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(A) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R^2$.Using the rotational analogue of Newton's second law,the torque

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$\frac{M R \omega}{2 t}$

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(A) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R^2$.Using the rotational analogue of Newton's second law,the torque $	au$ is given by $	au = I \alpha$,where $\alpha$ is the angular acceleration.The torque applied by the tangential force $F$ is $	au = F 	imes R$.Equating the two expressions for torque: $F R = \frac{1}{2} M R^2 \alpha$.Solving for angular acceleration: $\alpha = \frac{2 F}{M R}$.Since the disc comes to rest from an initial angular speed $\omega$ in time $t$,the magnitude of angular deceleration is $\alpha = \frac{\omega}{t}$.Substituting $\alpha$ into the equation: $\frac{2 F}{M R} = \frac{\omega}{t}$.Solving for $F$: $F = \frac{M R \omega}{2 t}$.

$A$ disc of mass $M \ kg$ and radius $R \ m$ is rotating at an angular speed of $\omega \ rad/s$ when the motor is switched off. Neglecting the friction at the axle,the force that must be applied tangentially to the wheel to bring it to rest in time $t$ is ...............

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