A wheel rotating at an angular velocity of 12 | vedclass

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(C) Initial angular velocity $\omega_0 = 1200 \ rpm = \frac{1200 	imes 2\pi}{60} \ rad \ s^{-1} = 40\pi \ rad \ s^{-1}$.Final angular velocity $\omeg

Vedclass · Accepted Answer

$314$

Vedclass · Answer

(C) Initial angular velocity $\omega_0 = 1200 \ rpm = \frac{1200 	imes 2\pi}{60} \ rad \ s^{-1} = 40\pi \ rad \ s^{-1}$.Final angular velocity $\omega = 0 \ rad \ s^{-1}$.Angular acceleration $\alpha = -4 \ rad \ s^{-2}$.Using the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha	heta$:$0^2 = (40\pi)^2 + 2(-4)	heta$$8	heta = 1600\pi^2$$	heta = 200\pi^2 \ rad$.Since one revolution corresponds to $2\pi \ rad$,the number of revolutions $N$ is given by:$N = \frac{	heta}{2\pi} = \frac{200\pi^2}{2\pi} = 100\pi$.Using $\pi \approx 3.14$,$N = 100 	imes 3.14 = 314$ revolutions.

$A$ wheel rotating at an angular velocity of $1200 \ rpm$ is slowed down at a constant angular acceleration of $4 \ rad \ s^{-2}$. How many revolutions will the wheel make before coming to rest?

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