A stationary horizontal disc is free to rotat | vedclass

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(B) The rotational kinetic energy $E_k$ of the disc is given by $E_k = K 	heta^2$. We know that the rotational kinetic energy is also expressed as $E

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

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(B) The rotational kinetic energy $E_k$ of the disc is given by $E_k = K 	heta^2$. We know that the rotational kinetic energy is also expressed as $E_k = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. Equating the two expressions: $\frac{1}{2} I \omega^2 = K 	heta^2$. Solving for $\omega^2$: $\omega^2 = \frac{2K}{I} 	heta^2$. Differentiating both sides with respect to time $t$: $\frac{d}{dt}(\omega^2) = \frac{d}{dt}(\frac{2K}{I} 	heta^2)$. $2 \omega \frac{d\omega}{dt} = \frac{2K}{I} \cdot 2 	heta \frac{d	heta}{dt}$. Since $\alpha = \frac{d\omega}{dt}$ and $\omega = \frac{d	heta}{dt}$,we have: $2 \omega \alpha = \frac{4K}{I} 	heta \omega$. Dividing both sides by $2 \omega$ (assuming $\omega 
eq 0$): $\alpha = \frac{2K}{I} 	heta$.

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

$A$ 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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