Physics - Rotation Question with Solution | TestHub

Since, we know that when a body rolls down from an inclined plane of height$h,$its velocity at the bottom of the plane,

$v=\sqrt{\frac{2gh}{1+\frac{K^2}{R^2}}}$

Where,$K=$radius of gyration of the body and$R=$radius of the body.

For the solid cylinder,

$K=\frac{R}{\sqrt{2}}$

$\therefore v=\sqrt{\frac{2gh}{1+\frac{\frac{R^2}{2}}{R^2}}}=\sqrt{\frac{2gh}{1+\frac{1}{2}}}=\sqrt{\frac{4gh}{3}}$