A solid cylinder and a solid sphere, having same mass M and radius R , roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be

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Aicharya · Accepted Answer

During pure rolling, the work done by the friction is zero. Therefore, we can apply conservation of mechanical energy.
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I {(\frac{v}{R})}^{2}$
$\Rightarrow m g h = \frac{1}{2} (m + \frac{I}{R^{2}}) v^{2}$
$\Rightarrow v = \sqrt{\frac{2 g h}{1 + \frac{I}{m R^{2}}}}$
For a solid cylinder $\frac{I}{m R^{2}} = \frac{1}{2}$ and for a solid sphere, $\frac{I}{m R^{2}} = \frac{2}{5}$
$\frac{v_{Cyl}}{v_{Sph}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + \frac{1}{2}}}$ $= \sqrt{\frac{14}{15}}$

A solid cylinder and a solid sphere, having same mass M and radius R , roll down the same inclined plane…

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