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(D) Using the third equation of rotational motion: $\omega^2 = \omega_0^2 - 2\alpha	heta$. Initially,the angular velocity becomes $\omega = \frac{\om

Vedclass · Accepted Answer

$12$

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(D) Using the third equation of rotational motion: $\omega^2 = \omega_0^2 - 2\alpha	heta$. Initially,the angular velocity becomes $\omega = \frac{\omega_0}{2}$ after $	heta_1 = 36 	imes 2\pi$ radians. Substituting these values: $(\frac{\omega_0}{2})^2 = \omega_0^2 - 2\alpha(36 	imes 2\pi)$. $\frac{\omega_0^2}{4} = \omega_0^2 - 144\pi\alpha$,which gives $144\pi\alpha = \frac{3\omega_0^2}{4}$,so $\alpha = \frac{3\omega_0^2}{576\pi} = \frac{\omega_0^2}{192\pi}$. Now,for the motion from $\omega = \frac{\omega_0}{2}$ to $\omega = 0$,let the additional rotations be $n$. The angle covered is $	heta_2 = n 	imes 2\pi$. Using $0^2 = (\frac{\omega_0}{2})^2 - 2\alpha(n 	imes 2\pi)$. $\frac{\omega_0^2}{4} = 2(\frac{\omega_0^2}{192\pi})(n 	imes 2\pi)$. $\frac{1}{4} = \frac{4n\pi}{192\pi} = \frac{n}{48}$. $n = \frac{48}{4} = 12$. Thus,the fan will make $12$ more rotations before coming to rest.

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?

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