Question

Find the radius of gyration of a uniform circular disc of radius $R=1\text{m}$ and mass $M=2\text{kg}$ about its axis passing through the edge and normal to the disc as shown in figure.

• A. $\sqrt{\frac{1}{2}}\text{m}$
• B. $\sqrt{\frac{5}{2}}\text{m}$
• C. $1\text{m}$
• D. $\sqrt{\frac{3}{2}}\text{m}$

Answer 1

The correct option is D $\sqrt{\frac{3}{2}}\text{m}$

Given:
Mass of disc $M=2\text{kg}$
Radius of disc $R=1\text{m}$

As we know, for circular disc
$I_{\text{com}}=\frac{M{R}^2}{2}$



Using parallel axes theorem,
$I=I_{\text{com}}+M{d}^2$ $\because$ Here, $d=R$
$=\frac{M{R}^2}{2}+M{R}^2$
$I=\frac{3}{2}M{R}^2$
We know, Radius of gyration $K=\sqrt{\frac{I}{M}}=\sqrt{\frac{\frac{3}{2}M{R}^2}{M}}=\sqrt{\frac{3}{2}}\times R$
$\Rightarrow K=\sqrt{\frac{3}{2}}\times 1\text{m}$
$\Rightarrow K=\sqrt{\frac{3}{2}}\text{m}$
So, radius of gyration of a uniform disc about its axis passing through the edge and normal to the disc is $\sqrt{\frac{3}{2}}\text{m}$.