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(C) The net torque $	au$ acting on the disc about its centre is the sum of the torques produced by the two forces.Since both forces are applied tange

Vedclass · Accepted Answer

$\frac{4 F}{M R}$

Vedclass · Answer

(C) The net torque $	au$ acting on the disc about its centre is the sum of the torques produced by the two forces.Since both forces are applied tangentially at a distance $R$ from the centre,each force produces a torque of magnitude $	au_i = F 	imes R$.Both forces create a torque in the same rotational direction (counter-clockwise),so the total torque is $	au_{net} = F R + F R = 2 F R$.The moment of inertia of a uniform disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.Using the rotational analogue of Newton's second law,$	au_{net} = I \alpha$,where $\alpha$ is the angular acceleration:$2 F R = (\frac{1}{2} M R^2) \alpha$Solving for $\alpha$:$\alpha = \frac{2 F R}{\frac{1}{2} M R^2} = \frac{4 F}{M R}$.

Two equal and opposite forces are applied tangentially to a uniform disc of mass $M$ and radius $R$ as shown in the figure. If the disc is pivoted at its centre and free to rotate in its plane,the angular acceleration of the disc is ............

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