QJEEM19Ph7612: JEE Main 2019 Physics - Class 11 JEEM19Ph7612 Solution: A stationary horizontal disc is free to rotate about its axis.…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

Kinetic energy KE =

\frac{1}{2} l ω^{2} = k θ^{2}

\Rightarrow ω^{2} = \frac{2 k θ^{2}}{l} \Rightarrow ω = \sqrt{\frac{2 k}{l}} θ

.... (A)

Differentiate (A) wrt time

\to

\frac{d ω}{d t} = α = \sqrt{\frac{2 k}{l}} (\frac{d θ}{d t})

\Rightarrow α = \sqrt{\frac{2 k}{l}} . \sqrt{\frac{2 k}{l}} θ {b y (1)}

\Rightarrow α = \frac{2 k}{l} θ

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