Question

A ring of mass $M$ and radius $R$ lies in the $x-y$ plane with its centre at the origin as shown. The mass distribution of the ring is non-uniform such that at any point $\text{P}$ 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. $M{R}^2$
• B. $\frac{1}{2}M{R}^2$
• C. $\frac{1}{2}\frac{M}{{\lambda }_0}R$
• D. $\frac{1}{\pi }\frac{M}{{\lambda }_0}R$

Answer 1

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

Given, $\lambda ={\lambda }_0{\cos }^2\theta$
Let us consider an elemental mass ${}^{\prime }d{m}^{\prime }$ of length ${}^{\prime }d{s}^{\prime }$. Even though mass distribution is non-uniform, each elemental mass chosen is at same distance ${}^{\prime }{R}^{\prime }$ from the centre.


$\therefore$ Moment of inertia of the ring about $z-$ axis is
$\left(d{m}_1\right)R^2+\left(d{m}_2\right)R^2+\left(d{m}_3\right)R^2+...+\left(d{m}_n\right)R^2$
$\Rightarrow I=[d{m}_1+d{m}_2+d{m}_3+...+d{m}_n]r^2$
$\Rightarrow I=M{R}^2$
i.e MOI of ring depends only on the total mass $M$ and radius $R$