2. Chapter 12-1 & 12-2

3. • Relations between s(t), v(t), and a(t) for general
INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION (Sections )
Today’s Objectives:
Students will be able to find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Relations between s(t), v(t), and a(t) for general
rectilinear motion
• Relations between s(t), v(t),
and a(t) when acceleration is constant
• Concept quiz
• Group problem solving
• Attention quiz

4. 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?

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

6. 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

7. The displacement of the particle is defined as its change in position.
POSITION AND DISPLACEMENT
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.

8. 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

9. The instantaneous acceleration is the time derivative of velocity.
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

10. 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.

11. 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 s
(1/2)a ds ) - (s 2a (v

12. Find: The distance the motorcycle travels before it stops.
EXAMPLE
Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it stops.
Plan: Establish the positive coordinate s in the direction the motorcycle is traveling. Since the acceleration is given as a function of time, integrate it once to calculate the velocity and again to calculate the position.

13. ò ò 1) Integrate acceleration to determine the velocity.
EXAMPLE (continued)
Solution:
1) Integrate acceleration to determine the velocity.
a = dv / dt => dv = a dt =>
=> v – vo = -3t2 => v = -3t2 + vo ò - = t o v dt dv ) 6 (
2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use vo = 27 m/s.
0 = -3t => t = 3 s
3) Now calculate the distance traveled in 3s by integrating the velocity using so = 0:
v = ds / dt => ds = v dt =>
=> s – so = -t3 + vot
=> s – 0 = (3)3 + (27)(3) => s = 54 m ò + - = t o s dt v ds ) 3 ( 2

14. Find: The speed at which ball B was thrown upward.
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. Apply the formulas for constant acceleration, with ac = ft/s2.

15. 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)ot + (1/2)act2
20 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t2) => t = s
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
20 ft = 5 + (vB)o(1.115) + (1/2)(-32.2)(1.115)2
=> (vB)o = 31.4 ft/s

16. Chapter 12-4 & 12-5

17. CURVILINEAR MOTION: RECTANGULAR COMPONENTS (Sections 12.4-12.5)
Today’s Objectives:
Students will be able to:
a) Describe the motion of a particle traveling along a curved path.
b) Relate kinematic quantities in terms of the rectangular components of the vectors.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• General curvilinear motion
• Rectangular components of kinematic vectors
• Concept quiz
• Group problem solving
• Attention quiz

18. APPLICATIONS
The path of motion of each plane in this formation can be tracked with radar and their 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 each plane at any instant?
Should they be the same for each aircraft?

19. APPLICATIONS (continued)
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?

20. POSITION AND DISPLACEMENT
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

21. 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!

22. 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.

23. RECTANGULAR COMPONENTS: POSITION
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

24. 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 = vxi + vyj + vzk
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.

25. 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 = axi + ayj + azk
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.

26. EXAMPLE
Given: The motion of two particles (A and B) is described by the position vectors
rA = [3t i + 9t(2 – t) j] m
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.

27. EXAMPLE (continued)
Solution:
1) The point of collision requires that rA = rB, so xA = xB and yA = yB .
x-components: 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
y-components: 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. Substituting this value into rA or rB yields
xA = xB = 6 m and yA = yB = 0

28. 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 = = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s . xA i + yA j • •
vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
Speed is the magnitude of the velocity vector.
vA = ( ) = 18.2 m/s
vB = ( ) = 6.71 m/s

29. GROUP PROBLEM SOLVING
Given: A particle travels along a path described by the parabola y = 0.5x2. The x-component of velocity is given by vx = (5t) ft/s. When t = 0, x = y = 0.
Find: The particle’s distance from the origin and the magnitude of its acceleration when t = 1 s.
Plan: Note that vx is given as a function of time.
1) Determine the x-component of position and acceleration by integrating and differentiating vx, respectively.
2) Determine the y-component of position from the parabolic equation and differentiate to get ay.
3) Determine the magnitudes of the position and acceleration vectors.

