A horizontal force F is applied at the centre of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is $μ$ . The centre of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct ?

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The magnitude of the frictional force on the object due to the ground is always $μ$ mg

For the same F, the value of a does not depends on whether the cylinder is solid or hollow

For a thin-walled hollow cylinder, $a = \frac{F}{2 m}$

For a solid cylinder, the maximum possible value of a is 2 $μ$ g

answer is B, D.

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Detailed Solution

$\begin{array}{l} F - f = ma \\ f \cdot R = Iα \\ a = Rα \\ FR = {mR}^{2} α + Iα \Rightarrow α = \frac{FR}{I + {mR}^{2}} \\ a = \frac{{FR}^{2}}{I + {mR}^{2}} \\ f = \frac{IF}{I + {mR}^{2}} = \frac{Ia}{R^{2}} \end{array}$

So we can say that value of a will depend on whether the cylinder is hollow or solid.

for solid cylinder $I = \frac{{mR}^{2}}{2}$

$a = \frac{{FR}^{2}}{\frac{3}{2} {mR}^{2}} = \frac{2 F}{3 m} \Rightarrow a = \frac{{fR}^{2}}{I} = \frac{{μmgR}^{2}}{\frac{{mR}^{2}}{2}} = 2 μg$

$f = μN = μmg$ is the limiting friction and not always the same value.
for hollow cylinder $a = \frac{{FR}^{2}}{2 {mR}^{2}}; I = {mR}^{2} = \frac{F}{2 m}$

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