A ring of mass M and radius R sliding with a | vedclass

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Question

(D) When the ring enters the rough surface,kinetic friction acts in the direction opposite to the velocity $v_0$. The force of friction is $f = \mu Mg

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) When the ring enters the rough surface,kinetic friction acts in the direction opposite to the velocity $v_0$. The force of friction is $f = \mu Mg$. This force causes a retardation in linear motion: $a = -\frac{f}{M} = -\mu g$. Also,this friction force provides a torque about the center of mass: $	au = fR = \mu MgR$. The angular acceleration is $\alpha = \frac{	au}{I} = \frac{\mu MgR}{MR^2} = \frac{\mu g}{R}$. Let $t$ be the time when pure rolling starts. At this time,the velocity of the center of mass is $v = v_0 - at = v_0 - \mu gt$ and the angular velocity is $\omega = \alpha t = \frac{\mu gt}{R}$. For pure rolling,$v = \omega R$,so $v_0 - \mu gt = (\frac{\mu gt}{R})R = \mu gt$. This gives $2\mu gt = v_0$,or $t = \frac{v_0}{2\mu g}$. The velocity at this time is $v = v_0 - \mu g(\frac{v_0}{2\mu g}) = v_0 - \frac{v_0}{2} = \frac{v_0}{2}$. Pure rolling starts when the point of contact becomes stationary relative to the surface. Thus,all statements are correct.

$A$ ring of mass $M$ and radius $R$ sliding with a velocity $v_0$ suddenly enters a rough surface where the coefficient of friction is $\mu$,as shown in the figure. Choose the correct statement$(s)$.

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