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 will be:

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

Answer 1

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

Given:
$\frac{1}{2}I{\omega }^2=k{\theta }^2$
$\Rightarrow {\omega }^2=\frac{2k{\theta }^2}{I}$
Differentiate with respect to $\theta$
$\Rightarrow 2\omega \frac{d\omega }{d\theta }=\frac{4k\theta }{I}$
$\Rightarrow 2\alpha =\frac{4k\theta }{I}\Rightarrow \alpha =\frac{2k\theta }{I}$