Question

A solid cylinder of mass M and radius R rolls without slipping down on inclined plane of length L and height h. What is the speed of its centre of mass, when the cylinder reaches its bottom?

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

Answer 1

Answer: (A)

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

Acceleration of a rolling body down on inclined plane given as
$a=\frac{gsin\theta }{1+\frac{K^2}{R^2}}$
For a solid cylinder
$I=M{K}^2=\frac{M{R}^2}{2}$
$\therefore \frac{K^2}{R^2}=\frac{1}{2}$
$\therefore a=\frac{gsin\theta }{1+\frac{1}{2}}=\frac{2}{3}gsin\theta$
Also $v^2=u_0^2+2as$
$=(0{)}^2+2\times \frac{2}{3}gsin\theta \times L$
$\frac{4}{3}g\times \frac{h}{L}\times L=\frac{4}{3}gh$
$v=\sqrt{\frac{4}{3}gh}$