Question

# Solid cylinder of mass $$M$$ and radius $$R$$ rolls down an inclined plane of height '$$h$$' and inclination '$$\theta$$'. The speed of its centre of mass, when the cylinder reaches the bottom of the inclined plane, is:

A
2gh
B
2gh3
C
4gh3
D
gh

Solution

## The correct option is C $$\sqrt{\dfrac{4gh}{3}}$$Using work energy theorem ,Loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy$$mgh = \dfrac {1}{2}mV^2 + \dfrac {1}{2} (\dfrac {m}{2}R^2) (\dfrac {V}{R})^2$$   $$mgh = \dfrac {3}{4}mV^2$$$$V = \sqrt {\dfrac {4gh}{3}}$$Physics

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