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(C) We assume the conservation of energy of the rolling body,i.e.,there is no loss of energy due to friction. The potential energy lost by the body in

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Solid sphere

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(C) We assume the conservation of energy of the rolling body,i.e.,there is no loss of energy due to friction. The potential energy lost by the body in rolling down the inclined plane $(=mgh)$ must be equal to the kinetic energy gained. Since the bodies start from rest,the kinetic energy gained is equal to the final kinetic energy of the bodies.The total kinetic energy $K$ of a body rolling without slipping is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $v$ is the final velocity of the center of mass,$k$ is the radius of gyration,and $R$ is the radius.Equating $K$ and $mgh$:$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$$v^2 = \frac{2gh}{1 + k^2/R^2}$$v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$For a ring,$k^2 = R^2$:$v_{	ext{ring}} = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{gh}$For a solid cylinder,$k^2 = R^2/2$:$v_{	ext{cylinder}} = \sqrt{\frac{2gh}{1 + 1/2}} = \sqrt{\frac{4gh}{3}}$For a solid sphere,$k^2 = 2R^2/5$:$v_{	ext{sphere}} = \sqrt{\frac{2gh}{1 + 2/5}} = \sqrt{\frac{10gh}{7}}$Comparing the velocities:$v_{	ext{ring}} = \sqrt{1.00gh}$$v_{	ext{cylinder}} = \sqrt{1.33gh}$$v_{	ext{sphere}} = \sqrt{1.43gh}$Thus,the solid sphere reaches the ground with the maximum velocity.

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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