Question

A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.

Answer 1

\frac{\lambda L^3}{8\pi^2}

Answer 2

1. Calculate the total mass (M) of the rod/ring:

Given linear mass density = λ Given length = L The total mass of the rod, which forms the ring, is: $M = \lambda L$

2. Relate the length of the rod to the radius (R) of the ring:

When the rod is bent into a ring, its length L becomes the circumference of the ring: $L = 2\pi R$ From this, the radius of the ring is: $R = \frac{L}{2\pi}$

3. Use the formula for the moment of inertia of a ring about its diameter:

For a thin ring of mass M and radius R, the moment of inertia about any of its diameters ($I_d$) is given by: $I_d = \frac{1}{2}MR^2$

4. Substitute M and R into the formula:

Substitute $M = \lambda L$ and $R = \frac{L}{2\pi}$ into the $I_d$ formula: $I_d = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2$ $I_d = \frac{1}{2} (\lambda L) \left(\frac{L^2}{4\pi^2}\right)$ $I_d = \frac{\lambda L^3}{8\pi^2}$

The moment of inertia of the ring about any of its diameter is $\frac{\lambda L^3}{8\pi^2}$.