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Q. Radius of gyration of disc of mass $ 50\,g $ and radius $2.5\, cm $ about an axis passing through its centre of gravity and perpendicular to the plane is

Punjab PMETPunjab PMET 2009System of Particles and Rotational Motion

Solution:

Here mass $M=50\, g$ and radius $R=-2.5\, cm$.
Required moment of inertia of the disc is given by
$I=\frac{MR^{2}}{2}=MK^{2}$
So, $K^{2}=\frac{R^{2}}{2}$
or $K=\frac{R}{\sqrt{2}}=\frac{2.5}{\sqrt{2}}=\frac{2.5\sqrt{2}}{2}$
$ =1.767 \, cm$
$= 1.77 \,cm$