulse & momentum relation: For gear A: M t − (F t) rA = IA wA For gear B: (F t) rB = IB wB  (F t) = (IB wB) / rB Combining the two equations yields: M t = IAwA + (rA/rB) IBwB Substituting from kinematics for wA= (rB/rA)wB, yields M t = wB [ (rB/rA) IA + (rA/rB) IB ] eqn (1) 531

where IA = mA (kA)2 = 10 (0.08)2 = kg·m2 IB = mB (kB)2 = 50 (0.15)2 = kg·m2 Using Eq. (1), M t = wB [ (rB/rA) IA + (rA/rB) IB ] 10 (5) = wB [ (0.2/0.1) (0.1/0.2) ] 50 = wB Therefore, wB = 72.4 rad/s and wA = (rB/rA) wB = (0.2/0.1) 72.4 = 144 rad/s 532

C) Depends on the case D) No clue!
ATTENTION QUIZ 1. If a slender bar rotates about end A, its angular momentum with respect to A is? A) (1/12) m l2 w B) (1/6) m l2 w C) (1/3) m l2 w D) m l2 w G l A w 2. As in the principle of work and energy, if a force does no work, it does not need to be shown on the impulse and momentum diagram/equation. A) False B) True C) Depends on the case D) No clue! Answers: 1. C 2. A 533

CONSERVATION OF MOMENTUM
Today’s Objectives: Students will be able to: Understand the conditions for conservation of linear and angular momentum. Use the condition of conservation of linear/ angular momentum. In-Class Activities: • Check Homework • Reading Quiz • Applications • Conservation of Linear and Angular Momentum • Concept Quiz • Group Problem Solving • Attention Quiz 534

1. If there are no external impulses acting on a body _____________.
READING QUIZ 1. If there are no external impulses acting on a body _____________. A) only linear momentum is conserved B) only angular momentum is conserved C) both linear momentum and angular momentum are conserved D) neither linear momentum nor angular momentum are conserved 2. If a rigid body rotates about a fixed axis passing through its center of mass, the body’s linear momentum is __________. A) constant B) zero C) m vG D) IG w Answer: C B 535

APPLICATIONS A skater spends a lot of time either spinning on the ice or rotating through the air. To spin fast, or for a long time, the skater must develop a large amount of angular momentum. If the skater’s angular momentum is constant, can the skater vary her rotational speed? How? The skater spins faster when the arms are drawn in and slower when the arms are extended. Why?

Conservation of angular momentum allows cats to land on their feet and divers to flip, twist, spiral and turn. It also helps teachers make their heads spin! Conservation of angular momentum makes water circle the drain faster as it gets closer to the drain.

CONSERVATION OF LINEAR MOMENTUM (Section 19.3)
Recall that the linear impulse and momentum relationship is L F dt = L2 or (m vG) F dt = (m vG)2 t2 t1 å ò If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, the linear momentum for a rigid body (or system) is constant, or conserved. So L1 = L2. This equation is referred to as the conservation of linear momentum. The conservation of linear momentum equation can be used if the linear impulses are small or non-impulsive.

ò CONSERVATION OF ANGULAR MOMENTUM
Similarly, if the sum of all the angular impulses due to external forces acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved (HG) MG dt = (HG)2 or IGw MG dt = IGw2 t2 t1 å ò The resulting equation is referred to as the conservation of angular momentum or (HG)1 = (HG)2 . If the initial condition of the rigid body (or system) is known, conservation of momentum is often used to determine the final linear or angular velocity of a body just after an event occurs.

• Establish the x, y, z inertial frame of reference and draw FBDs. • Write the conservation of linear momentum equation. • Write the conservation of angular momentum equation about a fixed point or at the mass center G. • Solve the conservation of linear or angular momentum equations in the appropriate directions. • If the motion is complicated, use of the kinematic equations relating the velocity of the mass center G and the angular velocity w may be necessary.

EXAMPLE Given: A 10 kg wheel (IG = kg·m2) rolls without slipping and does not rebound. Find: The minimum velocity, vG, the wheel must have to just roll over the obstruction at A. Plan: Since no slipping or rebounding occurs, the wheel pivots about point A. The force at A is much greater than the weight, and since the time of impact is very short, the weight can be considered non-impulsive. The reaction force at A is a problem as we don’t know either its direction or magnitude. This force can be eliminated by applying the conservation of angular momentum equation about A.

r ' m (vG)1 + IG w1 = r m (vG)2 + IG w2
EXAMPLE (continued) Solution: Impulse-momentum diagram: Conservation of angular momentum: (HA)1 = (HA)2 r ' m (vG)1 + IG w1 = r m (vG)2 + IG w2 ( ) 10 (vG) w1 = 0.2(10) (vG) w2 Kinematics: Since there is no slip, w = vG/r = 5 vG. Substituting and solving the momentum equation yields (vG)2 = (vG)1

Energy conservation equation : T2 + V2 = T3 + V3
EXAMPLE (continued) To complete the solution, conservation of energy can be used. Since it cannot be used for the impact (why?), it is applied just after the impact. In order to roll over the bump, the wheel must go to position 3 from 2. When (vG)2 is a minimum, (vG)3 is zero. Why? Energy conservation equation : T2 + V2 = T3 + V3 {½ (10) (vG)22 + ½ (0.156) w22 } + 0 = (0.03) Substituting w2 = 5 (vG)2 and (vG)2 = (vG)1 and solving yields (vG)1 = m/s 543

momentum of the bullet about A just before impact is ___________.
CONCEPT QUIZ 1. A slender rod (mass = M) is at rest. If a bullet (mass = m) is fired with a velocity of vb, the angular momentum of the bullet about A just before impact is ___________. A) 0.5 m vb2 B) m vb C) 0.5 m vb D) zero G 1.0 A w2 0.5 m 2. For the rod in question 1, the angular momentum about A of the rod and bullet just after impact will be ___________. A) m vb + M(0.5)w2 B) m(0.5)2w2 + M(0.5)2w2 C) m(0.5)2w2 + M(0.5)2w2 D) zero + (1/12) M w2 Answers: C 544

GROUP PROBLEM SOLVING Given: Two children(mA= mB= 30 kg) sit at the edge of the merry-go-round, which has a mass of 180 kg and a radius of gyration of kz = 0.6 m. Find: The angular velocity of the merry-go-round if A jumps off horizontally in the +t direction with a speed of 2 m/s, measured relative to the merry-go-round. Plan: Draw an impulse-momentum diagram. The conservation of angular momentum can be used to find the angular velocity.

Solution: Impulse-momentum diagram: Apply the conservation of angular momentum equation: = mA mB M 2 rad/s vA/M = 2 m/s  ∑(H)1 = ∑(H)2 180 (0.6)2 (2) + 2 × {(30) 2 (0.75) 2} = 180 (0.6)2 w + (30) w (0.75) 2 + (30) (0.75 w + 2) (0.75)  = w + 45 Solving yields w = 1.54 rad/s

A) all linear impulses sum to zero B) all angular impulses sum to zero
ATTENTION QUIZ 1. Using conservation of linear and angular momentum requires that _____________. A) all linear impulses sum to zero B) all angular impulses sum to zero C) both linear and angular impulses sum to zero D) None of the above 2. The angular momentum of a body about a point A that is the fixed axis of rotation but not the mass center (G) is _____________. A) IAw B) IG w C) rG (m vG) + IG w D) Both A & C Answers: 1. C 2. D 547

Rectilinear Kinematics

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Rectilinear Kinematics

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