The moment of inertia of a solid disc rotatin | vedclass

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(B) Let $M$ be the mass of each body. The moment of inertia of a disc about its diameter is $I_1 = \frac{MR_1^2}{4}$.The moment of inertia of a ring a

Vedclass · Accepted Answer

$4$

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(B) Let $M$ be the mass of each body. The moment of inertia of a disc about its diameter is $I_1 = \frac{MR_1^2}{4}$.The moment of inertia of a ring about its diameter is $I_2 = \frac{MR_2^2}{2}$.The moment of inertia of a solid sphere about its diameter is $I_3 = \frac{2MR_1^2}{5}$.According to the problem,$I_1 = 2.5 I_2$.$\frac{MR_1^2}{4} = 2.5 	imes \frac{MR_2^2}{2} \Rightarrow \frac{R_1^2}{4} = \frac{5}{2} 	imes \frac{R_2^2}{2} \Rightarrow \frac{R_1^2}{4} = \frac{5R_2^2}{4} \Rightarrow R_1^2 = 5R_2^2$.Now,we are given $I_3 = n I_2$.$\frac{2MR_1^2}{5} = n 	imes \frac{MR_2^2}{2}$.Substituting $R_1^2 = 5R_2^2$ into the equation:$\frac{2M(5R_2^2)}{5} = n 	imes \frac{MR_2^2}{2} \Rightarrow 2MR_2^2 = n 	imes \frac{MR_2^2}{2} \Rightarrow n = 4$.

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