Question

A ring having linear mass density $5\text{kg/m}$ and radius $0.5\text{m}$ is rotated about the axis passing through it's centre and perpendicular to it's plane. Find the moment of inertia ($\text{MI}$) of the ring about the required axis.

• A. ${\text{0.5 kg-m}}^2$
• B. ${\text{1 kg-m}}^2$
• C. $2.5\pi {\text{kg-m}}^2$
• D. $1.25\pi {\text{kg-m}}^2$

Answer 1

The correct option is D $1.25\pi {\text{kg-m}}^2$

Mass of ring,$\left(m\right)$ = linear mass density $\times$ circumference
$r=0.5\text{m}$
$\Rightarrow m=5\times \left(2\pi r\right)=5\times (2\times \pi \times 0.5)$
$\therefore m=5\pi \text{kg}$

$\text{MI}$ of ring about the required axis $Y{Y}^{\prime }$
as shown in figure is:
$I={mr}^2$
$I=\left(5\pi \right)(0.5{)}^2$
$\therefore I=1.25\pi {\text{kg-m}}^2$
So, moment of inertia of ring about the axis $Y{Y}^{\prime }$ is $1.25\pi {\text{kg-m}}^2$