Question

A solid cylinder of mass $m$ & radius $R$ rolls down inclined plane without slipping. The speed of its $C.M.$ when it reached the bottom is

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

Answer 1

Answer: (B)

$\sqrt{4gh/3}$

By energy conservation

$(K.E{)}_i+(P.E{)}_i=(K.E{)}_f+(P.E{)}_f$

$(K.E{)}_i=0,(P.E{)}_i=mgh,(P.E{)}_f=0$

$(K.E{)}_f=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}_{cm}^2$

Where $I$ (moment of inertia) = $\frac{1}{2}m{R}^2$ (for solid cylinder)

So $mgh=\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)\left(\frac{v_{cm}^2}{R^2}\right)+\frac{1}{2}m{v}_{cm}^2$

$\Rightarrow v_{cm}=\sqrt{4gh/3}$