Kinematics of Particles Chapter 11 Kinematics of Particles (Part 2)

Given the x-t curve, the v-t curve is equal to the x-t curve slope. 11.7 Graphical Solution of Rectilinear Motion Problem Given the x-t curve, the v-t curve is equal to the x-t curve slope. Given the v-t curve, the a-t curve is equal to the v-t curve slope.

Given the a-t curve, the change in velocity between t1 and t2 is equal to the area under the a-t curve between t1 and t2. Given the v-t curve, the change in position between t1 and t2 is equal to the area under the v-t curve between t1 and t2.

first moment of area under a-t curve with respect to t = t1 line. 11.8 Other Graphical Methods Moment-area method to determine particle position at time t directly from the a-t curve: using dv = a dt , first moment of area under a-t curve with respect to t = t1 line.

Method to determine particle acceleration from v-x curve: subnormal to v-x curve

Sample Problem 11.6 A subway car leaves station A; it gains speed at the rate of 1.2 m/s for 6 s and then at the rate of 1.8 m/s until it has reached the speed of 14.6 m/s. The car maintains the same speed until it approaches station B ; brakes are then applied, giving the car a constant deceleration and bringing it to a stop in 6 s. The total running time for A to B is 40 s. Draw the a-t, v-t , and x-t curves, and determine the distance stations A and B. 2

Solution Acceleration Time Curve Since the acceleration is either constant or zero, the a-t curve is made of horizontal straight-line segments. The values of t2 and a4 are determine as follows : The acceleration being negative, the corresponding area is below the t-axis this area represents a decrease in velocity.

Solution Velocity Time Curve Since the acceleration is either constant or zero, the v-t curve is made of straight line segments connecting the points determined above. Position Time Curve The points determined above should be joined by three arcs ofvparabola and one straight line segment. In constructing the x-t curve is equal to the value of v at that instant.

PROBLEMS

Solution A1= -12m/s A2= 8m/s a=( m/s) t(s) 2 6 10 14 -2

v=( m/s) t(s) A3 A4 A5 A6 A7 t(s) x(m) 4 -4 8 12 6 10 14 16 12 4 -8 -4 A3 A4 A5 A6 A7 12 6 10 14 16 12 4 -8 -4 10 14 t(s) x(m)

instantaneous velocity (vector) CURVILINEAR MOTION OF PARTICLES 11.9 Position Vector, Velocity, and Acceleration Particle moving along a curve other than a straight line is in curvilinear motion. Position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle. Consider particle which occupies position P defined by at time t and P’ defined by at t + Dt, instantaneous velocity (vector) instantaneous speed (scalar)

instantaneous acceleration (vector) Consider velocity of particle at time t and velocity at t + Dt, In general, acceleration vector is not tangent to particle path and velocity vector.

Let be a vector function of scalar variable u, 11.10 Derivatives of Vector Functions Let be a vector function of scalar variable u, Derivative of vector sum, Derivative of product of scalar and vector functions, Derivative of scalar product and vector product,

11.11 Rectangular Components of Velocity And Acceleration When position vector of particle P is given by its rectangular components, Velocity vector, Acceleration vector,

Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile, with initial conditions, Integrating twice yields Motion of projectile could be replaced by two independent rectilinear motions. Motion in horizontal direction is uniform. Motion in vertical direction is uniformly accelerated.

Differentiating twice, velocity of B relative to A. 11.12 Motion Relative to a Frame in Translation Designate one frame as the fixed frame of reference. All other frames not rigidly attached to the fixed reference frame are moving frames of reference. Position vectors for particles A and B with respect to the fixed frame of reference Oxyz are Vector joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and Differentiating twice, velocity of B relative to A. acceleration of B relative to A. Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A.

Sample Problem 11.7 A projectile is fired from the edge of a 150m cliff with an initial velocity of 180 m/s at an angle of 30º with the horizontal. Neglecting air resistance, find the horizontal distance from the gun to the point where the projectile strikes the ground. the greatest elevation above the ground reached by the projectile.

