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(A) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.For a solid sphere rotating about its diameter,the moment of inertia is $I_

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.For a solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} MR^2$.Let the angular speed of the sphere be $\omega_s = \omega$.Thus,the kinetic energy of the sphere is $K_s = \frac{1}{2} (\frac{2}{5} MR^2) \omega^2 = \frac{1}{5} MR^2 \omega^2$.For a disc rotating about an axis passing through its centre and perpendicular to its plane,the moment of inertia is $I_d = \frac{1}{2} MR^2$.The angular speed of the disc is given as $\omega_d = 2\omega$.Thus,the kinetic energy of the disc is $K_d = \frac{1}{2} (\frac{1}{2} MR^2) (2\omega)^2 = \frac{1}{4} MR^2 (4\omega^2) = MR^2 \omega^2$.The ratio of the kinetic energy of the disc to that of the sphere is $\frac{K_d}{K_s} = \frac{MR^2 \omega^2}{\frac{1}{5} MR^2 \omega^2} = 5$.Therefore,the ratio is $5: 1$.

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