This chapter extends mechanics from point particles to extended rigid bodies, introducing the centre of mass, torque, angular momentum, moment of inertia, and the rotational versions of Newton’s laws and kinetic energy. It ties everything together with rolling motion, where a body translates and rotates at once. It is a high-yield NEET topic that blends conceptual reasoning with numerical problems, especially on rolling, conservation of angular momentum, and moment of inertia.
Table of Contents
- Key Concepts - Centre of mass, vector product, torque, angular momentum, moment of inertia, rolling
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Centre of Mass
The centre of mass (CM) of a body or system is the single point at which the whole mass can be taken to be concentrated for describing its translational motion. When you flip a spanner in the air, the CM follows a smooth parabola even though the spanner tumbles wildly.
For a system of particles, the position of the centre of mass is:
R = (m₁r₁ + m₂r₂ + … + mₙrₙ) / (m₁ + m₂ + … + mₙ) = Σmᵢrᵢ / M
- For a two-particle system on a line: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂).
- For symmetric uniform bodies (sphere, ring, rod, disc), the CM lies at the geometric centre.
- The CM may lie outside the body - for example, at the centre of a ring or a horseshoe.
2. Motion of the Centre of Mass
The centre of mass moves as if the total external force acted on the total mass concentrated there. Internal forces between particles cancel in pairs (Newton’s third law) and never shift the CM.
M·a_cm = F_external, and total momentum P = M·v_cm
- If the net external force is zero, v_cm is constant - the CM moves with uniform velocity (or stays at rest).
- This is why, when a shell explodes in mid-air, the fragments’ CM keeps following the original parabolic path.
3. Vector (Cross) Product of Two Vectors
The vector product (or cross product) of two vectors gives a third vector perpendicular to both. It is the maths behind torque and angular momentum.
A × B = AB sin θ n̂, where n̂ is given by the right-hand rule.
- Magnitude: |A × B| = AB sin θ (equals the area of the parallelogram formed by A and B).
- It is anti-commutative: A × B = −(B × A).
- î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ; and î × î = ĵ × ĵ = k̂ × k̂ = 0.
- If A and B are parallel (θ = 0°), the cross product is zero.
4. Angular Velocity and Angular Acceleration
Angular velocity (ω) is the rate of change of angular displacement. It is a vector along the axis of rotation, given by the right-hand rule.
ω = dθ/dt, and linear velocity v = ω × r (so v = ωr in magnitude).
Angular acceleration (α) is the rate of change of angular velocity:
α = dω/dt
- SI units: ω in rad/s, α in rad/s².
- All particles of a rigid body share the same ω and α, but different linear speeds (greater for points farther from the axis).
5. Torque (Moment of Force)
Torque is the turning effect of a force about an axis - the rotational analogue of force. A longer spanner turns a bolt more easily because torque depends on the distance from the axis.
τ = r × F, so τ = rF sin θ = F × (perpendicular distance)
- SI unit: N·m. It is a vector quantity.
- Torque is maximum when the force is applied perpendicular to the position vector (θ = 90°).
- The rotational form of Newton’s second law is τ = Iα.
6. Angular Momentum
Angular momentum (L) is the rotational analogue of linear momentum - the “quantity of rotation” a body has.
L = r × p = r × mv, and for a rigid body L = Iω
- SI unit: kg·m²/s (or J·s). It is a vector quantity.
- The relation between torque and angular momentum is τ = dL/dt - the rotational version of F = dp/dt.
7. Conservation of Angular Momentum
If the net external torque on a system is zero, its total angular momentum remains constant.
If τ_ext = 0, then L = Iω = constant, so I₁ω₁ = I₂ω₂
This is why a spinning ice-skater speeds up when she pulls her arms in: reducing I forces ω to increase so that Iω stays fixed. A diver curling into a ball to spin faster uses the same principle.
