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(B) Using the kinematic equation for rotational motion: $\omega^2 = \omega_0^2 - 2\alpha	heta$,where $	heta = 2\pi n$.For the first $36$ rotations,t

Vedclass · Accepted Answer

$12$

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(B) Using the kinematic equation for rotational motion: $\omega^2 = \omega_0^2 - 2\alpha	heta$,where $	heta = 2\pi n$.For the first $36$ rotations,the angular velocity reduces to half: $\left(\frac{\omega_0}{2}ight)^2 = \omega_0^2 - 2\alpha(2\pi 	imes 36)$.$\frac{\omega_0^2}{4} = \omega_0^2 - 144\pi\alpha \implies 144\pi\alpha = \frac{3}{4}\omega_0^2 \implies \alpha = \frac{3\omega_0^2}{576\pi}$.Now,let the total number of rotations to come to rest be $n'$.Using $0 = \omega_0^2 - 2\alpha(2\pi n')$,we get $n' = \frac{\omega_0^2}{4\pi\alpha}$.Substituting $\alpha$: $n' = \frac{\omega_0^2}{4\pi} 	imes \frac{576\pi}{3\omega_0^2} = \frac{576}{12} = 48$ rotations.The additional rotations required are $n' - 36 = 48 - 36 = 12$ rotations.

When a ceiling fan is switched off,its angular velocity reduces to $50\%$ while it makes $36$ rotations. How many more rotations will it make before coming to rest? (Assume uniform angular retardation)

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