Three bodies: a ring,a solid cylinder,and a s | vedclass

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(A) Let the radius of gyration of the body be $K$,mass be $m$,height be $h$,and radius be $R$.According to the law of conservation of mechanical energ

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(A) Let the radius of gyration of the body be $K$,mass be $m$,height be $h$,and radius be $R$.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 $I = mK^2$ and $\omega = v/R$,we have:$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 $v_{	ext{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.For a solid cylinder,$K^2 = \frac{R^2}{2}$,so $v_{	ext{cylinder}} = \sqrt{\frac{2gh}{1+0.5}} = \sqrt{\frac{4gh}{3}} \approx 1.15\sqrt{gh}$.For a solid sphere,$K^2 = \frac{2}{5}R^2$,so $v_{	ext{sphere}} = \sqrt{\frac{2gh}{1+0.4}} = \sqrt{\frac{10gh}{7}} \approx 1.19\sqrt{gh}$.Comparing the velocities,the ring has the minimum velocity at the bottom of the inclined plane.

Three bodies: a ring,a solid cylinder,and a solid sphere,roll down an inclined plane without slipping. They start from rest. Which of the bodies reaches the bottom of the plane with the minimum velocity?

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