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Question

(D) Let the initial angular velocity be $\omega_0$ and the final angular velocity be $\omega = \frac{\omega_0}{2}$. The angular displacement for $36$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Let the initial angular velocity be $\omega_0$ and the final angular velocity be $\omega = \frac{\omega_0}{2}$. The angular displacement for $36$ rotations is $	heta_1 = 36 	imes 2\pi = 72\pi 	ext{ rad}$.Using the equation of motion $\omega^2 = \omega_0^2 + 2\alpha	heta$:$(\frac{\omega_0}{2})^2 = \omega_0^2 + 2\alpha(72\pi)$$\frac{\omega_0^2}{4} - \omega_0^2 = 144\pi\alpha$$-\frac{3\omega_0^2}{4} = 144\pi\alpha \implies \alpha = -\frac{3\omega_0^2}{576\pi} = -\frac{\omega_0^2}{192\pi}$.Now,for the second part,the fan starts with $\omega_i = \frac{\omega_0}{2}$ and comes to rest $(\omega_f = 0)$:$0^2 = (\frac{\omega_0}{2})^2 + 2\alpha	heta_2$$0 = \frac{\omega_0^2}{4} + 2(-\frac{\omega_0^2}{192\pi})	heta_2$$\frac{\omega_0^2}{4} = \frac{\omega_0^2}{96\pi}	heta_2$$	heta_2 = \frac{96\pi}{4} = 24\pi 	ext{ rad}$.Number of rotations = $\frac{	heta_2}{2\pi} = \frac{24\pi}{2\pi} = 12$ rotations.

When a ceiling fan is switched off,its angular velocity falls to half while it makes $36$ rotations. How many more rotations will it make before coming to rest? (Assume uniform angular retardation)

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