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 $k{\theta }^2$ (where $k$ is a constant). 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

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

Given, kinetic energy $=k{\theta }^2$
Kinetic energy of a rotating body about its axis $=\frac{1}{2}I{\omega }^2$
where, $I$ is moment of inertia and $\omega$ is angular velocity
$\therefore \frac{1}{2}I{\omega }^2=k{\theta }^2$
${\omega }^2=\frac{2k{\theta }^2}{I}$
$\Rightarrow \omega =\sqrt{\frac{2k}{I}}\theta$ ... (i)
Differentiating the above equation w.r.t. time on both sides, we get
$\frac{d\omega }{dt}=\sqrt{\frac{2k}{I}}.\frac{d\theta }{dt}=\sqrt{\frac{2k}{I}}.\omega$ $[\because \omega =\frac{d\theta }{dt}]$
$\therefore$ Angular acceleration,
$\alpha =\frac{d\omega }{dt}=\sqrt{\frac{2k}{I}}.\omega =\sqrt{\frac{2k}{I}}.\sqrt{\frac{2k}{I}}\theta$ [using Eq. (i)]
or $\alpha =\frac{2k}{I}\theta$