Question

A solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. The angular velocity of the cylinder when it reaches the bottom of the plane will be

• A. $\frac{2}{R}\sqrt{gh}$
• B. $\frac{2}{R}\sqrt{\frac{gh}{2}}$
• C. $\frac{2}{R}\sqrt{\frac{gh}{3}}$
• D. $\frac{1}{2R}\sqrt{gh}$

Answer 1

The correct option is C $\frac{2}{R}\sqrt{\frac{gh}{3}}$


[since there is no relative motion at point of contact, work done by friction $W_f=0$ in pure rolling]
Applying mechanical energy conservation,

$\text{Loss in PE}=\text{Gain in}K{E}_{\text{Trans}}+\text{Gain in}K{E}_{\text{Rot}}$
or $Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$
$\Rightarrow Mgh=\frac{1}{2}M{R}^2{\omega }^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$
[$\because v=R\omega$ for pure rolling]
$\Rightarrow Mgh=\frac{3}{4}M{R}^2{\omega }^2$
$\Rightarrow {\omega }^2=\frac{4gh}{3R^2}$
$\therefore \omega =\frac{2}{R}\sqrt{\frac{gh}{3}}$