Systems of Particles and Rotational Motion — NEET Notes

Rigid body — and the two motions it can have

A rigid body is an idealised system of particles in which the distance between every pair of particles is fixed. No real body is perfectly rigid, but for wheels, beams, tops, planets and molecules the deformation under ordinary forces is negligible. The motion of any rigid body splits cleanly into two kinds: pure translation, in which every particle has the same velocity at every instant, and pure rotation about a fixed axis, in which every particle moves in a circle whose centre lies on the axis. Rolling motion is a superposition of both — and the key insight of this chapter is that any unfixed rigid body's motion can be decomposed into a translation of its centre of mass plus a rotation about it. Throughout, "rotation" means rotation about a fixed axis unless we say otherwise.

Centre of mass — the one point that obeys Newton's laws

Throw a hammer spinning across the room. The head wobbles, the handle loops wildly, but one point inside the hammer follows a clean parabola. That point is the centre of mass — the mass-weighted mean of the positions of all particles in the system. For two particles on a line, X = (m₁x₁ + m₂x₂)/(m₁ + m₂); generalised to n particles in three dimensions,

R_cm = (Σ m_i r_i) / M

Where M = Σ m_i is the total mass

For two equal masses, the CM is the midpoint; for three equal masses, the centroid of the triangle. NEET 2020 tested the two-particle case directly (5 kg, 10 kg, 1 m rod → 67 cm from the 5 kg mass); NEET 2022 repeated it with 10 kg and 20 kg on a 10 m rod, answer 20/3 m. For a continuous body, R_cm = (1/M)∫r dm, and symmetry usually solves the problem: rod → midpoint, disc → centre, sphere → centre, cone → axis at one-quarter height from base. The CM need not lie inside the material — for a ring it sits in empty space. When a body has a hole, apply the negative-mass trick: treat the cut-out as negative mass.

Motion of the centre of mass & linear momentum of a system

The reason the centre of mass matters is contained in a single equation. If we differentiate the definition twice with respect to time and multiply by total mass, we obtain

Total linear momentum of a system: P = M v_cm. Differentiate and only external forces survive (internal forces cancel in Newton's-third-law pairs): dP/dt = F_ext. When F_ext = 0, momentum is conserved. NEET 2022's exploding shell (mass ratio 2 : 2 : 1, initially at rest) used this: the two equal fragments fly off perpendicular at v, contributing momenta 2mv each; resultant 2√2 mv must be balanced by the third (mass m) fragment, giving v′ = 2√2 v. The same logic explains why a projectile that explodes mid-flight has a centre of mass that keeps following the original parabola: the explosion is internal; gravity, the only external force, is unchanged.

Vector product — the algebraic tool for rotation

Rotation lives in three dimensions, and the natural product of two vectors that captures perpendicular geometry is the cross product. For two vectors a and b separated by angle θ (the smaller of the two angles, less than 180°):

a × b = |a| |b| sin θ n̂

Magnitude ab sin θ; direction by right-hand rule

The cross product is not commutative — reversing the order reverses the sign: a × b = −(b × a). It vanishes when the vectors are parallel or antiparallel (sin θ = 0). The unit-vector triplets obey the cyclic rule: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ. In determinant form,

a × b = (a_yb_z − a_zb_y) î + (a_zb_x − a_xb_z) ĵ + (a_xb_y − a_yb_x) k̂

The cross product in Cartesian components

Every rotational quantity here — torque, angular momentum, the link between linear and angular velocity — is a cross product. NEET 2020 asked for the torque from F = 3ĵ N at r = 2k̂ m: τ = 2k̂ × 3ĵ = −6î N·m. NEET 2018 took the harder version, with F = 4î + 5ĵ − 6k̂ at (2, 0, −3) about (2, −2, −2): r = 2ĵ − k̂, and the determinant expansion gives τ = −7î − 4ĵ − 8k̂. The vanishing case — a × a = 0 — is the hidden reason dL/dt = τ has a "v × p = 0" step.

Angular velocity and its link to linear velocity

For a rigid body rotating about a fixed axis, every particle traces a circle whose centre lies on the axis. The angular velocity ω = dθ/dt is the same for every particle of the body — that is what makes it a property of the body, not of an individual particle. ω is a vector along the axis, sense by right-hand screw rule. Angular acceleration α = dω/dt; for fixed-axis rotation ω and α are collinear. NEET 2023 asked a one-line conceptual question: along which direction does the angular acceleration of a body moving along a circle point? Along the axis of rotation, not tangential, not radial.

