System of Particles &
Rotational Motion
From Centre of Mass to Rolling Motion — every concept built analytically, not by rote.
1. Centre of Mass (COM)
COM is the weighted average position of all particles. JEE Advanced often tests COM of a system where a piece is removed — this is a high-yield trap. Master the "removal" method.
Definition & Formula
For a system of N particles with masses m₁, m₂, …, mₙ at positions r₁, r₂, …, rₙ:
COM is not always inside the body (e.g., hollow ring). It is the point where entire mass can be assumed to act for translational purpose only — not for rotation!
COM of Continuous Bodies
By symmetry, COM is at the geometric centre.
This is a common NEET & JEE question. Derived via integration.
Note: Less than semicircular ring — mass is distributed differently.
Solid hemisphere COM = 3R/8. Remember both!
Students often confuse COM of hollow hemisphere (R/2) with solid hemisphere (3R/8). In exams, the body type is specified — read carefully. Getting this wrong loses 4 marks in JEE.
COM of Removed Part (Cavity Method)
If a piece of mass m₂ at position r₂ is removed from original body (mass M, COM at R):
For cavity problems, always treat the complete body first. Find its COM. Then subtract the removed part using the formula above. This method avoids integration and saves 3+ minutes in JEE.
Motion of COM
If Fext = 0, then vcm = constant (or zero). This is Newton's 1st law for a system.
JEE Advanced often gives a problem where a man walks on a boat (or similar). The key: in absence of external force, COM doesn't move. Set up: m₁Δx₁ + m₂Δx₂ = 0. This is the master equation for all such problems. The boat and man displacements are NOT equal — they're inversely proportional to masses.
2. Angular Quantities
The beauty of this topic is the analogy with linear motion. Every linear quantity has an angular counterpart. Master the analogy table — it saves derivation time in exams.
The Analogy Table
| Linear Quantity | Symbol | Angular Counterpart | Symbol | Relation |
|---|---|---|---|---|
| Displacement | s | Angular Displacement | θ | s = rθ |
| Velocity | v | Angular Velocity | ω | v = rω |
| Acceleration | a | Angular Acceleration | α | at = rα |
| Mass | m | Moment of Inertia | I | — |
| Force | F | Torque | τ | — |
| Momentum (p) | mv | Angular Momentum | L | L = Iω |
| KE = ½mv² | — | KE = ½Iω² | — | — |
| Work = F·s | — | Work = τ·θ | — | — |
Angular Velocity (ω)
Angular Acceleration (α)
Angular quantities are vectors. Direction is given by the Right-Hand Rule (curl fingers in direction of rotation, thumb points along ω). For a body rotating anticlockwise when viewed from above, ω points upward (along +z).
Components of Acceleration in Circular Motion
Due to change in magnitude of velocity. Points along tangent.
Due to change in direction. Points toward centre.
If angular speed is constant (uniform circular motion), α = 0, so aₜ = 0. But centripetal acceleration still exists! Many students say "acceleration = 0" for uniform circular motion — completely wrong. The velocity is changing in direction, so acceleration is non-zero.
Vector Product (Cross Product)
The cross product is fundamental to torque and angular momentum.
- A × B = −(B × A) [Anti-commutative]
- A × A = 0
- î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ
- ĵ × î = −k̂ (reverse order, reverse sign)
3. Torque & Angular Momentum
Torque (τ)
Torque is the "turning effect" of a force. It's the rotational analogue of force.
The perpendicular distance from the axis to the line of action of force is called the "moment arm" or "lever arm." Torque = Force × Lever Arm. This is the key insight — a force at the hinge produces ZERO torque (r = 0 or θ = 0). A force along the radial direction also produces zero torque (sinθ = 0).
Analogous to F = ma. This is Newton's 2nd Law for rotation.
Body either at rest or rotating uniformly. This is rotational equilibrium.
Torque depends on the choice of axis. If a question says "torque about point O," you must use position vector from O. If the axis changes, the torque changes. Always specify the axis when calculating torque.
Angular Momentum (L)
Angular momentum of a particle moving in a straight line about an external point is NOT zero (unless the line passes through that point). L = mvd, where d = perpendicular distance from the point to the line of motion. This is a classic NEET + JEE trap.
Law of Conservation of Angular Momentum
Pulls arms in → I decreases → ω increases (spin faster). L = Iω = constant.
Gravity acts through the sun (no torque). L conserved. Sweeps equal areas in equal time (Kepler's 2nd law).
JEE favourite. Bullet embeds in disc. L conserved, but KE is NOT conserved (inelastic). Calculate ω after collision from L.
Person throws ball, platform rotates to conserve L. Classic NEET numerical.
In any problem involving angular momentum conservation: (1) Identify if net external torque = 0. (2) Write L_initial = L_final. (3) Expand L = Iω for each part. Bullets, person-on-disc, and gymnast problems all follow this template. Recognise the type — it saves 2 minutes.
