A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its k

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6.System of Particles and Rotational Motion

hard

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 $

(JEE MAIN-2019)

Solution

$Kinetic\,energy\,KE = \frac{1}{2}I{\omega ^2} = K{\theta ^2}$

$ \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 $

Standard 11

Physics

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