The moment of inertia of a disc about an axis | vedclass

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(A) The moment of inertia of a disc about an axis passing through its centre of mass and perpendicular to its plane is $I = \frac{MR^2}{2}$.Using the

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$2$

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(A) The moment of inertia of a disc about an axis passing through its centre of mass and perpendicular to its plane is $I = \frac{MR^2}{2}$.Using the parallel axis theorem,$I_{axis} = I_{cm} + Mh^2$.For an axis tangential to the rim,the distance from the centre is $h = R$. Therefore,the moment of inertia $I_1$ is:$I_1 = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2$.For an axis passing through a point midway between the centre and the rim,the distance from the centre is $h = R/2$. Therefore,the moment of inertia $I_2$ is:$I_2 = \frac{MR^2}{2} + M(R/2)^2 = \frac{MR^2}{2} + \frac{MR^2}{4} = \frac{3}{4} MR^2$.The ratio of the two moments of inertia is:$\frac{I_1}{I_2} = \frac{\frac{3}{2} MR^2}{\frac{3}{4} MR^2} = \frac{3}{2} 	imes \frac{4}{3} = \frac{2}{1}$.Thus,the ratio is $2:1$.

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