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 non uniform such that, at any point P on the ring, the mass per unit length is given by λ=λ0 cos 2 θ (where λ0 is a positive constant). Then the moment of inertia of the ring about z -axis is:

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 non uniform such that, at any point P on the ring, the mass per unit length is given by λ=λ0 cos 2 θ (where λ0 is a positive constant). Then the moment of inertia of the ring about z -axis is: (A) M R2 (B) (1/2) M R2 (C) (1/2) (M/λ0) R (D) (1/π) (M/λ0) R

Tardigrade Admin · Accepted Answer

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 same distance

R

from origin.

∴

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}

$M R^{2}$

$\frac{1}{2} M R^{2}$

$\frac{1}{2} \frac{M}{λ _{0}} R$

$\frac{1}{π} \frac{M}{λ _{0}} R$

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 same distance $R$ from origin.
$∴$ MI of ring about $z$ -axis is
$= d m_{1} R^{2} + d m_{2} R^{2} + \dots + d m_{n} R^{2} = M R^{2}$

MR2

21​MR2

21​λ0​M​R

π1​λ0​M​R

Divide the ring into infinitely small lengths of mass dmi​. Even though mass distribution is non-uniform, each mass dmi​ is at same distance R from origin. ∴ MI of ring about z -axis is =dm1​R2+dm2​R2+…+dmn​R2=MR2

$M R^{2}$

$\frac{1}{2} M R^{2}$

$\frac{1}{2} \frac{M}{λ _{0}} R$

$\frac{1}{π} \frac{M}{λ _{0}} R$

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 same distance $R$ from origin.
$∴$ MI of ring about $z$ -axis is
$= d m_{1} R^{2} + d m_{2} R^{2} + \dots + d m_{n} R^{2} = M R^{2}$