A stationary horizontal disc is free to rotat | vedclass

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Question

$Kinetic\,energy\,KE = \frac{1}{2}I{\omega ^2} = K{	heta ^2}$

$ \Rightarrow {\omega ^2} = \frac{{2k{	heta ^2}}}{I} \Rightarrow \omega  = \sq

Vedclass · Accepted Answer

$\frac {2K}{I}\,	heta $

Vedclass · Answer

$Kinetic\,energy\,KE = \frac{1}{2}I{\omega ^2} = K{	heta ^2}$

$ \Rightarrow {\omega ^2} = \frac{{2k{	heta ^2}}}{I} \Rightarrow \omega  = \sqrt {\frac{{2k}}{I}} 	heta \,\,\,\,\,\,\,\,\,\,\,...\left( A ight)$

$Differentiate\,\left( A ight)\,wrt\,time\, 	o $

$\frac{{d\omega }}{{dt}} = \alpha  = \sqrt {\frac{{2k}}{I}} \left( {\frac{{d	heta }}{{dt}}} ight)$

$ \Rightarrow \alpha  = \sqrt {\frac{{2k}}{I}} .\sqrt {\frac{{2k}}{I}} \,	heta \left\{ {by\,\left( 1 ight)} ight\}$

$ \Rightarrow \alpha  = \frac{{2k}}{I}	heta $

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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