Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 'I'. The ratio of moment of inertia about a parallel axis tangential to its rim to passing through a point midway between the centre and the rim is

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Aicharya · Accepted Answer

$\mathrm{I}=\frac{\mathrm{MR}^2}{2} \Rightarrow$ M.I. of disc about an axis passing through centre of mass By parallel axis theorem, M.I for axis tangential to rim, $\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$ M.I for axis passing through a point midway between centre and $\operatorname{rim}(h=R / 2)$, $\begin{aligned} & \mathrm{I}_2=\frac{\mathrm{MR}^2}{2}+\frac{\mathrm{MR}^2}{4}=\frac{3}{4} \mathrm{MR}^2 \\ \therefore \quad & \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\frac{3}{2} \mathrm{MR}^2}{\frac{3}{4} \mathrm{MR}^2}=\frac{2}{1} \Rightarrow 2: 1 \end{aligned}$

Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 'I'.…

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