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 $\rightarrow$
$\frac{d\omega}{dt} = \propto = $\sqrt{\frac{2k}{I}} \bigg( \frac{d\theta}{dt}\bigg)$
$\Rightarrow \, \, \propto$ = $\sqrt{\frac{2k}{I}} . \sqrt{\frac{2k}{I}} \theta \{ by (1) \}$
$\Rightarrow \, \, \propto$ = $\frac{2 k }{I} \theta $