Question

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $\theta$, where $\theta$ is the angle by which it has rotated, is given as $k{\theta }^2$. If its moment of inertia is $I$ then the angular acceleration of the disc is :

• A. $\frac{k}{2I}\theta$
• B. $\frac{k}{I}\theta$
• C. $\frac{k}{4I}\theta$
• D. $\frac{2k}{I}\theta$

Answer 1

Answer: (D)

Kinetic energy KE = $\frac{1}{2}I{\omega }^2=k{\theta }^2$
$\Rightarrow {\omega }^2=\frac{2k{\theta }^2}{I}\Rightarrow \omega =\sqrt{\frac{2k}{I}\theta }...(1)$
Differentiate (1) wrt time $\to$
$\frac{d\omega }{dt}=\propto =$\sqrt{\frac{2k}{I}} \bigg( \frac{d\theta}{dt}\bigg)$<br/>$\Rightarrow {\,} {\,} \propto$=$\sqrt{\frac{2k}{I}} . \sqrt{\frac{2k}{I}} \theta \{ by (1) \}$<br/>$\Rightarrow {\,} {\,} \propto$=$\frac{2 k }{I} \theta $