Rotational motion is one of the highest-yield topics in NEET Physics, contributing 3–4 questions every year. It extends linear mechanics to rigid bodies rotating about a fixed axis and introduces new quantities — torque, moment of inertia, angular momentum, and rolling kinetic energy — each with direct parallels to their linear counterparts.

Centre of Mass (CM)

The centre of mass is the unique point where the total mass of a system can be considered concentrated for translational purposes. For a system of discrete particles, its x-coordinate is $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$ with SI unit metres [L]. For continuous bodies, $x_{cm} = \frac{\int x\, dm}{\int dm}$. Memorise the standard positions: uniform rod (L/2 from either end), triangular lamina (h/3 from base), semicircular ring (2R/π from the centre of the full circle), semicircular disc (4R/3π from centre), and uniform hemisphere (3R/8 from the flat face).

Moment of Inertia (MI)

The moment of inertia $I = \sum m_i r_i^2$ [M $L^{2}$ $T^{0}$] (kg $m^{2}$) measures how mass is distributed relative to the rotation axis. It is the rotational analogue of mass. Key standard values: ring about its axis $I = MR^2$; disc about its axis $I = \frac{1}{2}MR^2$; solid sphere about a diameter $I = \frac{2}{5}MR^2$; hollow sphere about a diameter $I = \frac{2}{3}MR^2$; uniform rod about its centre $I = \frac{ML^2}{12}$; uniform rod about one end $I = \frac{ML^2}{3}$. The radius of gyration $K = \sqrt{I/M}$ [L] (m) is the equivalent distance from the axis at which all mass could be concentrated.

Parallel and Perpendicular Axis Theorems

The Parallel Axis Theorem $I = I_{cm} + Md^2$ shifts the axis from the centre of mass by distance $d$ and applies to any body (2D or 3D). The Perpendicular Axis Theorem $I_z = I_x + I_y$ applies only to planar (2D/flat) bodies such as discs, rings, and laminas — never to spheres or cylinders. This distinction is a perennial NEET trap.

Torque and Angular Momentum

Torque (rotational force) is $\vec{\tau} = \vec{r} \times \vec{F}$, magnitude $\tau = rF\sin\theta$ [M $L^{2}$ $T^{-2}$] (N m). Angular momentum is $L = I\omega = mvr\sin\theta$ [M $L^{2}$ $T^{-1}$] (kg $m^{2}$/s). Newton's second law for rotation: $\tau = \frac{dL}{dt}$. When the net external torque is zero, angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$. Classic example: an ice skater pulling her arms in reduces her moment of inertia and therefore spins faster. Crucially, kinetic energy is not conserved in this process — it increases because the skater does internal muscular work.

Rolling Without Slipping

A body rolling without slipping satisfies the constraint $v_{cm} = \omega R$ (contact-point velocity is exactly zero). Total kinetic energy is the sum of translational and rotational parts:

$KE_{total} = \frac{1}{2}mv^2\!\left(1 + \frac{K^2}{R^2}\right)$

On an inclined plane, the acceleration of a rolling body is:

$a = \frac{g\sin\theta}{1 + K^2/R^2}$

A smaller $K^2/R^2$ ratio means higher acceleration. The rolling race ranking on any incline (regardless of mass or radius): solid sphere (2/5) > disc (1/2) > hollow sphere (2/3) > ring (1). The solid sphere always wins and the ring always comes last.

Three Solved Numericals

1. Disc tangent-in-plane MI: Use perpendicular axis theorem to find $I_d = MR^2/4$, then parallel axis theorem to get $I_{tangent} = 5MR^2/4 = 0.625$ kg $m^{2}$ for $M = 2$ kg, $R = 0.5$ m.
2. Rolling velocities from height $h = 2$ m: disc reaches $v = 5.16$ m/s, ring reaches $v = 4.47$ m/s — disc wins.
3. Ice-skater: $I_i = 6$ kg $m^{2}$, $\omega_i = 2$ rev/s; $I_f = 2$ kg $m^{2}$, so $\omega_f = 6$ rev/s; kinetic energy triples.

NEET Exam Focus

Always state which theorem you are using and verify its applicability. Check dimensional consistency. In rolling problems, never mix up $v = \omega R$ (rolling) with $v = R\alpha$ (purely rotational acceleration). In angular momentum questions, confirm the net external torque is truly zero before invoking conservation.