System of Particles & Rotational Motion

An extended body or a group of particles can be treated, for many purposes, as if its entire mass were concentrated at a single point — the centre of mass (COM). For a system of particles, the position of the COM is the mass-weighted average of the positions: $x_{cm} = \dfrac{\sum m_i x_i}{\sum m_i}$, and similarly for $y$ and $z$. The COM lies closer to the heavier particles, exactly as a see-saw balances nearer the heavier child.

For symmetric, uniform bodies the COM sits at the geometric centre: at the centre of a uniform rod, disc, sphere or ring. Notably, the COM can lie outside the material of the body — for a ring or a hollow shell it is at the empty centre. Recognising these standard positions saves time in NEET.

The power of the COM idea is in how it moves. The COM of a system moves as though all the external force acted on the total mass placed there: $\vec{F}_{\text{ext}} = M\vec{a}_{cm}$. Internal forces — however complicated — never shift the COM, because they cancel in action-reaction pairs. So a wrench spinning through the air, or an exploding shell, has a COM that follows a simple path (a straight line or a parabola) even while the pieces tumble or scatter.

This leads to a clean conservation rule. If no external force acts, the velocity of the COM is constant; if the system starts at rest, the COM stays put. When a bomb at rest explodes, the fragments fly apart but their COM remains exactly where the bomb was. When a person walks from one end of a stationary, frictionless boat to the other, the boat slides back so that the COM of person-plus-boat does not move. Spotting that the COM cannot move is often the whole solution to a NEET problem.

To make a body rotate, a force alone is not enough — where it acts matters. The turning effect of a force is the torque (moment of force), $\tau = r\,F\sin\theta = rF_{\perp}$, where $r$ is the distance from the axis to the point of application and $\theta$ the angle between $\vec{r}$ and $\vec{F}$. Torque is largest when the force is applied perpendicular to the position vector and far from the axis — which is why a long spanner and a push at its end loosen a nut most easily. Its SI unit is the newton-metre.

The rotational analogue of linear momentum is angular momentum, $L = I\omega$ (or $L = mvr$ for a particle), where $I$ is the moment of inertia and $\omega$ the angular velocity. Just as force changes linear momentum, torque changes angular momentum: $\tau = \dfrac{dL}{dt}$. This is the rotational form of Newton's second law.

The most striking consequence is the conservation of angular momentum: when the net external torque is zero, $L = I\omega$ stays constant. So if a body's moment of inertia decreases, its angular speed must rise to keep $L$ fixed. A spinning skater pulling in their arms speeds up, and a diver tucking into a ball rotates faster — both are NEET-favourite illustrations of $I_1\omega_1 = I_2\omega_2$.

For a body at rest, NEET also tests the conditions for equilibrium. A rigid body is in equilibrium when two conditions hold together: the net external force is zero (translational equilibrium) and the net external torque about any axis is zero (rotational equilibrium). The familiar 'principle of moments' for a balanced see-saw or beam — clockwise moments equal anticlockwise moments — is simply the second condition written out. Setting both sums to zero solves problems on ladders, hinged beams and balanced rods.

Moment of inertia $I$ is the rotational counterpart of mass — it measures a body's resistance to a change in its rotational motion. For a single particle, $I = mr^{2}$; for an extended body it is $I = \sum m_i r_i^{2}$, summed over all mass elements at their distances $r_i$ from the axis. Crucially, $I$ depends not just on the mass but on how that mass is distributed relative to the axis: mass far from the axis contributes much more (because of the $r^{2}$), which is why a flywheel has its mass at the rim.

Because $I$ depends on the axis, the same body has different moments of inertia about different axes. NEET expects the standard results: for a thin rod of length $L$ about its centre, $I = \tfrac{1}{12}ML^{2}$; for a ring about its central axis, $I = MR^{2}$; for a disc, $I = \tfrac{1}{2}MR^{2}$; for a solid sphere, $I = \tfrac{2}{5}MR^{2}$; and for a hollow sphere, $I = \tfrac{2}{3}MR^{2}$. A related quantity is the radius of gyration $k$, defined by $I = Mk^{2}$ — the distance from the axis at which the whole mass could be concentrated to give the same $I$.

Two theorems let us find $I$ about new axes without re-summing. The parallel axis theorem, $I = I_{cm} + Md^{2}$, gives the moment of inertia about any axis parallel to one through the COM, a distance $d$ away. The perpendicular axis theorem, valid only for plane laminae, states $I_z = I_x + I_y$ — the moment about an axis perpendicular to the lamina equals the sum about two perpendicular axes in its plane.

These theorems are NEET workhorses. For example, the moment of inertia of a disc about a diameter follows from the perpendicular axis theorem ($I_x = I_y$, so each is $\tfrac{1}{2}I_z = \tfrac{1}{4}MR^{2}$), and the moment of a rod about one end follows from the parallel axis theorem ($\tfrac{1}{12}ML^{2} + M(L/2)^{2} = \tfrac{1}{3}ML^{2}$). Memorising the standard $I$ values and applying these two theorems handles most of the chapter's calculations.

Rotational dynamics mirrors linear dynamics term for term. The rotational form of Newton's second law is $\tau = I\alpha$, where $\alpha$ is the angular acceleration — the direct analogue of $F = ma$. The kinematic equations carry over too, with angle $\theta$, angular velocity $\omega$ and angular acceleration $\alpha$ replacing $s$, $v$ and $a$: $\omega = \omega_0 + \alpha t$, and so on. A rotating body also stores rotational kinetic energy $KE_{rot} = \tfrac{1}{2}I\omega^{2}$, the counterpart of $\tfrac{1}{2}mv^{2}$.

Rolling motion combines translation of the centre of mass with rotation about it. For rolling without slipping, the contact point is momentarily at rest, which gives the key constraint $v = R\omega$ (and $a = R\alpha$). The total kinetic energy of a rolling body is the sum of both parts: $KE = \tfrac{1}{2}Mv^{2} + \tfrac{1}{2}I\omega^{2}$.

Using $v = R\omega$ and $I = Mk^{2}$, the rolling kinetic energy becomes $KE = \tfrac{1}{2}Mv^{2}\left(1 + \dfrac{k^{2}}{R^{2}}\right)$. The factor $k^{2}/R^{2}$ depends only on the shape, not the mass or radius — a hollow body (larger $k^{2}/R^{2}$) stores a bigger share of its energy in rotation than a solid one. This single idea answers the classic NEET question of which body rolls fastest down an incline.

When several bodies roll from rest down the same incline, the one with the smallest $k^{2}/R^{2}$ reaches the bottom first, because less of its energy is tied up in rotation. The order is therefore: solid sphere ($\tfrac{2}{5}$) > disc/solid cylinder ($\tfrac{1}{2}$) > ring/hollow cylinder ($1$). Remarkably, this order is independent of mass and radius — all solid spheres reach the bottom together, ahead of all discs, which beat all rings. Recognising this ranking, rather than computing each case, is the efficient NEET approach.