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?                 

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, \[Mgh=\frac{1}{2}M{{v}^{2}}+\frac{1}{4}\,M{{v}^{2}}\] \[orMgh=\frac{3}{4}M{{v}^{2}}\] \[or{{v}^{2}}=\frac{4}{3}gh\] \[orv=\sqrt{\frac{4}{3}gh}\]                 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.