In the figure shown,the plank is being pulled | vedclass

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(D) Let $v_{cm}$ be the velocity of the centre of mass of the cylinder and $\omega$ be its angular velocity.At the contact point between the cylinder

Vedclass · Accepted Answer

the speed of the centre of mass of the cylinder is $v/2$.

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(D) Let $v_{cm}$ be the velocity of the centre of mass of the cylinder and $\omega$ be its angular velocity.At the contact point between the cylinder and the plank (bottom point),the velocity of the cylinder must match the velocity of the plank because there is no slipping. Thus,$v_{cm} - R\omega = v$ (assuming $v_{cm}$ is to the right and $\omega$ is clockwise).At the contact point between the cylinder and the top string (top point),the velocity of the cylinder must match the velocity of the string. Since the string is fixed to the wall,its velocity is $0$. Thus,$v_{cm} + R\omega = 0$.Adding these two equations: $(v_{cm} - R\omega) + (v_{cm} + R\omega) = v + 0$,which gives $2v_{cm} = v$,so $v_{cm} = v/2$.Subtracting the equations: $(v_{cm} - R\omega) - (v_{cm} + R\omega) = v - 0$,which gives $-2R\omega = v$,so $\omega = -v/(2R)$ (the negative sign indicates counter-clockwise rotation).Therefore,the speed of the centre of mass of the cylinder is $v/2$.

In the figure shown,the plank is being pulled to the right with a constant speed $v$. If the cylinder does not slip,then:

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