Question 602: Centre of Mass and Rotational Motion

Question

The ratio of radii of gyration of a ring to a disc (both circular) of same radii and mass, about a tangential axis perpendicular to the plane is

Accepted Answer

$$\begin{aligned}
&	ext { The radius of gyration is given by }\
&K=\sqrt{\frac{I}{M}} \Rightarrow K \propto \sqrt{1}
\end{aligned}$$
Moment of inertia of ring about tangential axis is calculated by parallel axis theorem, $$I_{	ext {ring }}=I_c+M R^2 =M R^2+M R^2=2 M R^2$$
Similarly,
$$\begin{aligned}
& I_{	ext {disc }}=I_c+M R^2=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2 \
& 	herefore \quad \frac{K_{	ext {ring }}}{K_{	ext {disc }}}=\sqrt{\frac{I_{	ext {ring }}}{I_{	ext {disc }}}}=\sqrt{\frac{2 M R^2}{\frac{3}{2} M R^2}}=\frac{2}{\sqrt{3}}
\end{aligned}$$

Answer

$$\frac{\sqrt{3}}{\sqrt{2}}$$

Answer

$$\frac{2}{\sqrt{5}}$$

Answer

$$\frac{\sqrt{2}}{1}$$

Answer

$$\frac{2}{\sqrt{3}}$$

Centre of Mass and Rotational Motion

The ratio of radii of gyration of a ring to a disc (both circular) of same radii and mass, about a tangential axis perpendicular to the plane is

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