Question

A ring of mass $M$ and radius $R$ lies in $x-y$ plane with its centre at origin as shown. The mass distribution of ring is nonuniform such that at any point 𝑃 on the ring, the mass per unit length is given by $\lambda ={\lambda }_0{\cos }^2\theta$ where ${\lambda }_0$ is a positive constant). Then the moment of inertia of the ring about $z$-axis is:

• A. $\frac{1}{2}\frac{M}{{\lambda }_0}R$
• B. $M{R}^2$
• C. $\frac{1}{2}M{R}^2$
• D. $\frac{1}{\pi }\frac{M}{{\lambda }_0}R$

Answer 1

The correct option is B $M{R}^2$

Divide the ring into infinitely small lengths of mass $d{m}_i$. Even though mass distribution is non-uniform, each mass $d{m}_i$ is at the same distance $R$ from origin.
$\therefore MI$ of ring about $z$-axis is

$=d{m}_1{R}^2+d{m}_2{R}^2+\dots +d{m}_n{R}^2=M{R}^2$

Final Answer: (a)