JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(\frac {K}{I}\,\theta \)
- B \(\frac {K}{2I}\,\theta \)
- C \(\frac {K}{4I}\,\theta \)
- D \(\frac {2K}{I}\,\theta \)
Answer & Solution
Correct Answer
(D) \(\frac {2K}{I}\,\theta \)
Step-by-step Solution
Detailed explanation
\(Kinetic\,energy\,KE = \frac{1}{2}I{\omega ^2} = K{\theta ^2}\) \( \Rightarrow {\omega ^2} = \frac{{2k{\theta ^2}}}{I} \Rightarrow \omega = \sqrt {\frac{{2k}}{I}} \theta \,\,\,\,\,\,\,\,\,\,\,...\left( A \right)\) \(Differentiate\,\left( A \right)\,wrt\,time\, \to \)…
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