30. ò GROUP PROBLEM SOLVING (continued) Solution: 1) x-components:
Velocity: vx = x = dx/dt = (5t) ft/s
Position: = => x = (5/2)t2 = (2.5t2) ft
Acceleration: ax = x = vx = d/dt (5t) = 5 ft/s2 •• ò x dx t dt 5 •
2) y-components:
Position: y = 0.5x2 = 0.5(2.5t2)2 = (3.125t4) ft
Velocity: vy = dy/dt = d (3.125t4) /dt = (12.5t3) ft/s
Acceleration: ay = vy = d (12.5t3) /dt = (37.5t2) ft/s2

31. GROUP PROBLEM SOLVING (continued)
3) The distance from the origin is the magnitude of the position vector:
r = x i + y j = [2.5t2 i t4 j] ft
At t = 1 s, r = (2.5 i j) ft
Distance: d = r = ( ) 0.5 = 4.0 ft
The magnitude of the acceleration vector is calculated as:
Acceleration vector: a = [5 i t2 j ] ft/s2
Magnitude: a = ( )0.5 = 37.8 ft/s2

32. Chapter 12-6

33. MOTION OF A PROJECTILE (Section 12.6)
Today’s Objectives:
Students will be able to analyze the free-flight motion of a projectile.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Kinematic equations for projectile motion
• Concept quiz
• Group problem solving
• Attention quiz

34. APPLICATIONS
A kicker should know at what angle, q, and initial velocity, vo, 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?

35. APPLICATIONS (continued)
A fireman wishes to know the maximum height on the wall he can project water from the hose. At what angle, q, should he hold the hose?

36. CONCEPT OF PROJECTILE MOTION
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., 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.

37. 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)?

38. 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) – ½g(t)2
vy2 = voy2 – 2g(y – yo)
For any given problem, only two of these three equations can be used. Why?

39. Example 1
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)
x g x
vo cos θ 2 vo cos θ – 2
( ) ( ) ( )
By substituting for t:

40. ( ) Simplifying the last equation, we get:
Example 1 (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.

41. Given: Snowmobile is going 15 m/s at point A.
Example 2
Given: Snowmobile is going 15 m/s at point A.
Find: The horizontal distance it travels (R) and the time in the air.
Solution:
First, place the coordinate system at point A. Then write the equation for horizontal motion.
+ xB = xA + vAxtAB and vAx = 15 cos 40° m/s
Now write a vertical motion equation. Use the distance equation.
+ yB = yA + vAytAB – 0.5gctAB2 vAy = 15 sin 40° m/s
Note that xB = R, xA = 0, yB = -(3/4)R, and yA = 0.
Solving the two equations together (two unknowns) yields
R = 19.0 m tAB = 2.48 s

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

43. GROUP PROBLEM SOLVING (continued)
Solution:
Motion in x-direction:
Using xB = xA + vox(tAB) =
tAB=
(4/5)100
vA (cos 25)
88.27 vA
-64 = 0 + vA(sin 45) 80
vA (cos 25)
– ½ (9.81)
88.27 vA 2
Motion in y-direction:
Using yB = yA + voy(tAB) – ½ g(tAB)2
vA = m/s

44. Chapter 12-7

45. NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)
CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)
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, if any
• Reading quiz
• Applications
• Normal and tangential components of velocity and acceleration
• Special cases of motion
• Concept quiz
• Group problem solving
• Attention quiz

46. APPLICATIONS
Cars traveling along a clover-leaf interchange experience an acceleration due to a change in speed 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?

47. APPLICATIONS (continued)
A motorcycle travels up a hill for which the path can be approximated by a function y = f(x).
If the motorcycle starts from rest and increases its speed at a constant rate, how can we determine its velocity and acceleration at the top of the hill?
How would you analyze the motorcycle's “flight” at the top of the hill?

48. NORMAL AND TANGENTIAL COMPONENTS
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.

49. 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.

50. 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 = vut where v = s = ds/dt .
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.

51. 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 = vut + (v2/r)un = atut + anun.
After mathematical manipulation, the acceleration vector can be expressed as:

52. ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
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

53. 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.

54. SPECIAL CASES OF MOTION (continued)
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vot + (1/2)(at)ct2
v = vo + (at)ct
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

55. 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 x un.
There is no motion, thus no velocity or acceleration, in the binomial direction.

