Question

A stationary horizontal disc is free to rotate about its axis. When 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:
$\begin{aligned} & a)\dfrac{k}{2I}\theta \\\ & b)\dfrac{k}{I}\theta \\\ & c)\dfrac{k}{4I}\theta \\\ & d)\dfrac{2k}{I}\theta \\\ \end{aligned}$

Answer 1

It is given that the above disc is free to rotate about its axis when torque is applied to it. It is also given to us the kinetic energy of the disc varies with $\theta$governed by the function $k{{\theta }^{2}}$. Hence we will first write the expression for work done by the torque applied in terms of moment of inertia and the angular acceleration and accordingly obtain the value of angular acceleration.
Formula used:
$\tau =I\alpha$
$W=\tau \theta$

Complete answer:
In the above figure we see that a disc is free to rotate about its axis of rotation passing through the centre. On application of torque ($\tau$) (tangential force)the disc will start rotating. If the angular displacement covered by the disc on application of torque is $\theta$, then the work done (W)by this torque is equal to,
$W=\tau \theta ...(1)$
If the body has moment of inertia I and moves with angular acceleration $\alpha$on application of torque, we can write that $\tau =I\alpha$. In the question it is given to us that when torque is applied on the disc its kinetic energy is given as $k{{\theta }^{2}}...(2)$. But this energy is equal to the work done by the torque. Hence equating equation 1 and 2 we get,
$\begin{aligned} & W=\tau \theta =k{{\theta }^{2}} \\\ & \Rightarrow \tau \theta =k{{\theta }^{2}}\text{, }\because \tau =I\alpha \\\ & \Rightarrow I\alpha \theta =k{{\theta }^{2}} \\\ & \Rightarrow \alpha =\dfrac{k\theta }{I} \\\ \end{aligned}$

So, the correct answer is “Option B”.

Note:
It is to be noted that the body is free to move about its axis. Therefore we can say that there is no energy lost to overcome the barrier in order to make the disc rotate. Hence the work done by the torque is entirely converted to kinetic energy of the disc.