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

Vedclass · Accepted Answer

$1:5$

Vedclass · Answer

(D) The rotational kinetic energy is given by $E = \frac{1}{2} I \omega^2$.For the solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} m R^2$.For the solid cylinder rotating about its geometrical axis,the moment of inertia is $I_c = \frac{1}{2} m R^2$.Given that the angular speed of the cylinder is twice that of the sphere,we have $\omega_c = 2 \omega_s$.The ratio of their kinetic energies is:$\frac{E_{sphere}}{E_{cylinder}} = \frac{\frac{1}{2} I_s \omega_s^2}{\frac{1}{2} I_c \omega_c^2} = \frac{I_s \omega_s^2}{I_c (2 \omega_s)^2} = \frac{I_s}{4 I_c}$.Substituting the values of $I_s$ and $I_c$:$\frac{E_{sphere}}{E_{cylinder}} = \frac{\frac{2}{5} m R^2}{4 	imes \frac{1}{2} m R^2} = \frac{2/5}{2} = \frac{1}{5}$.Thus,the ratio is $1:5$.

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