Newton's 2nd Law for Rotation
The equation τ = dL/dt is MORE general than τ = Iα. It holds even when I changes (like a skater). In problems where mass redistributes (like a sliding bead on a rotating rod), use τ = dL/dt, not τ = Iα. This distinction separates toppers from average students.
4. Moment of Inertia (I)
Moment of inertia is the rotational analogue of mass. It quantifies how difficult it is to change a body's angular velocity.
Unlike mass, MI depends on the axis of rotation AND the distribution of mass. A solid sphere has I = 2MR²/5 about its diameter, but I = 7MR²/5 about a tangential axis. Always specify the axis!
Standard Moments of Inertia
| Body | Axis | MI (I) |
|---|---|---|
| Thin rod (length L) | Through centre ⊥ length | ML²/12 |
| Thin rod (length L) | Through one end ⊥ length | ML²/3 |
| Circular Ring (radius R) | Through centre ⊥ plane | MR² |
| Circular Ring (radius R) | Diameter | MR²/2 |
| Circular Disc (radius R) | Through centre ⊥ plane | MR²/2 |
| Circular Disc (radius R) | Diameter | MR²/4 |
| Solid Sphere (radius R) | Diameter | 2MR²/5 |
| Hollow Sphere (radius R) | Diameter | 2MR²/3 |
| Solid Cylinder (radius R) | Own axis | MR²/2 |
| Hollow Cylinder (radius R) | Own axis | MR² |
| Rectangular plate (a×b) | Through centre ⊥ plane | M(a²+b²)/12 |
Parallel Axis Theorem
where d = distance between parallel axes (one must pass through COM).
The parallel axis theorem adds Md² to the COM axis MI. Since Md² ≥ 0, the COM axis gives the MINIMUM MI. The axis through COM is always the most efficient for rotation. JEE often asks: "about which axis is MI minimum?" → Always through COM.
Perpendicular Axis Theorem
Valid only for planar (2D) bodies. If x and y are in the plane, z is perpendicular to the plane.
Perpendicular axis theorem applies ONLY to 2D (planar) bodies — disc, ring, rectangular plate. NEVER apply it to 3D bodies like sphere or cylinder. This is a guaranteed mistake-maker in JEE Main.
For a disc: I_diameter = MR²/4. Using perpendicular axis: I_perpendicular = I_x + I_y = MR²/4 + MR²/4 = MR²/2. ✓ This confirms the formula. Use perpendicular axis to cross-verify your answers quickly.
5. Equilibrium of Rigid Bodies
A rigid body is in equilibrium when it has no translational or rotational acceleration.
Vector sum of all external forces = 0. Body does NOT accelerate linearly.
Net torque about any point = 0. Body does NOT have angular acceleration.
For rotational equilibrium, you can choose ANY point to calculate torques — the net will still be zero. Choose the point that eliminates the most unknowns (usually the point where an unknown force acts, making its torque zero). This is the "lever arm trick" that solves 80% of equilibrium problems.
Types of Equilibrium
COM is at lowest point. Displaced slightly → restoring force. Returns to original position. E.g., ball in a bowl.
COM is at highest point. Displaced slightly → moves further away. E.g., pencil balanced on tip.
COM stays at same height. Displaced → stays in new position. E.g., ball on flat surface.
Toppling Condition
A body topples when the vertical line through its COM falls OUTSIDE the base area. For a block on an incline, find the critical angle θ where the COM is directly above the edge of the base. This is a classic JEE Advanced problem type.
6. Rolling Motion
Rolling motion is the superposition of pure translation (of COM) + pure rotation (about COM). This is where most students lose marks in JEE. The key: the contact point has ZERO velocity in pure rolling.
Condition for Pure Rolling
Think of rolling as: velocity of any point = translational velocity (v_cm) + rotational velocity (ωr). At contact point: v_cm (forward) + ωR (backward) = 0 → v_cm = ωR. At top point: v_cm + ωR = 2v_cm (maximum velocity).
Velocity of Different Points
Momentarily at rest (no slipping). This is what makes rolling different from sliding.
Moves with translational velocity only.
Maximum velocity. Catches most students off-guard.
Kinetic Energy of Rolling Body
| Body | k²/R² | % Rotational KE |
|---|---|---|
| Hollow Cylinder (Ring) | 1 | 50% |
| Solid Cylinder (Disc) | 1/2 | 33.3% |
| Hollow Sphere | 2/3 | 40% |
| Solid Sphere | 2/5 | 28.6% |
Rolling on Inclined Plane
Smaller k²/R² → higher acceleration → reaches bottom first.
Solid sphere > Solid cylinder > Hollow sphere > Hollow cylinder. Remember: S–S–H–H. The solid sphere always wins. Regardless of mass or radius — only shape matters for pure rolling on incline.
The acceleration formula a = g sinθ/(1 + k²/R²) is for pure rolling WITHOUT friction acting externally. If the body slides instead of rolls, friction is kinetic and the problem changes entirely. Always check if "rolling without slipping" is stated.