Q.(a) Derive the expression for moment of inertia of a uniform ring about an axis passing through the centre and perpendicular to the plane. OR

Moment of inertia ($I$) is the rotational analogue of mass: just as mass measures a body's inherent resistance to a change in its state of linear motion (inertia), moment of inertia measures a body's resistance to a change in its state of rotational motion, for rotation about a specific axis. Its SI unit is kg m$^2$, and its dimension is $[ML^2]$.

For a single point mass $m_i$ at perpendicular distance $r_i$ from a fixed axis, the moment of inertia about that axis is $I=m_ir_i^2$. For a rigid body made up of many such point masses, the moment of inertia about the axis is the sum over all of them:

Key contrast with mass. Ordinary mass is (to excellent approximation, ignoring relativistic effects) a fixed, invariant property of a body. Moment of inertia is emphatically not invariant — it depends not only on how much mass a body has, but crucially on exactly how that mass is distributed relative to the chosen axis. The same rigid body has a different moment of inertia for every different axis of rotation one might choose, even axes lying entirely outside the physical body itself.

Computing $I$ for a continuous body. Treating an infinitesimally small mass element $dm$, at perpendicular distance $r$ from the axis, as a point mass gives $dI=(dm)r^2$; integrating over the whole body gives the moment of inertia of any continuous, uniformly-distributed bulk object:

This integral, carried out with the mass element $dm$ expressed via the object's linear density $\lambda=M/\ell$ (for a rod), surface density $\sigma=M/(\pi R^2)$ (for a disc), or similar, yields the standard results for common shapes:

• Uniform rod (mass $M$, length $\ell$), axis through the center, perpendicular to the rod: $I=\dfrac{1}{12}M\ell^2$; axis through one end: $I=\dfrac{1}{3}M\ell^2$.
• Uniform ring (mass $M$, radius $R$), axis through the center, perpendicular to the plane: $I=MR^2$ (every mass element is at the same distance $R$ from the axis, so the integral is trivial). …