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Question

As shown in the figure, a solid cylinder of mass $$m$$ and radius $$R$$ starts rolling down an inclined plane of inclination $$\theta$$. Sufficient friction is present for the pure rolling of the cylinder. Determine the speed of its centre of mass when the cylinder has fallen a height $$H = 1\ m$$. 
(Take $$g = 10\ m/s^2$$)
1968097_5b571753e3254635a0b3d756b6fc3905.png


A
203m/s
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B
10m/s
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C
103m/s
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D
403m/s
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Solution

The correct option is C $$\sqrt{\dfrac{40}{3}} m/s$$
The situation is shown in the figure:-

For no slipping there will be no loss of mechanical energy.

Energy at position $$1$$ is $$E_{1} = mgH$$

Energy at position $$2$$ is $$E_{2} = \dfrac{1}{2} mV^{2}_{cm} + \dfrac{1}{2} I_{cm} \omega^{2} $$ 

$$\because \omega = \dfrac{V_{cm}}{R} $$    (For pure rolling) 

$$I_{cm} = \dfrac{mR^{2}}{2} $$ 
$$\Rightarrow E_{2} = \dfrac{3}{4} m V_{cm}^{2} $$ 

From the law of conservation of energy:-

$$E_{1} = E_{2} \Rightarrow mgH = \dfrac{3}{4} m V_{cm}^{2} \Rightarrow V_{cm} = \sqrt{\dfrac{4}{3} gH} = \sqrt{\dfrac{40}{3}} m/s $$

Correct Option - (D) 

2045688_1968097_ans_d2e0aca070ac4a7aae6a56c0e1a88560.png

Physics

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