Question

A solid cylinder of radius $r$ rolls down on an inclined plane without slipping. The speed of the centre of mass when it reaches the bottom, is

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

Answer 1

Answer: (D)

From law of conservation of energy
Translational kinetic energy + rotational
kinetic energy $=$ potential energy of fall
$\therefore \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$ ...(i)
where $I$ is moment of inertia, $\omega$ the angular velocity and $v$ the linear velocity.
For a solid cylinder $I=\frac{1}{2}m{r}^2$ ... (ii)
and $\omega =\frac{v}{r}$ ...(iii)
From Eqs. (i), (ii) and (iii), we get
$\frac{1}{2}m{v}^2+\frac{1}{4}m{r}^2\times \frac{v^2}{r^2}=mgh$
$\Rightarrow v=\sqrt{\frac{4gh}{3}}$