You are sitting on the edge of a merry-go-round of radius 1.50 m. When you are a

Question

You are sitting on the edge of a merry-go-round of radius 1.50 m. When you are at the eastern-most point of your motion, you have angular velocity of 36 rpm and you start to drag your foot on the ground. By the time you have made half a revolution, you are moving with angular velocity of 19 rpm. Assuming constant angular acceleration, what are the magnitudes and direction of your linear acceleration when it is a) at the western-most point of its trajectory' after making half of a revolution? b) When and where does the merry-go-round stop?

Explanation / Answer

1 rpm = 2pi/60 rad/s = 0.105

linear velocity at eastern most point v1 = R*w1 = 1.5*36*pi/30 = 1.8*pi m/s

linear velocity at western most point v2 = R*w2 = 1.5*19*pi/30 = 0.95pi m/s

distance rotated = 1/2 rev = 2*pi*R/2 = pi*R

v2^2 - v1^2 = 2*a*s

(0.95*pi)^2 - (1.8*pi)^2 = 2*a*pi*R

pi*(0.95^2-1.8^2) = 2*a*1.5

a = -0.78*pi m/s^2

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part b

vf = 0

vf = vi + a*t

0 = 1.8*pi - (0.8pi*t)

t = 2.25 s <<<<=======ANSWER

wf^2 - w1^2 = 2*alpha*theta

alpha = a/R

0 - (36*pi/30)^2 = 2*0.78pi/1.5*theta

theta = 4.35 rad = 0.7 rev

after making 0.7 rev it stops <<<=======ANSWER

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