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 $\theta ,\text{where}\theta$ 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. $\frac{2k}{I}\theta$
• C. $\frac{k}{I}\theta$
• D. $\frac{k}{4I}\theta$

Answer 1

The correct option is B $\frac{2k}{I}\theta$

Given kinetic energy is $E=k{\theta }^2$
Work done by torque is responsible for change in kinetic energy. Hence,
$\therefore \tau =\frac{dE}{d\theta }$
$\Rightarrow I\alpha =2k\theta$ (Where $\tau =I\alpha )$
$\Rightarrow \alpha =\frac{2k\theta }{I}$
Hence option $\left(D\right)$ is correct.