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 \(k\theta^2\). If its moment of inertia is \(I\) then the angular acceleration of the disc is :
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 \(k\theta^2\). If its moment of inertia is \(I\) then the angular acceleration of the disc is :
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Work-energy theorem for rotation: \(\tau\,d\theta=d(KE)\), then \(\alpha=\tau/I\).
Updated On: Nov 5, 2026
- \(\frac{k}{4I}\theta\)
- \(\frac{k}{I}\theta\)
- \(\frac{k}{2I}\theta\)
- \(\frac{2k}{I}\theta\)
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The Correct Option is D
Solution and Explanation
By the work-energy theorem, \(\tau\,d\theta=d(KE)\Rightarrow\tau=\frac{d(k\theta^2)}{d\theta}=2k\theta\).
Angular acceleration \(\alpha=\frac{\tau}{I}=\frac{2k\theta}{I}\).
Answer: (4) \(\frac{2k}{I}\theta\)
Angular acceleration \(\alpha=\frac{\tau}{I}=\frac{2k\theta}{I}\).
Answer: (4) \(\frac{2k}{I}\theta\)
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