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(D) Using the rotational kinematic equation: $\omega^2 = \omega_0^2 - 2\alpha	heta$,where $	heta$ is the angular displacement.For the first $n$ rota

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$\frac{n}{3}$

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(D) Using the rotational kinematic equation: $\omega^2 = \omega_0^2 - 2\alpha	heta$,where $	heta$ is the angular displacement.For the first $n$ rotations,the angular displacement is $	heta_1 = 2\pi n$. The final angular velocity is $\omega_1 = \omega_0/2$.So,$(\omega_0/2)^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \omega_0^2/4 = \omega_0^2 - 4\pi n\alpha \implies 4\pi n\alpha = 3\omega_0^2/4 \implies 2\alpha = \frac{3\omega_0^2}{8\pi n}$.Let the additional rotations be $n'$. The angular displacement is $	heta_2 = 2\pi n'$. The final angular velocity is $0$.Using $0^2 = (\omega_0/2)^2 - 2\alpha(2\pi n')$,we get $0 = \omega_0^2/4 - 2\alpha(2\pi n')$.Substituting $2\alpha$: $0 = \omega_0^2/4 - \left(\frac{3\omega_0^2}{8\pi n}ight)(2\pi n') \implies \frac{\omega_0^2}{4} = \frac{3\omega_0^2 n'}{4n} \implies 1 = \frac{3n'}{n} \implies n' = \frac{n}{3}$.

$A$ disc is rotating with an angular velocity $\omega_0$. $A$ constant retarding torque is applied on it to stop the disc. The angular velocity becomes $(\omega_0/2)$ after $n$ rotations. How many more rotations will it make before coming to rest?

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