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(C) According to the law of conservation of mechanical energy,the potential energy at the top equals the sum of translational and rotational kinetic e

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Solid sphere

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(C) According to the law of conservation of mechanical energy,the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the body rolls without slipping,$\omega = v/R$ and $I = mk^2$,where $k$ is the radius of gyration.$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$For a ring,$k^2 = R^2$,so $v_{ring} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh} \approx 1.00 \sqrt{gh}$.For a solid cylinder,$k^2 = R^2/2$,so $v_{cylinder} = \sqrt{\frac{2gh}{1+1/2}} = \sqrt{\frac{4gh}{3}} \approx 1.15 \sqrt{gh}$.For a solid sphere,$k^2 = 2R^2/5$,so $v_{sphere} = \sqrt{\frac{2gh}{1+2/5}} = \sqrt{\frac{10gh}{7}} \approx 1.19 \sqrt{gh}$.Comparing the values,the solid sphere has the maximum velocity at the bottom of the inclined plane.

Three bodies,a ring,a solid cylinder,and a solid sphere,roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

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