  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 $$g = 10\ m/s^2$$) A
203m/s  B
10m/s  C
103m/s  D
403m/s  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) Physics

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