A stationary horizontal disc is free to rotat | vedclass

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(D) The rotational kinetic energy is given by $KE = \frac{1}{2} I \omega^2 = K 	heta^2$. From this,we have $\omega^2 = \frac{2 K 	heta^2}{I}$,which

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

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(D) The rotational kinetic energy is given by $KE = \frac{1}{2} I \omega^2 = K 	heta^2$. From this,we have $\omega^2 = \frac{2 K 	heta^2}{I}$,which implies $\omega = \sqrt{\frac{2 K}{I}} 	heta$. Differentiating both sides with respect to time $t$,we get $\frac{d\omega}{dt} = \alpha = \sqrt{\frac{2 K}{I}} \left(\frac{d	heta}{dt}ight)$. Since $\frac{d	heta}{dt} = \omega$,we substitute $\omega = \sqrt{\frac{2 K}{I}} 	heta$ into the expression: $\alpha = \sqrt{\frac{2 K}{I}} \cdot \left(\sqrt{\frac{2 K}{I}} 	hetaight) = \frac{2 K}{I} 	heta$.

$A$ 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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