A solid cylinder of mass M rolls without slipping (pure rolling) down an inclined plane making an angle 30^o with the horizontal. The frictional force between the cylinder and the incline is …..

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$\frac{Mg}{6}$

$\frac{Mg}{3}$

$\frac{Mg}{5}$

$\frac{Mg}{4}$

answer is D.

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Detailed Solution

While rolling down an inclined plane,
Frictional force, $f = \frac{MgSin θ}{1 + (\frac{R^{2}}{K^{2}})}$
For a solid cylinder, $\frac{K^{2}}{R^{2}} = \frac{1}{2} \Rightarrow \frac{R^{2}}{K^{2}} = 2$
Put $θ = 30^{\circ} then f = \frac{mg}{6}$

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