A merry-go-round makes 120 revolutions per mi | vedclass

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(C) Given: Initial frequency $n_1 = 120 \ rev/min = 2 \ rev/s$. Initial angular velocity $\omega_0 = 2\pi n_1 = 2\pi(2) = 4\pi \ rad/s$. Final angular

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$2\pi \ s$,$2\pi$ revolutions

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(C) Given: Initial frequency $n_1 = 120 \ rev/min = 2 \ rev/s$. Initial angular velocity $\omega_0 = 2\pi n_1 = 2\pi(2) = 4\pi \ rad/s$. Final angular velocity $\omega = 0$. Angular retardation $\alpha = -2 \ rad \ s^{-2}$.Using the equation of motion $\omega = \omega_0 + \alpha t$:$0 = 4\pi + (-2)t \implies 2t = 4\pi \implies t = 2\pi \ s$.Now,to find the number of revolutions,we first find the total angular displacement $	heta$:$	heta = \omega_0 t + \frac{1}{2} \alpha t^2 = (4\pi)(2\pi) + \frac{1}{2}(-2)(2\pi)^2 = 8\pi^2 - 4\pi^2 = 4\pi^2 \ rad$.The number of revolutions $N = \frac{	heta}{2\pi} = \frac{4\pi^2}{2\pi} = 2\pi$ revolutions.

$A$ merry-go-round makes $120$ revolutions per minute. $A$ child sitting on it starts crying,so the merry-go-round is slowed down with a retardation of $2 \ rad \ s^{-2}$. How long will it take for the merry-go-round to come to a stop? How many revolutions will it make before coming to a stop?

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