Question

A large carousel at an amusement park is brought to rest with a friction brake. If the ride operator applies a constant force to the brake and it takes 8 complete revolutions for the ride to come to rest, how many revolutions did it take to slow to 1/2 of the original angular speed?

Answer 1

here,

initially

the angle covered before stopping , theta1 = 8 * (2*pi)

let the initial speed be w0 and angular accelration be alpha

using third equation of motion

w^2 - w0^2 = 2 * alpha * theta

alpha = - w0^2 /( 2 * 8 * (2pi))

and

when w = w0/2

let the angle covered be theta'

using third equation of motion

w^2 - w0^2 = 2 * alpha * theta'

(w0/2)^2 - w0^2 = 2 * (- w0^2 /( 2 * 8 * (2pi)) ) * theta'

(1/2)^2 - 1^2 = (- 1 /( 8 * (2pi)) ) * theta'

theta' = 6 * (2pi)

so the number of revolutions are 6 rev