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.