A solid cylinder is rolling down a rough inclined plane of inclination . Then [2006]

Question

A solid cylinder is rolling down a rough inclined plane of inclination $θ$ . Then [2006]

Smart Achievers · Accepted Answer

(3, 4)

The sliding friction acts in the opposite direction of $m g \sin θ$ to oppose the relative motion. Because of frictional force the cylinder rolls.

Hence frictional force aids rotation but hinders translational motion.

Applying $F_{net} = m a$ along the direction of inclined plane, $m g \sin θ - f = m a_{c},$

where $a_{c}$ is acceleration of centre of mass of the cylinder

$∴$ $f = m g \sin θ - m a_{c} ...(i)$

$But a_{c} = \frac{g \sin θ}{1 + \frac{I_{c}}{m R^{2}}} = \frac{g \sin θ}{1 + \frac{m R^{2} / 2}{m R^{2}}} = \frac{2}{3} g \sin θ$

$∴$ $f = \frac{m g \sin θ}{3}$

Clearly, if $θ$ is reduced, frictional force is reduced.

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