From a semi-circular disc of mass $M$ and radius $R_{2}$ . A semi-circular disc of radius $R_{1}$ is removed as shown in the figure. If the mass of original uncut disc is $M$ , find the momentum of inertia of residual disc about an axis passing through centre $O$ and perpendicular to the plane of the disc.

\frac{M}{{2 R}_{2}^{2}} (R_{2}^{4} - R_{1}^{4})

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\frac{M}{R_{2}^{2}} (R_{2}^{4} - R_{1}^{4})

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\frac{M}{R_{2}^{2}} (\frac{R_{2}^{4}}{4} - R_{1}^{4})

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\frac{M}{{2 R}_{2}^{2}} (R_{2}^{4} - \frac{R_{1}^{4}}{4})

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Solution

The correct option is A $\frac{M}{{2 R}_{2}^{2}} (R_{2}^{4} - R_{1}^{4})$
$m o m e n t o f i n e r t i a o f t h e o r i g i n a l s e m i c i r c u l a r d i s c t h r o u g h c e n t r e a n d p e r p e n d i c u l a r t o t h e p l a n e o f d i s c = \frac{M {R_{2}}^{2}}{2} p l a n a r m a s s d e n s i t y = \frac{2 M}{π {R_{2}}^{2}} S o t h e m a s s o f s e m i c i r c u l a r o f r a d i u s R_{1} = \frac{M {R_{1}}^{2}}{{R_{2}}^{2}} m o m e n t o f i n e r t i a o f t h e s e m i c i r c u l a r d i s c o f r a d i u s R_{2} = \frac{M {R_{1}}^{4}}{2 {R_{2}}^{2}} m o m e n t o f i n e r t i a o f t h e r e m a i n i n g p o r t i o n = \frac{M {R_{2}}^{2}}{2} - \frac{M {R_{1}}^{4}}{2 {R_{2}}^{2}} = \frac{M}{2 {R_{2}}^{2}} ({R_{2}}^{4} - {R_{1}}^{4})$

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