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(B) The total kinetic energy $(K_{total})$ of a body rolling without slipping is the sum of translational kinetic energy $(K_{trans})$ and rotational

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Hollow sphere

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(B) The total kinetic energy $(K_{total})$ of a body rolling without slipping is the sum of translational kinetic energy $(K_{trans})$ and rotational kinetic energy $(K_{rot})$.$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Given that $K_{rot} = 40\%$ of $K_{total}$,it implies $K_{rot} = 0.4 K_{total}$,which means $K_{trans} = 0.6 K_{total}$.Therefore,the ratio $\frac{K_{rot}}{K_{trans}} = \frac{0.4}{0.6} = \frac{2}{3}$.Substituting $K_{rot} = \frac{1}{2}I\omega^2$ and $K_{trans} = \frac{1}{2}mv^2$ with $v = R\omega$:$\frac{\frac{1}{2}I\omega^2}{\frac{1}{2}m(R\omega)^2} = \frac{2}{3} \implies \frac{I}{mR^2} = \frac{2}{3}$.Thus,$I = \frac{2}{3}mR^2$.This moment of inertia corresponds to a hollow sphere.

$A$ body is rolling without slipping on a horizontal plane. If the rotational kinetic energy of the body is $40\%$ of the total kinetic energy,then the body might be:

Similar Questions

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