Question

A solid cylinder of mass M and radius R rolls down an inclined plane without slipping. The speed of its centre of mass when it reaches the bottom is (h is the height of inclined plane) :

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

Answer 1

Answer: (B)

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

By conservation of energy

$\frac{1}{2}m{\nu }^2+\frac{1}{2}I{\omega }^2=Mgh$

but $V=R\omega$ ; $I=\frac{1}{2}M{R}^2$

$\Rightarrow \frac{1}{2}m{\nu }^2+\frac{1}{2}\left(\frac{1}{2}m{\nu }^2\right)=Mgh$

$\frac{3}{4}m{\nu }^2=Mgh$

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