Question

A circular ring of radius $5\text{m}$ has moment of inertia of $100{\text{kg-m}}^2$ about its natural axis. This ring is converted into a circular disc having same moment of inertia about the disc’s natural axis. What is the radius of the circular disc?

• A. $10\text{m}$
• B. $\sqrt{2}\text{m}$
• C. $5\sqrt{2}\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is C $5\sqrt{2}\text{m}$

Radius of ring $r=5\text{m}$
Moment of inertia of ring $=I_{\text{Ring}}=100{\text{kg-m}}^2$
$\Rightarrow I_{Ring}=M{r}^2\Rightarrow 100=M\times 5^2$
$\Rightarrow M=\frac{100}{25}=\text{4 kg}$

When the ring is converted into a disc, the mass remains same.
$\therefore$ Mass of disc $;M=4\text{kg}$
Given: Moment of inertia of disc is also same.
$\therefore I_{disc}={\text{100 kg-m}}^2$
$\Rightarrow I_{disc}=\frac{M{R}^2}{2}$
$\Rightarrow 100=\frac{4\times R^2}{2}\Rightarrow R^2=\frac{200}{4}=50$
$\Rightarrow R=5\sqrt{2}\text{m}$

Therefore the radius of the disc is $5\sqrt{2}\text{m}$.