question_answer

A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom? [AIPMT 2003]

A) \[\sqrt{\frac{4}{3}gh}\]

B) \[\sqrt{4gh}\]

C) \[\sqrt{2gh}\]

D) \[\sqrt{\frac{3}{4}gh}\]

Correct Answer: A

Solution :

The situation is shown in the figure.

Potential energy of cylinder at the top will be converted into rotational kinetic energy and translational kinetic energy. So, energy conservation gives.

\[Mgh=\frac{1}{2}M{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}\]

\[=\frac{1}{2}M{{v}^{2}}+\frac{1}{2}\frac{M{{R}^{2}}}{2}\frac{{{v}^{2}}}{{{R}^{2}}}\left( \because \,\,{{I}_{cylinder}}=\frac{M{{R}^{2}}}{2} \right)\]

So,       \[Mgh=\frac{1}{2}M{{v}^{2}}+\frac{1}{4}\,M{{v}^{2}}\]

or         \[Mgh=\frac{3}{4}M{{v}^{2}}\]

or         \[{{v}^{2}}=\frac{4}{3}gh\]

or         \[v=\sqrt{\frac{4}{3}gh}\]

Note:    In a collision of two bodies whether it is perfectly elastic or inelastic, linear momentum is always conserved but kinetic energy need not to be conserved.