56. Plan: The boat starts from rest (v = 0 when t = 0).
EXAMPLE PROBLEM
Given: Starting from rest, a motorboat travels around a circular path of r = 50 m at a speed that increases with time,
v = (0.2 t2) m/s.
Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 3 s.
Plan: The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 3s using v(t).
2) Calculate the tangential and normal components of acceleration and then the magnitude of the acceleration vector.

57. 2) The acceleration vector is a = atut + anun = vut + (v2/r)un.
EXAMPLE (continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is given by v = (0.2t2) m/s. At t = 3s:
v = 0.2t2 = 0.2(3)2 = 1.8 m/s
2) The acceleration vector is a = atut + anun = vut + (v2/r)un. .
Tangential component: at = v = d(.2t2)/dt = 0.4t m/s2
At t = 3s: at = 0.4t = 0.4(3) = 1.2 m/s2 .
Normal component: an = v2/r = (0.2t2)2/(r) m/s2
At t = 3s: an = [(0.2)(32)]2/(50) = m/s2
The magnitude of the acceleration is
a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2

58. Find: The magnitude of the plane’s acceleration when it is at point A.
GROUP PROBLEM SOLVING
Given: A jet plane travels along a vertical parabolic path defined by the equation y = 0.4x2. At point A, the jet has a speed of 200 m/s, which is increasing at the rate of 0.8 m/s2.
Find: The magnitude of the plane’s acceleration when it is at point A.
Plan: 1) The change in the speed of the plane (0.8 m/s2) is the tangential component of the total acceleration.
2) Calculate the radius of curvature of the path at A.
3) Calculate the normal component of acceleration.
4) Determine the magnitude of the acceleration vector.

59. GROUP PROBLEM SOLVING (continued)
Solution:
1) The tangential component of acceleration is the rate of increase of the plane’s speed, so at = v = 0.8 m/s2. .
2) Determine the radius of curvature at point A (x = 5 km):
dy/dx = d(0.4x2)/dx = 0.8x, d2y/dx2 = d (0.8x)/dx = 0.8
At x =5 km, dy/dx = 0.8(5) = 4, d2y/dx2 = 0.8
=> r = ________________ = [1 + (4)2]3/2/(0.8) = km
]3/2
(dy/dx)2 1 [ + 2
d2y/dx
3) The normal component of acceleration is
an = v2/r = (200)2/(87.62 x 103) = m/s2
4) The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5 = [(0.8)2 + (0.457)2]0.5 = m/s2

60. Chapter 12-8
Omitted

61. CURVILINEAR MOTION: CYLINDRICAL COMPONENTS (Section 12.8)
Today’s Objectives:
Students will be able to determine velocity and acceleration components using cylindrical coordinates.
In-Class Activities:
Check homework, if any
Reading quiz
Applications
Velocity Components
Acceleration Components
Concept quiz
Group problem solving
Attention quiz
Omitted

62. APPLICATIONS
The cylindrical coordinate system is used in cases where the particle moves along a 3-D curve.
In the figure shown, the boy slides down the slide at a constant speed of 2 m/s. How fast is his elevation from the ground changing (i.e., what is z )? .
Omitted

63. APPLICATIONS (continued)
A polar coordinate system is a 2-D representation of the cylindrical coordinate system.
Omitted
When the particle moves in a plane (2-D), and the radial distance, r, is not constant, the polar coordinate system can be used to express the path of motion of the particle.

64. POSITION (POLAR COORDINATES)
Omitted
We can express the location of P in polar coordinates as r = rur. 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.

65. VELOCITY (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
Omitted

66. 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 – rq2)ur + (rq + 2rq)uθ .. . .
The magnitude of acceleration is a = (r – rq2)2 + (rq + 2rq)2
The term (r – rq2) is the radial acceleration or ar.
The term (rq + 2rq) is the transverse acceleration or aq ..
Omitted

67. 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:
Omitted
Velocity: vP = rur + rquθ + zuz
Acceleration: aP = (r – rq2)ur + (rq + 2rq)uθ + zuz .. .