The linear velocity of a particle at position r from the origin (on the axis) is

Torque and angular momentum

Force changes linear momentum; the rotational analogue is torque, defined for a particle at position vector r acted on by force F as

τ = r × F

Moment of force about the origin

Magnitude τ = rF sin θ = r_⊥F, where r_⊥ = r sin θ is the moment arm. The torque vanishes if the line of action of F passes through the origin — which is why pushing a door on the hinge line accomplishes nothing while pushing at the far edge swings it open instantly. Dimensions M L² T⁻² (same as energy, but torque is a vector); SI unit N m.

Angular momentum of a particle is the rotational analogue of linear momentum:

L = r × p

Where p = m v is linear momentum

Magnitude |L| = mvr sin θ = mvr_⊥. SI unit kg m² s⁻¹ (= J s). Like torque, L depends on the choice of origin. A particle moving in a straight line at constant velocity has constant angular momentum mvr_⊥ about any external point — r_⊥ is fixed, no external torque acts. Differentiating L = r × p with the product rule gives dL/dt = v × mv + r × F. The first term vanishes (v parallel to itself); the second is τ. So:

Conservation of angular momentum

Setting τ_ext = 0 in dL/dt = τ_ext gives L = constant. This is the law of conservation of angular momentum, and it is among the most powerful tools in this chapter. The classic demonstrations all reduce to "I changes, ω changes inversely, but Iω stays the same":

If τ_ext = 0 then I₁ ω₁ = I₂ ω₂.

The skater, the diver, the collapsing star — same equation.

An ice skater pulls her arms in: I falls, ω rises. A neutron star collapses from a million-km radius to ten km, I drops by 10¹⁰, rotation rate goes from once a month to many times per second. A diver tucking up triples ω. All governed by the same equation. NEET 2018 tested this directly: a solid sphere rotating freely in free space has its radius increased at fixed mass — which quantity stays constant? Only L (no external torque). NEET 2017 applied the law to flywheel coupling: two discs of equal I rotating at ω₁ and ω₂, joined face-to-face, share ω′ = (ω₁ + ω₂)/2 by L-conservation; energy loss to internal friction is I(ω₁ − ω₂)²/4 — the canonical demonstration that L-conservation does not imply KE-conservation.

Equilibrium of a rigid body

A particle is in equilibrium when the vector sum of forces on it is zero. A rigid body has a second requirement, because forces can also produce rotation. Mechanical equilibrium of a rigid body demands two simultaneous conditions:

A body can be in partial equilibrium — translationally yes, rotationally no, or vice versa. A pair of equal and opposite forces with different lines of action is a couple: ΣF = 0, but the moments add — pure rotation without translation. Opening a bottle lid, the compass needle in Earth's magnetic field, the steering wheel — all couples. The moment of a couple is independent of the chosen origin (NCERT example 6.7).

The principle of moments applies rotational equilibrium to levers: taking moments about the fulcrum, d₁F₁ = d₂F₂, i.e. load arm × load = effort arm × effort. The ratio F₁/F₂ is the mechanical advantage; when the effort arm exceeds the load arm, MA > 1 and a small effort lifts a large load (seesaw, crowbar, pliers, balance beam). NEET 2021 applied this to a 200 cm, 500 g rod balanced on a wedge at the 40 cm mark, with 2 kg at the 20 cm mark and unknown m at 160 cm. Equating moments: 2 × 0.2 = 0.5 × 0.6 + m × 1.2 → m = 1/12 kg.

The centre of gravity is the point at which total gravitational torque on the body is zero. In a uniform gravitational field, CG coincides with the CM. NEET 2017 tested this — the correct statement was that CM is the point at which total gravitational torque on the body vanishes, which is precisely CG in the uniform-field approximation.

Moment of inertia — the rotational analogue of mass

When a rigid body rotates about a fixed axis with angular velocity ω, every particle of mass m_i at perpendicular distance r_i from the axis has linear speed v_i = r_iω and kinetic energy ½m_ir_i²ω². The total kinetic energy is

K = ½ (Σ m_i r_i²) ω² = ½ I ω²

Where I = Σ m_i r_i² is the moment of inertia

I = Σm_ir_i² (or ∫r² dm for a continuous body) is the moment of inertia. Compare ½Iω² with ½mv², τ = Iα with F = ma: I plays the role of mass in rotation. Unlike mass, I depends on the axis chosen — the same disc has different I about its diameter and its central perpendicular axis. SI unit kg m²; dimensions M L².

