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Question

If a circular concentric hole is made in a disc, then about its own geometrical axis

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Solution

If M be the mass of solid disc and m be the mass of remaining disc, Then
$m = \frac{M}{π b^{2}} \times π (b^{2} - a^{2}) = \frac{M (b^{2} - a^{2})}{b^{2}}$
Now for the solid disc $I = \frac{1}{2} {Mb}^{2}$
and for the hollow disc $I^{'} = \frac{1}{2} m (a^{2} + b^{2}) = \frac{M (b^{2} - a^{2}) (b^{2} + a^{2})}{2 b^{2}}$
$∴ I^{'} < I$
K = Radius of gyration of solid disc = $\sqrt{\frac{I}{M}}$ and $K' =$ Radius of gyration of hollow disc $= \sqrt{\frac{I'}{m}}$
$∴ K = \sqrt{\frac{\frac{1}{2} {Mb}^{2}}{M}} = \frac{b}{\sqrt{2}} and K' = \sqrt{\frac{b^{2} + a^{2}}{2}} > K$

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