Question

A ring of radius $5\text{m}$ and linear mass density $\lambda =\frac{0.1}{\pi }\text{kg/m}$ is spinning about an axis passing through its COM perpendicular to its plane. Find the moment of inertia of the ring about this axis.

• A. $25{\text{kg-m}}^2$
• B. $42{\text{kg-m}}^2$
• C. $13{\text{kg-m}}^2$
• D. $31{\text{kg-m}}^2$

Answer 1

The correct option is A $25{\text{kg-m}}^2$

Given, radius of ring $\left(r\right)=5\text{m}$
Linear mass density $\left(\lambda \right)=\frac{0.1}{\pi }\text{kg/m}$
$\therefore$ Mass of the ring $\left(m\right)=\lambda \times \left(\text{circumference of the ring}\right)$
$=\lambda \times 2\pi r$
$=\frac{0.1}{\pi }\times 2\pi \times 5$
$=1\text{kg}$

Moment of inertia about $z$ axis as shown in figure
$I_z=m{r}^2=1\times (5{)}^2=25{\text{kg-m}}^2$