The radius of gyration k is defined by I = Mk² — the distance at which a point mass M would have the same I. For a disc, k = R/√2 about the central axis and R/2 about a diameter; their ratio √2 was NEET 2022's direct question. A flywheel concentrates mass near the rim to maximise I (the r² weighting rewards you for it), which is why it smooths out fluctuations in engine torque.

Theorems of parallel and perpendicular axes

Memorising six rows of a table won't carry you through every problem. NEET routinely asks for the moment of inertia about an axis the table doesn't list — through an edge of a rod, through a point at the rim of a disc, perpendicular to a holed disc. Two theorems generate every such answer from a table entry.

The same negative-mass logic resolved NEET 2016: a disc with a half-radius hole cut so its rim passes through the centre. Mass removed M/4; I of cut-out about disc centre = MR²/32 + MR²/16 = 3MR²/32 (parallel-axis); remaining I = MR²/2 − 3MR²/32 = 13MR²/32. NEET 2021's variant (90° sector removed from a ring): 3MR²/4.

The perpendicular-axis theorem applies to planar bodies only — rings, discs, thin plates. For a uniform disc, I_x = I_y by symmetry, so I_z = 2I_diameter, giving I_diameter = MR²/4 from I_z = MR²/2. The theorem fails for solid spheres, cylinders, or any 3-D body — always check the lamina condition.

Kinematics of rotational motion about a fixed axis

For uniform angular acceleration, the kinematic equations of rotation are direct copies of the linear ones, with θ, ω, α replacing x, v, a:

ω = ω₀ + α t  ·  θ = θ₀ + ω₀t + ½αt²  ·  ω² = ω₀² + 2α(θ − θ₀)

Three SUVAT equations for rotation

NEET 2022 used the first equation: a flywheel from 1200 to 3120 rpm in 16 s gives α = 4π rad/s². NEET 2016 used the third with a 50 cm disc at α = 2 rad/s² for 2 s; centripetal acceleration Rω² = 8 m/s² dominates over tangential Rα = 1 m/s², giving net ≈ 8 m/s². A related 2016 problem extracted tangential acceleration 0.1 m/s² from KE after two revolutions.

Dynamics of rotational motion about a fixed axis

The linear–rotational analogy is so complete that it can be set out in two parallel columns. Every translational quantity has a rotational counterpart, every translational equation has a rotational twin.

Work–energy theorem in rotational form: W = ½Iω² − ½Iω₀². NEET 2017 used τ = Iα directly: a rope on a hollow cylinder (I = MR², M = 3 kg, R = 0.4 m), pulled with 30 N gives τ = 12 N·m, α = 12/0.48 = 25 rad/s². NEET 2018 used W = ΔK_rot: three bodies of equal mass and radius at the same ω require W ∝ I to stop, so W_ring > W_disc > W_sphere. Power follows from dW = τ dθ: P = τω, mirroring P = Fv.

Rolling motion — the synthesis

Rolling without slipping is the canonical combination of translation and rotation, and it is the single most heavily tested topic in this chapter. Take a wheel of radius R rolling along a flat surface without slipping. The centre of mass moves with velocity v_cm; the wheel rotates with angular velocity ω. The "no slipping" condition fixes the relationship between them.

The kinetic energy of a rolling body is the sum of translational and rotational pieces — there is no double-counting if you measure both about the centre of mass:

For a body rolling down an incline of angle θ from rest through height h, energy conservation gives v² = 2gh / (1 + k²/R²) and the acceleration along the incline is a = g sin θ / (1 + k²/R²). The same denominator (1 + k²/R²) controls both. NEET 2016 used this to compare a disc and a sphere released from the same height; the sphere (k²/R² = 2/5) reaches the bottom first because its denominator (7/5) is smaller than the disc's (3/2).

NEET 2018 reframed it: K_t/(K_t + K_r) for a solid sphere = 1/(1 + k²/R²) = 5/7. Same physics, different wording. Three further facts: static friction at the contact point does no work (zero velocity there); friction is essential to begin rolling (no rolling on frictionless ice); the topmost point of a rolling body moves at 2v_cm, the bottommost at zero — tracing a cycloid.