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(C) Using the conservation of mechanical energy,the total potential energy at the top is converted into translational and rotational kinetic energy at

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$21 : 28 : 30$

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(C) Using the conservation of mechanical energy,the total potential energy at the top is converted into translational and rotational kinetic energy at the bottom:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the objects roll without slipping,$\omega = v/R$ and $I = mk^2$,where $k$ is the radius of gyration.$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(v/R)^2 = \frac{1}{2}mv^2(1 + k^2/R^2)$Translational kinetic energy $K_t = \frac{1}{2}mv^2 = \frac{mgh}{1 + k^2/R^2}$.For a ring,$k^2 = R^2 \implies K_{ring} = \frac{mgh}{1+1} = \frac{mgh}{2}$.For a coin (disc),$k^2 = R^2/2 \implies K_{coin} = \frac{mgh}{1+1/2} = \frac{2mgh}{3}$.For a solid ball (sphere),$k^2 = 2R^2/5 \implies K_{ball} = \frac{mgh}{1+2/5} = \frac{5mgh}{7}$.Taking the ratio $K_{ring} : K_{coin} : K_{ball} = \frac{1}{2} : \frac{2}{3} : \frac{5}{7}$.Multiplying by the $LCM$ of denominators $(42)$: $21 : 28 : 30$.

Starting from rest,at the same time,a ring,a coin (disc),and a solid ball of the same mass roll down an incline without slipping. The ratio of their translational kinetic energies at the bottom will be:

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