Given kinetic energy as a function of angular displacement

The rotational kinetic energy is given by $K = k\theta^2.$

Use the relation between rotational kinetic energy and angular speed

For a rotating disc, $K = \frac{1}{2}I\omega^2.$

Hence, $\frac{1}{2}I\omega^2 = k\theta^2.$

So, $\omega^2 = \frac{2k}{I}\theta^2.$

Relate angular acceleration to angular velocity and angular displacement

We use $\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta}\frac{d\theta}{dt} = \omega\frac{d\omega}{d\theta}.$

Also, $\omega\frac{d\omega}{d\theta} = \frac{1}{2}\frac{d(\omega^2)}{d\theta}.$

Therefore, $\alpha = \frac{1}{2}\frac{d(\omega^2)}{d\theta}.$

Differentiate

Since $\omega^2 = \frac{2k}{I}\theta^2,$ we get $\frac{d(\omega^2)}{d\theta} = \frac{4k}{I}\theta.$

Thus, $\alpha = \frac{1}{2}\cdot \frac{4k}{I}\theta = \frac{2k}{I}\theta.$

Match with the options

$\alpha = \frac{2k}{I}\theta$

This corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Derived answer: D

They match.