At time = 0.00 seconds an amusement park carousel with a diameter of 8.00m takes

Question

At time = 0.00 seconds an amusement park carousel with a diameter of 8.00m takes 12.0s to complete a revolution before it begins to slow down. At the end of the ride, it slows down smoothly, taking 2.70 rev to come to a stop.

A) What is the angular speed of the carousel at time = 0.00 seconds in rad/s?

B) What is the linear speed at time = 0.00 seconds of a horse located a distance of 3.00 m from the center?

C) what is the magnitufe of the angular acceleration during the time the carousel is slowing down?

Explanation / Answer

period is T = 2*pi*r/v

12 = 2*3.142*4/v

v = 2.09 m/sec

angular speed is w = v/r = 2.09/4 =0.5225 rad/s

B) v = r*w = 3*0.5225 = 1.5675 m/sec

C) angular accelaration alpha

angular displacement is theta = 2.7*2*3.142 = 16.96 rad

using wf^2-wi^2 = 2*alpha*theta

0^2-0.5225^2 = 2*alpha*16.96

alpha = -0.00804 rad/s^2

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