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(D) Let $F$ be the friction force acting at point $P$. The acceleration of the center of mass of the cylinder is $a_{cm} = \frac{F}{M}$.The angular ac

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$\frac{Ma}{3}$

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(D) Let $F$ be the friction force acting at point $P$. The acceleration of the center of mass of the cylinder is $a_{cm} = \frac{F}{M}$.The angular acceleration $\alpha$ of the cylinder about its center is given by $	au = I\alpha$,where $	au = F \cdot r$ and $I = \frac{Mr^2}{2}$.Thus,$F \cdot r = \frac{Mr^2}{2} \alpha \Rightarrow \alpha = \frac{2F}{Mr}$.Since the cylinder does not slip on the rug,the acceleration of the point $P$ on the cylinder must be equal to the acceleration of the rug,$a$.The acceleration of point $P$ is $a_P = a_{cm} + \alpha r$ (in the direction of $a$).Therefore,$a = \frac{F}{M} + \left(\frac{2F}{Mr}ight)r = \frac{F}{M} + \frac{2F}{M} = \frac{3F}{M}$.Solving for $F$,we get $F = \frac{Ma}{3}$.

Consider a cylinder of mass $M$ resting on a rough horizontal rug that is pulled out from under it with acceleration $a$ perpendicular to the axis of the cylinder. What is the friction force $F_{friction}$ at point $P$? It is assumed that the cylinder does not slip.

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