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

• A. $\sqrt{\left(2gh\right)}$
• B. $\sqrt{\frac{4}{3}gh}$
• C. $\sqrt{\frac{3}{4}gh}$
• D. $\sqrt{\left(4g/h\right)}$

Answer 1

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

Work-energy theorem, $Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{w}^2$

For solid cylinder, $I=\frac{1}{2}M{R}^2$

Also $v=Rw$ (due to pure rolling)

Thus $Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}\times \frac{1}{2}M{R}^2{w}^2$

$⟹v=\sqrt{\frac{4}{3}gh}$