Question

A circular cylinder of radius r and mass m, starts from the top of an inclined plane and rolls down without any slip. fter its center moves to a point having a vertical height h as shown below, the velocity of the center of mass, with g as acceleration due to gravity is

• A. $\sqrt{\frac{gh}{3}}$
• B. $\sqrt{\frac{2gh}{3}}$
• C. $\sqrt{\frac{4gh}{3}}$
• D. $\sqrt{2gh}$

Answer 1

The correct option is C $\sqrt{\frac{4gh}{3}}$

$K.E_{\text{translation}}+K.E_{\text{rotation}}=mgh$

$\frac{1}{2}m{v}^2+\frac{1}{2}\times I.{\omega }^2=mgh$

$\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{m{r}^2}{2}.{\omega }^2=mgh$

$\frac{3}{4}m{v}^2=mgh$

$v=\sqrt{\frac{4gh}{3}}$