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. |
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