Solution The vertical and the horizontal motion will be considered separately. Vertical Motion . Uniformly Acceleration Motion Choosing the positive sence of the y axis upward and placing the origin O at the gun, we have Substituting into the equations of uniformly accelerated motion, we have Horizontal Motion. Uniform Motion Choosing the positive sense of the x axis to the right, we have Substituting into the equation of uniform motion, we obtain

t = 19.91 s x = 3100 m a. Horizontal Distance When the projectile strikes the ground, we have Carrying this value into Eq. (2) for the vertical motion, we write t = 19.91 s Carrying t=19.91 s into Eq. (4) for the horizontal motion, we obtain x = 3100 m b. Greatest Elevation When the projectile reaches its greatest elevation, we have Vy=0 ; carrying this value into Eq. (3) for the vertical motion, we write Greatest Elevation above ground = 150 m + 143 m = 563 m

Sample Problem 11.8 A projectile is fired with an initial velocity of 224 m/s at a target B located 610 m above the gun A and at a horizontal distance of 3658 m. Neglecting air resistance, determine the value of the firing angle α

Solution The horizontal and the vertical motion will be considered separately. Horizontal motion . Placing the origin of the coordinate axes at the gun, we have Substituting into the equation of uniform horizontal motion, we obtain The time required for the projectile to move through a horizontal distance of 3658 m is obtained by setting x equal to 3658 m Vertical Motion Substituting into the equation of uniformly accelerated vertical motion, we obtain t 2

Solution

PROBLEMS

30º

50 m 3 m 13 m h Vo

50 m 3 m 13 m h Vo

50 m 3 m 13 m h Vo

11.13 Tangential and Normal Components Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components. are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, and is the angle between them.

With the velocity vector expressed as the particle acceleration may be written as but After substituting, Tangential component of acceleration reflects change of speed and normal component reflects change of direction. Tangential component may be positive or negative. Normal component always points toward center of path curvature.

Relations for tangential and normal acceleration also apply for particle moving along space curve. Plane containing tangential and normal unit vectors is called the osculating plane. Normal to the osculating plane is found from Acceleration has no component along binormal.

The particle velocity vector is 11.14 Radial and Transverse Components When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP. The particle velocity vector is Similarly, the particle acceleration vector is

When particle position is given in cylindrical coordinates, it is convenient to express the velocity and acceleration vectors using the unit vectors Position vector, Velocity vector, Acceleration vector,

Sample Problem 11.10 VA = 96 km/h 762 m A motorist is traveling on a curved section of highway of radius 762 m at the speed of 96 km/h. The motorist suddenly applies the brake, causing the automobile to slow down at a constant rate. knowing that after 8s the speed has been reduced to 72 km/h, determine the acceleration of the automobile immediately after the brakes have been applied.

Tangential Component of Acceleration Tangential Component of Acceleration. First the speeds are expressed in m/s VA = 96 km/h 762 m Since the automobile slows down at a constant rate, we have Normal component of acceleration. Immediately after the brakes have been applied, the speed is still 26.7 m/s, and we have

Magnitude and Direction of Acceleration. The magnitude and direction of the resultant a of the components an and at are VA = 96 km/h 762 m

Sample Problem 11.12 The rotation of the 0.9m arm about OA about O is defined by q = 0.15t2 where q is expressed in radians and t in seconds. Collar B slides along the arm in such a way that its distance from O is r = 0.9 - 0.12t2 where r is expressed in meters and t in seconds. After the arm OA has rotated through 30o, determine the total velocity of the collar, the total acceleration of the collar, the relative acceleration of the collar with respect to the arm.

Time t at which θ=30º. Substituting θ=30º =0 Time t at which θ=30º. Substituting θ=30º =0.524 rad into the expression for θ, we obtain Equation of Motion. Substituting t=1.869 s in the expressions for r,θ,and their first and second derivatives, we have a. Velocity of B We obtain the values of vr and vθwhen t = 1.869 s

b. Acceleration of B. We obtain c. Acceleration of B with Respect to Arm OA.. We note that the motion of the collar with respect to the arm is rectilinear and defined by the coordinate r. We write

PROBLEMS

Problem 11.136 At the instant shown, race car A is passing race car B with a relative velocity of 1m/s. Knowing that the speeds of both cars are constant and that the relative acceleration of car A with respect to car B is 0.25 m/s directed toward the center of curvature, determine the speed of car A the speed of car B 2

Problem 11.142 91.4 m 137.2 m

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