68. Omitted Given: r = 5 cos(2q) (m) q = 3t2 (rad/s) qo = 0
EXAMPLE
Given: r = 5 cos(2q) (m)
q = 3t2 (rad/s)
qo = 0
Find: Velocity and acceleration at q = 30°.
Plan: Apply chain rule to determine r and r and evaluate at q = 30°. . . .. t ò o =
Solution: q = q dt = 3t2 dt = t3
At q = 30°, q = = t3. Therefore: t = s.
q = 3t2 = 3(0.806)2 = 1.95 rad/s 6 p .
Omitted

69. EXAMPLE (continued)
q = 6t = 6(0.806) = rad/s2
r = 5 cos(2q) = 5 cos(60) = 2.5m
r = -10 sin(2q)q = -10 sin(60)(1.95) = m/s
r = -20 cos(2q)q2 – 10 sin(2q)q
= -20 cos(60)(1.95)2 – 10 sin(60)(4.836) = -80 m/s2 .. .
Omitted
Substitute in the equation for velocity
v = rur + rquθ
v = ur + 2.5(1.95)uθ
v = (16.88)2 + (4.87)2 = m/s .

70. Omitted Substitute in the equation for acceleration:
EXAMPLE (continued)
Substitute in the equation for acceleration:
a = (r – rq2)ur + (rq + 2rq)uθ
a = [-80 – 2.5(1.95)2]ur + [2.5(4.836) + 2(-16.88)(1.95)]uθ
a = ur – 53.7uθ m/s2
a = (89.5)2 + (53.7)2 = m/s2 .. .
Omitted

71. Omitted Given: The car’s speed is constant at 1.5 m/s.
GROUP PROBLEM SOLVING
Given: The car’s speed is constant at 1.5 m/s.
Find: The car’s acceleration (as a vector).
Hint: The tangent to the ramp at any point is at an angle
f = tan-1( ) = °
Also, what is the relationship between f and q? 12
2p(10)
Omitted
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 cosf and vz = z = v sinf .

72. GROUP PROBLEM SOLVING (continued)
Therefore: q = ( ) = rad/s
q = 0
v cosf r . ..
vz = z = v sinf = m/s
z = 0
r = r = 0 . ..
Omitted .. .
a = (r – rq2)ur + (rq + 2rq)uθ + zuz
a = (-rq2)ur = -10(0.147)2ur = ur m/s2

73. Chapter 12-9

74. ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section 12.9)
Today’s Objectives:
Students will be able to relate the positions, velocities, and accelerations of particles undergoing dependent motion.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Define dependent motion
• Develop position, velocity, and acceleration relationships
• Concept quiz
• Group problem solving
• Attention quiz

75. APPLICATIONS
The cable and pulley system shown here can be used to modify the speed of block B relative to the speed of the motor. It is important to relate the various motions in order to determine the power requirements for the motor and the tension in the cable.
If the speed of the cable coming onto the motor pulley is known, how can we determine the speed of block B?

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

77. DEPENDENT MOTION
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.

78. 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 arc CD on the pulley.

79. 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 .

80. 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.

81. DEPENDENT MOTION EXAMPLE (continued)
The position coordinates are related by the equation
2sB + h + sA = l
Where l is the total cord length minus the lengths of the red segments.
Since l 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 the sign convention!

82. 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 = l
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.

83. 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!

84. Find: The speed of block B.
EXAMPLE PROBLEM
Given: In the figure on the left, the cord at A is pulled down with a speed of 8 ft/s.
Find: The speed of block B.
Plan: There are two cords involved in the motion in this example. The position of a point on one cord must be related to the position of a point on the other cord. There will be two position equations (one for each cord).

85. • sA can be defined to the center of the pulley above point A.
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 relating positions on the two cords. Note that pulley C relates the motion of the two cords. sA sC sB
DATUM
• Define the datum line through the top pulley (which has a fixed position).
• sA can be defined to the center of the pulley above 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.

