The ratio of kinetic energies of two spheres | vedclass

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Question

(A) For a sphere rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy:$K = K_{tr} + K_{rot}

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(A) For a sphere rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy:$K = K_{tr} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the moment of inertia for a solid sphere is $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$,we have:$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$Given that the velocities of the centers of mass are equal $(v_1 = v_2)$,the ratio of kinetic energies is:$\frac{K_1}{K_2} = \frac{\frac{7}{10}m_1v_1^2}{\frac{7}{10}m_2v_2^2} = \frac{m_1}{m_2}$Given $\frac{K_1}{K_2} = \frac{2}{1}$,it follows that $\frac{m_1}{m_2} = \frac{2}{1}$.Thus,the ratio of their masses is $2:1$.

The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$,then the ratio of their masses will be

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