Two discs of the same thickness but different | vedclass

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Question

(B) The moment of inertia $I$ of a disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.Sinc

Vedclass · Accepted Answer

$3:1$

Vedclass · Answer

(B) The moment of inertia $I$ of a disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.Since the mass $M$ is related to density $\rho$,thickness $t$,and radius $R$ by $M = \rho \cdot V = \rho \cdot (\pi R^2 t)$,we can express the radius as $R^2 = \frac{M}{\pi t \rho}$.Substituting this into the moment of inertia formula: $I = \frac{1}{2}M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.Given that the mass $M$ and thickness $t$ are constant for both discs,we have $I \propto \frac{1}{\rho}$.Therefore,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \frac{\rho_2}{\rho_1}$.Given the ratio of densities $\rho_1 : \rho_2 = 1 : 3$,we have $\frac{\rho_2}{\rho_1} = \frac{3}{1}$.Thus,$\frac{I_1}{I_2} = 3:1$.

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