Physics 170 Week 8, Lecture 3 http://www.phas.ubc.ca/∼gordonws/170 Physics 170 203 Week 8, Lecture 3 1 Textbook Chap

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Physics 170 Week 8, Lecture 3 http://www.phas.ubc.ca/∼gordonws/170

Physics 170 203 Week 8, Lecture 3

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Textbook Chapter 12: Section 12.9-10

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Learning Goals: • After today’s lecture, students will be able to relate the positions, velocities and accelerations of two particles undergoing dependent motion – primarily problems with pulleys. • Students will be able to understand translating frames of reference and use translating frames of reference to analyze relative motion.

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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.

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Dependent motion cont’d

In this example, 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 `tot , the position coordinates are related mathematically by the equation sA (t) + `CD + sB (t) = `tot Here, `CD is the length of cord passing over the arc CD of the pulley. Here, we are allowing for the blocks to move by allowing sA and sB to depend on time. `tot and `CD do not depend on time.

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Dependent motion cont’d

The velocities of the block can be related by differentiating the position equation, taking into account d d that dt `tot = 0 and dt `CD = 0. s˙ A (t) + s˙ B (t) = 0 which tells us that vA (t) = −vB (t) . Acceleration can be found by differentiating the velocity equation, aA (t) = −aB (t) .

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Example: Determine the speed of block A if block B has an upward speed of 6 ft/s.

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Solution: We begin by defining the positionss of the components that can move from fixed, reference, “datum” positions. There are the variables sA (t) and sB (t) depicted in the figure.

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Solution cont’d: We then write an equation for the total length of the cord sA (t) + 3sB (t) = constant Note that, in this equation, we have not worried much about the total length or other contributions to the length that cannot change with time. Physics 170 203 Week 8, Lecture 3

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Solution cont’d: Now, we take the derivative of the equation for the length of the cord by time to get sA + 3sB = constant → s˙ A + 3s˙ B = 0 vA (t) = −3vB (t)

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Solution cont’d: We have derived the following formula which relates the velocities of components of the system, vA = −3vB . We are told that vB = −6 f t/s Then vA = −3vB = (−3)(−6 f t/s) and finally vB = 18 f t/s

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Dependent Motion Proceedure: These proceedures can be used to relate the dependent motion of particles moving along rectilinear paths (only the magnitudes of velocities and accelerations change, not their line of direction). • Define position coordinates from fixed datum lines, along the path of each particle which can move. • Relate the position coordinates of the particles to the cord length. Segments of cord that do not change in length during the motion may be left out. There should be one such relation for every cord in the system. • Differentiate the equations for positions to get equations for velocities and accelerations.

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Example: Determine the speed with which the block B rises if the end of the cord at A is pulled down with a speed of 2m/s.

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Solution: We begin by identifying the position of all components of the system which will move, the hand at A by sA (t), the pulley at C by sC (t) and the pulley at E by sB (t).

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Solution cont’d: We then write an equation for the length of each cord, sA (t) − sC (t) + sB (t) − sC (t) + sB (t) = constant sC (t) + sB (t) = constant

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Solution cont’d: We take derivatives of the cord equations, d d (sA − sC + sB − sC + sB ) = 0 , (sC + sB ) = 0 dt dt to get

vA + 2vB − 2vC = 0 and

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vB + vC = 0

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Solution cont’d: Solve the second of the two equations for velocities, vA + 2vB − 2vC = 0 and vB + vC = 0 , to eliminate vC : vC = −vB . Plug this into the other equation and get vB = − 14 vA = −0.5 m/s so, finally, vA = −0.5m/s .

