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Q.
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is
UPSEEUPSEE 2009System of Particles and Rotational Motion
Solution:
For a disc moment of inertia about a tangential axis in its own plane $=\frac{5}{4} M R^{2}$
$\therefore M_{1} K_{1}^{2}=\frac{5}{4} M_{1} R^{2}$
$\Rightarrow K_{1}=\frac{\sqrt{5}}{2} R$
Now, for a ring moment of inertia about a tangential axis in its own plane $=\frac{3}{2} M_{2} R^{2}$
$\therefore M_{2} K_{2}^{2}=\frac{3}{2} M_{2} R^{2} $
$ \Rightarrow K_{2}=\sqrt{\frac{3}{2}} R $
$\therefore \frac{K_{1}}{K_{2}}=\frac{\sqrt{5}}{\sqrt{6}}$