A solid sphere of mass 2 kg and radius 0.5 | vedclass

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Question

(B) For a solid sphere rolling without slipping, the moment of inertia about its center of mass is $I = \frac{2}{5}MR^2$.The rotational kinetic energy

Vedclass · Accepted Answer

$2$

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(B) For a solid sphere rolling without slipping, the moment of inertia about its center of mass is $I = \frac{2}{5}MR^2$.The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{5}MR^2\omega^2$.Since the sphere is rolling without slipping, the velocity of the center of mass is $v = R\omega$, so $\omega = \frac{v}{R}$.Substituting $\omega$ into the rotational kinetic energy expression: $K_{rot} = \frac{1}{5}MR^2(\frac{v}{R})^2 = \frac{1}{5}Mv^2$.The translational kinetic energy is $K_{trans} = \frac{1}{2}Mv^2$.The ratio of rotational to translational kinetic energy is $\frac{K_{rot}}{K_{trans}} = \frac{\frac{1}{5}Mv^2}{\frac{1}{2}Mv^2} = \frac{2}{5}$.Thus, the ratio is $2: 5$.

$A$ solid sphere of mass $2 \ kg$ and radius $0.5 \ m$ is rolling without slipping on a horizontal surface. The ratio of the rotational and translational kinetic energies of the sphere is (in $: 5$)

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