8. Equilibrium of a Rigid Body
A rigid body is in mechanical equilibrium when both its linear acceleration and angular acceleration are zero. This needs two conditions to be satisfied together.
- Translational equilibrium: the net external force is zero (ΣF = 0).
- Rotational equilibrium: the net external torque is zero (Στ = 0).
A couple is a pair of equal, opposite, non-collinear forces; it produces pure rotation with zero net force. The principle of moments for a lever states: load × load arm = effort × effort arm.
9. Moment of Inertia
Moment of inertia (I) is the rotational analogue of mass - it measures a body’s resistance to a change in its rotational motion. It depends not just on mass but on how that mass is distributed about the axis.
I = Σmᵢrᵢ² = ∫r² dm
- SI unit: kg·m². It is a scalar (for a fixed axis).
- Rotational kinetic energy: KE = ½Iω².
- The same body can have different I about different axes - mass farther from the axis means larger I.
10. Radius of Gyration
The radius of gyration (k) is the distance from the axis at which the whole mass of the body can be assumed concentrated to give the same moment of inertia.
I = Mk², so k = √(I/M)
- SI unit: metre. It depends on the axis of rotation and on mass distribution.
- For a thin rod about its centre, k = L/√12; for a solid sphere about a diameter, k = √(2/5)·R.
11. Theorems of Perpendicular and Parallel Axes
These two theorems let you find the moment of inertia about a new axis from a known one - saving you long integrations in the exam.
Perpendicular Axes Theorem (for plane laminae only)
The moment of inertia of a planar body about an axis perpendicular to its plane equals the sum of its moments of inertia about two mutually perpendicular axes in the plane, meeting at that point.
I_z = I_x + I_y
Parallel Axes Theorem (for any body)
The moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass plus Md², where d is the distance between the axes.
I = I_cm + Md²
12. Moments of Inertia of Standard Bodies
You must memorise these standard results - they appear directly in numericals and rolling problems.
| Body | Axis | Moment of Inertia |
|---|---|---|
| Thin rod (length L) | Through centre, ⊥ to rod | ML²/12 |
| Thin rod (length L) | Through one end, ⊥ to rod | ML²/3 |
| Ring (radius R) | Through centre, ⊥ to plane | MR² |
| Disc (radius R) | Through centre, ⊥ to plane | ½MR² |
| Solid sphere (radius R) | About a diameter | (2/5)MR² |
| Hollow sphere (radius R) | About a diameter | (2/3)MR² |
| Solid cylinder (radius R) | About its own axis | ½MR² |
13. Kinematics and Dynamics of Rotational Motion
Rotation about a fixed axis obeys equations identical in form to the linear equations of motion - just swap the linear quantities for their angular partners.
| Linear | Rotational |
|---|---|
| v = u + at | ω = ω₀ + αt |
| s = ut + ½at² | θ = ω₀t + ½αt² |
| v² = u² + 2as | ω² = ω₀² + 2αθ |
| F = ma | τ = Iα |
| P = Fv | P = τω |
Work done by a torque is W = τθ, and rotational power is P = τω.
14. Rolling Motion
Rolling without slipping combines translation of the centre of mass with rotation about the CM. The contact point is momentarily at rest, which is the condition v_cm = ωR.
The total kinetic energy of a rolling body splits into two parts:
KE = ½Mv_cm² + ½Iω² = ½Mv²(1 + k²/R²)
For a body rolling down an incline of angle θ without slipping, the acceleration is:
a = g sin θ / (1 + k²/R²)
- Smaller k²/R² → larger acceleration. A solid sphere (k²/R² = 2/5) beats a disc (½), which beats a ring (1).
- This is why a solid sphere reaches the bottom of a ramp first, regardless of mass or radius.
