JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Hard
- MCQ
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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Answer & Solution
Correct Answer
(D) \(\frac {2K}{I}\,\theta \)
Step-by-step Solution
Detailed explanation
\(Kinetic\,energy\,KE = \frac{1}{2}I{\omega ^2} = K{\theta ^2}\)
\( \Rightarrow {\omega ^2} = \frac{{2k{\theta ^2}}}{I} \Rightarrow \omega = \sqrt {\frac{{2k}}{I}} \theta \,\,\,\,\,\,\,\,\,\,\,...\left( A \right)\)
\(Differentiate\,\left( A \right)\,wrt\,time\, \to \)
\(\frac{{d\omega }}{{dt}} = \alpha = \sqrt {\frac{{2k}}{I}} \left( {\frac{{d\theta }}{{dt}}} \right)\)
\( \Rightarrow \alpha = \sqrt {\frac{{2k}}{I}} .\sqrt {\frac{{2k}}{I}} \,\theta \left\{ {by\,\left( 1 \right)} \right\}\)
\( \Rightarrow \alpha = \frac{{2k}}{I}\theta \)
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