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 $\mathrm{k} \theta^{2}$. If its moment of inertia is I then the angular acceleration of the disc is:

(D) $\frac{2 \mathrm{k}}{\mathrm{I}} \theta$

(A) $\frac{\mathrm{k}}{\mathrm{I}} \theta$

(B) $\frac{\mathrm{k}}{2 \mathrm{I}} \theta$

(C) $\frac{\mathrm{k}}{4 \mathrm{I}} \theta$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Rotational Mechanics

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

θ

, where

θ

is the angle by which it has rotated, is given as

k θ^{2}

. If its moment of inertia is I then the angular acceleration of the disc is:

Answer: (D)

The rotational kinetic energy is

\frac{1}{2} I ω^{2}

, and we are told this equals

k θ^{2}

. So

ω^{2} = \frac{2 k}{I} θ^{2}

, which gives

ω = \frac{2 k}{I} θ

. To get the angular acceleration, differentiate

ω

with respect to time:

α = \frac{d ω}{d t} = \frac{2 k}{I} \frac{d θ}{d t} = \frac{2 k}{I} ω

. Now put

ω = \frac{2 k}{I} θ

back in, and the two square roots multiply to give

α = \frac{2 k}{I} θ

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option