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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
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B
2gh3
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C
4gh3
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D
gh
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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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