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(D) The retarding torque is constant,so the angular retardation $\alpha$ is also constant.Using the rotational kinematic equation $\omega^2 = \omega_0

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$\frac{n}{3}$

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(D) The retarding torque is constant,so the angular retardation $\alpha$ is also constant.Using the rotational kinematic equation $\omega^2 = \omega_0^2 - 2\alpha	heta$:For the first $n$ rotations,the angular displacement is $	heta_1 = 2\pi n$. The final angular velocity is $\frac{\omega_0}{2}$.$\left(\frac{\omega_0}{2}ight)^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \frac{\omega_0^2}{4} = \omega_0^2 - 4\pi n\alpha \implies 4\pi n\alpha = \frac{3\omega_0^2}{4} \implies \alpha = \frac{3\omega_0^2}{16\pi n} \quad ...(i)$For the next $	heta_2$ rotations (where $	heta_2 = 2\pi n'$),the disc comes to rest $(\omega = 0)$:$0^2 = \left(\frac{\omega_0}{2}ight)^2 - 2\alpha(2\pi n') \implies 0 = \frac{\omega_0^2}{4} - 4\pi n'\alpha \implies 4\pi n'\alpha = \frac{\omega_0^2}{4} \quad ...(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{4\pi n'\alpha}{4\pi n\alpha} = \frac{\omega_0^2 / 4}{3\omega_0^2 / 4} \implies \frac{n'}{n} = \frac{1}{3} \implies n' = \frac{n}{3}$.Thus,the disc will make $\frac{n}{3}$ more rotations before coming to rest.

$A$ 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 $\frac{\omega_0}{2}$ after $n$ rotations. How many more rotations will it make before coming to rest?

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