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(B) The moment of inertia of a solid sphere about its diameter is $I_S = \frac{2}{5} M R^2$. The moment of inertia of a solid cylinder about its geome

Vedclass · Accepted Answer

$1: 5$

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(B) The moment of inertia of a solid sphere about its diameter is $I_S = \frac{2}{5} M R^2$. The moment of inertia of a solid cylinder about its geometrical axis is $I_C = \frac{1}{2} M R^2$. Let $\omega_C$ be the angular speed of the cylinder and $\omega_S$ be the angular speed of the sphere. Given that $\omega_S = \frac{\omega_C}{2}$. The rotational kinetic energy is given by $K.E. = \frac{1}{2} I \omega^2$. The ratio of the kinetic energy of the sphere to that of the cylinder is: $\frac{K.E._S}{K.E._C} = \frac{\frac{1}{2} I_S \omega_S^2}{\frac{1}{2} I_C \omega_C^2} = \frac{I_S}{I_C} 	imes \left( \frac{\omega_S}{\omega_C} ight)^2$. Substituting the values: $\frac{K.E._S}{K.E._C} = \frac{\frac{2}{5} M R^2}{\frac{1}{2} M R^2} 	imes \left( \frac{\omega_C / 2}{\omega_C} ight)^2 = \frac{2/5}{1/2} 	imes \left( \frac{1}{2} ight)^2 = \frac{4}{5} 	imes \frac{1}{4} = \frac{1}{5}$. Thus,the ratio is $1: 5$.

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