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 \(F_{friction}\) at point \(P\) ? It is assumed that the cylinder does not slip

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Correct Answer

(D) \(\frac{Ma}{3}\)

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Detailed explanation

For pure, rolling,

\(\begin{array}{l}
{a_{og}} = {a_{op}} + {a_{pg}}\\
\,\,\,\,\,\,\,\, = - {a_{po}} + {a_{pg}}\\
\,\,\,\,\,\,\,\, = - \alpha r + a\\
\Rightarrow \frac{F}{M} = - \frac{{F \times r}}{{\frac{{M{r^2}}}{2}}} + a\\
\Rightarrow \frac{{3F}}{M} = a \Rightarrow F = \frac{{Ma}}{3}
\end{array}\)

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