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Relative position, velocity, acceleration The position of particle B with respect to particle A is ~rBA (t) = ~rB (t) − ~rA (t) The velocity of particle B with respect to particle A is ~vBA (t) = ~vB (t) − ~vA (t) =

d d ~rB (t) − ~rA (t) dt dt

The acceleration of particle B with respect to particle A is d d2 d d2 ~aBA (t) = ~aB (t) − ~aA (t) = ~vB (t) − ~vA (t) = 2 ~rB (t) − t ~rA (t) dt dt dt dt Also, change in position (velocity) is the integral of velocity (acceleration) Z t2 Z t2 ~rBA (t2 ) − ~rBA (t1 ) = dt ~vBA , ~vBA (t2 ) − ~vBA (t1 ) = dt ~aBA t1

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Example: Two planes, A and B, are flying at the same altitude. If their velocities are vA = 600 km/h and vB = 500 km/h such that the angle between their straight-line courses is θ = 75 degrees, determine the velocity of plane B with respect to plane A.

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Solution: Assume planes are in xy-plane. Then, orienting the coordinate axes such that plane A is traveling in the negative x-direction, we find h i vA = (600 km/h) cos(75) ˆi − sin(75) ˆj , vB = −(500 km/h) ˆi

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Solution cont’d: To find the velocity of B relative to A, we form h i ~vBA = ~vB −~vA = −(500 km/h) ˆi−(600 km/h) cos(75) ˆi − sin(75) ˆj Finally ~vBA = (−655 km/h) ˆi + (−580 km/h) ˆj

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Example: At the instant shown, cars A and B are traveling at speeds of 30 m/h and 20 mi/h, respectively. If A is increasing its speed at 400 mi/h2 whereas the speed of B is decreasing at 800 mi/h2 , determine the velocity and acceleration of B with respect to A.

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Solution: We assume that the ground is the xy-plane and that car A is traveling in the negative x-direction. In its case, its velocity is ~vA = −(30 m/h)ˆi and its acceleration is ~aA = −(400 mi/h2 ) ˆi .

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Solution cont’d: Car B is traveling on a curved trajectory. Its tangential and normal unit vectors are u ˆt = − sin 30 ˆi + cos 30 ˆj , u ˆn = cos 30 ˆi + sin 30ˆj The velocity is ~vB = vB u ˆt , ~aB = v˙ B or

2 vB u ˆt + u ˆn ρB

h i ~vB = (20 mi/h) − sin 30 ˆi + cos 30 ˆj h i 2 ~aB = −(800 mi/hr ) − sin 30 ˆi + cos 30 ˆj i (20 mi/hr)2 h + cos 30 ˆi + sin 30ˆj (0.3 mi)

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Solution cont’d: Given that ~vA = −(30 m/h) ˆi , ~aA = −(400 mi/h2 ) ˆi h i ~vB = (20 mi/h) − sin 30 ˆi + cos 30 ˆj h i 2 ~aB = −(800 mi/hr ) − sin 30 ˆi + cos 30 ˆj i (20 mi/hr)2 h cos 30 ˆi + sin 30ˆj + (0.3 mi) The relative velocity is ~vBA = ~vB −~vA = [(30mi/h) − (20 mi/h) sin 30] ˆi+(20 mi/h) cos 30 ˆj ~vBA = (20 mi/h) ˆi + (17.3 mi/h) ˆj .

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Solution cont’d: The relative acceleration is ~aBA

·

h i = ~aB − ~aA = −(800 mi/hr ) − sin 30 ˆi + cos 30 ˆj 2

i (20 mi/hr)2 h + cos 30 ˆi + sin 30ˆj + (400 mi/h2 ) ˆi (0.3 mi) 2

(20 mi/hr) = (800 mi/hr ) sin 30 + cos 30 + (400 mi/h2 (0.3 mi) ¸ · 2 (20 mi/hr) 2 sin 30 ˆj + −(800 mi/hr ) cos 30 + (0.3 mi) 2

¸ ˆi

~a = (1950 mi/h2 ) ˆi + (−26.2 mi/h2 ) ˆj .

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For the next lecture, please read Textbook Chapter 13: Section 13.1

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