86. 2vA + 4vB = 0 => vB = - 0.5vA = - 0.5(8) = - 4 ft/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: sA sC sB
DATUM
Cord 1: 2sA + 2sC = l1
Cord 2: sB + (sB – sC) = l2
3) Eliminating sC between the two equations, we get
2sA + 4sB = l1 + 2l2
4) Relate velocities by differentiating this expression. Note that l1 and l2 are constant lengths.
2vA + 4vB = => vB = - 0.5vA = - 0.5(8) = - 4 ft/s
The velocity of block B is 4 ft/s up (negative sB direction).

87. Find: The speed of block B.
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.

88. 3) Differentiate to relate velocities: vA + 2vB + 2vC = 0
GROUP PROBLEM SOLVING (continued)
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 + 2sC = l sA sC sB
DATUM
3) Differentiate to relate velocities:
vA + 2vB + 2vC = 0
4 + 2vB + 2(-2) =0
vB = 0

89. Chapter 12-10

90. RELATIVE MOTION ANALYSIS (Section 12.10)
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, if any
• Reading quiz
• Applications
• Relative position, velocity and acceleration
• Vector & graphical methods
• Concept quiz
• Group problem solving
• Attention quiz

91. APPLICATIONS
When you try to hit a moving object, the position, velocity, and acceleration of the object must be known. 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?

92. APPLICATIONS (continued)
When fighter jets take off or land on an aircraft carrier, the velocity of the carrier becomes an issue.
If the aircraft carrier travels at 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?

93. RELATIVE POSITION
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.

94. 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 .

95. 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.
aB/A = aB – aA or
aB = aA + aB/A

96. 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 Cartesian vectors and the resulting scalar 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?

97. LAWS OF SINES AND COSINES
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 review, their formulations are provided below. a b c C B A
Law of Sines: C c B b A a
sin =
Law of Cosines: A bc c b a
cos 2 - + = B ac C ab

98. EXAMPLE
Given: vA = 600 km/hr
vB = 700 km/hr
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.

99. vB/A = vB – vA = (- 1191.5 i + 344.1 j ) km/hr
EXAMPLE (continued)
Solution:
vA = 600 cos 35 i – 600 sin 35 j
= (491.5 i – j ) km/hr
vB = -700 i km/hr
a) Vector Method:
vB/A = vB – vA = ( i j ) km/hr hr km v A B 2 .
1240 ) 1
344 ( 5
1191 / = +
where - 16
tan q

100. Note that the vector that measures the tip of B relative to A is vB/A.
EXAMPLE (continued)
b) Graphical Method:
145
vB = 700 km/hr
vA = 600 km/hr
vB/A
Note that the vector that measures the tip of B relative to A is vB/A.
Law of Cosines: - + =
cos )
600 )(
700 ( 2 / A B v q hr km .
1240
Law of Sines: q
sin )
145°
sin( / A B v = or 1 . 16

101. GROUP PROBLEM SOLVING y x
Given: vA = 10 m/s
vB = 18.5 m/s
at)A = 5 m/s2
aB = 2 m/s2
Find: vA/B
aA/B
Plan: Write the velocity and acceleration vectors for A and B and determine vA/B and aA/B by using vector equations.
Solution:
The velocity of A is:
vA = 10 cos(45)i – 10 sin(45)j = (7.07i – 7.07j) m/s

102. GROUP PROBLEM SOLVING (continued)
The velocity of B is:
vB = 18.5i (m/s)
The relative velocity of A with respect to B is (vA/B):
vA/B = vA – vB = (7.07i – 7.07j) – (18.5i) = i – 7.07j
or vB/A = (11.43)2 + (7.07)2 = m/s
q = tan-1( ) = ° q
7.07
11.43

103. GROUP PROBLEM SOLVING (continued)
The acceleration of A is:
aA = (at)A + (an)A = [5 cos(45)i – 5 sin(45)j]
+ [-( ) sin(45)i – ( ) cos(45)j]
aA = 2.83i – 4.24j (m/s2)
102
100
The acceleration of B is:
aB = 2i (m/s2)
The relative acceleration of A with respect to B is:
aA/B = aA – aB = (2.83i – 4.24j) – (2i) = 0.83i – 4.24j
aA/B = (0.83)2 + (4.24)2 = m/s2
b = tan-1( ) = 78.9° b
4.24
0.83