A same torque is applied to a disc and a ring | vedclass

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(C) The torque $	au$ is related to the moment of inertia $I$ and angular acceleration $\alpha$ by the equation $	au = I\alpha$. Since the torque $

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the disc will rotate with greater angular frequency.

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(C) The torque $	au$ is related to the moment of inertia $I$ and angular acceleration $\alpha$ by the equation $	au = I\alpha$. Since the torque $	au$ is the same for both,the angular acceleration is given by $\alpha = \frac{	au}{I}$. The moment of inertia of a ring is $I_{	ext{ring}} = MR^2$ and for a disc is $I_{	ext{disc}} = \frac{1}{2}MR^2$. Since $I_{	ext{ring}} > I_{	ext{disc}}$,the angular acceleration of the disc will be greater than that of the ring $(\alpha_{	ext{disc}} > \alpha_{	ext{ring}})$. Consequently,the disc will rotate with a greater angular frequency (or angular velocity) compared to the ring after the same time interval.

$A$ same torque is applied to a disc and a ring of equal mass and radii. Then:

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