Rotational Motion Question 52
Question: 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]
Options:
A) $ \sqrt{\frac{4}{3}gh} $
B) $ \sqrt{4gh} $
C) $ \sqrt{2gh} $
D) $ \sqrt{\frac{3}{4}gh} $
Show Answer
Answer:
Correct Answer: A
Solution:
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}Mv^{2}+\frac{1}{2}I{{\omega }^{2}} $
$ =\frac{1}{2}Mv^{2}+\frac{1}{2}\frac{MR^{2}}{2}\frac{v^{2}}{R^{2}}$
$( \because I _{cylinder}=\frac{MR^{2}}{2} ) $
So, $ Mgh=\frac{1}{2}Mv^{2}+\frac{1}{4}Mv^{2} $
or $ Mgh=\frac{3}{4}Mv^{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.
