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Question

(C) Let the initial angular velocity be $\omega_0$ and uniform angular retardation be $\alpha$. Using the equation of rotational motion $\omega^2 = \o

Vedclass · Accepted Answer

$\frac{16 n}{15}$

Vedclass · Answer

(C) Let the initial angular velocity be $\omega_0$ and uniform angular retardation be $\alpha$. Using the equation of rotational motion $\omega^2 = \omega_0^2 - 2\alpha	heta$,where $	heta = 2\pi n$. At time $t$,$\omega = \frac{\omega_0}{4}$. So,$(\frac{\omega_0}{4})^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \frac{\omega_0^2}{16} = \omega_0^2 - 4\pi n\alpha$. $4\pi n\alpha = \omega_0^2(1 - \frac{1}{16}) = \frac{15\omega_0^2}{16}$. Thus,$2\alpha = \frac{15\omega_0^2}{32\pi n}$. Now,for the fan to come to rest,final angular velocity $\omega_f = 0$. Let $n'$ be the total revolutions from switch off to rest. $0^2 = \omega_0^2 - 2\alpha(2\pi n')$. $2\alpha(2\pi n') = \omega_0^2$. Substituting $2\alpha$: $(\frac{15\omega_0^2}{32\pi n})(2\pi n') = \omega_0^2$. $\frac{15n'}{16n} = 1 \implies n' = \frac{16n}{15}$.

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