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.You need to login to perform this action.
You will be redirected in
3 sec