Moment of intertia

Question

From a disc of radius $R$and mass $M$ a circular hole of diameter$R$ whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Moderate

Solution

$\text{Mass per unit area of disc}=\frac{M}{\pi {R}^{2}}$

Mass of removed portion of disc,

${M}^{\text{'}}=\frac{M}{\pi {R}^{2}}\times \pi {\left(\frac{R}{2}\right)}^{2}=\frac{M}{4}$

Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,

${I}_{O}^{\text{'}}={I}_{{O}^{\text{'}}}+{M}^{\text{'}}{d}^{2}$

$=\frac{1}{2}\times \frac{M}{4}\times {\left(\frac{R}{2}\right)}^{2}+\frac{M}{4}\times {\left(\frac{R}{2}\right)}^{2}=\frac{M{R}^{2}}{32}+\frac{M{R}^{2}}{16}=\frac{3M{R}^{2}}{32}$

When portion of disc would not have been removed, the moment of inertia of complete disc about centre O is

${I}_{O}=\frac{1}{2}M{R}^{2}$

So, moment of inertia of the disc with removed portion is

$I={I}_{O}-{I}_{O}^{\text{'}}=\frac{1}{2}M{R}^{2}-\frac{3M{R}^{2}}{32}=\frac{13M{R}^{2}}{32}$

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