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Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 6–8 marks | Centre of mass, torque, moment of inertia, conservation of angular momentum |
| JEE Main / Advanced | 2–4 questions (heavy weight) | Rolling motion, moment of inertia (theorems), angular momentum, pulley with mass |
| NEET | 2–3 questions | Moment of inertia of standard bodies, torque, rotational KE, conservation of L |
[TABLE: Question-type split - VSA (1 mark): definitions, units, standard MI values; SA (2–3 marks): torque/angular momentum numericals, theorems of axes; LA (5 marks): conservation of angular momentum, rolling on an incline derivation.]
Important Definitions
| Term | Definition |
|---|---|
| Centre of mass | Point where the total mass of a system is taken to be concentrated: R = Σmᵢrᵢ/M |
| Vector product | A × B = AB sin θ n̂ - a vector perpendicular to both, by the right-hand rule |
| Angular velocity | Rate of change of angular displacement: ω = dθ/dt (a vector along the axis) |
| Torque | Turning effect of a force: τ = r × F = rF sin θ |
| Angular momentum | Rotational analogue of momentum: L = r × p = Iω |
| Conservation of angular momentum | L = Iω is constant when net external torque is zero |
| Moment of inertia | Resistance to change in rotation: I = Σmᵢrᵢ² |
| Radius of gyration | Distance k where I = Mk², so k = √(I/M) |
| Parallel axes theorem | I = I_cm + Md² |
| Rolling condition | Rolling without slipping: v_cm = ωR |
Solved Examples
Example 1
Two particles of masses 2 kg and 3 kg are placed at x = 0 m and x = 5 m. Find the position of their centre of mass.
Answer: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2×0 + 3×5)/(2 + 3) = 15/5 = 3 m from the origin.
Example 2
A force F = (2î + 3ĵ) N acts at a point r = (4î + ĵ) m. Find the torque about the origin.
Answer: τ = r × F = (4î + ĵ) × (2î + 3ĵ) = (4×3 − 1×2)k̂ = (12 − 2)k̂ = 10 k̂ N·m.
Example 3
A wheel of moment of inertia 0.5 kg·m² rotates at 20 rad/s. Find its angular momentum and rotational kinetic energy.
Answer: L = Iω = 0.5 × 20 = 10 kg·m²/s. KE = ½Iω² = ½ × 0.5 × 20² = 100 J.
Example 4
A disc of mass 2 kg and radius 0.5 m rotates about its central axis. Find its moment of inertia and radius of gyration.
Answer: I = ½MR² = ½ × 2 × 0.5² = 0.25 kg·m². k = √(I/M) = √(0.25/2) = √0.125 ≈ 0.354 m.
Example 5
A skater spinning at 3 rad/s with I = 6 kg·m² pulls in her arms, reducing I to 2 kg·m². Find her new angular speed.
Answer: By conservation of angular momentum, I₁ω₁ = I₂ω₂. ω₂ = (6 × 3)/2 = 18/2 = 9 rad/s.
Example 6
A solid sphere rolls without slipping down an incline of angle 30°. Find its acceleration. (g = 10 m/s², k²/R² = 2/5)
Answer: a = g sin θ/(1 + k²/R²) = (10 × 0.5)/(1 + 2/5) = 5/(7/5) = 25/7 ≈ 3.57 m/s².
Important Questions for Board Exams
1-Mark Questions (VSA)
- Define moment of inertia and give its SI unit.
- Why does a spinning skater speed up when she folds her arms in?
- Can the centre of mass of a body lie outside the body? Give one example.
- What is the value of î × ĵ?
- State the rolling-without-slipping condition.
2–3-Mark Questions (SA)
- State and explain the perpendicular axes theorem. To which bodies does it apply?
- Define torque and angular momentum, and write the relation between them.
- Derive the relation τ = Iα starting from the definition of torque.
- A ring, a disc and a solid sphere of equal mass and radius roll down the same incline. Which reaches the bottom first and why?
5-Mark Questions (LA)
- State the parallel axes theorem and use it to find the moment of inertia of a thin rod about an axis through one end perpendicular to its length.
- State and prove the law of conservation of angular momentum from τ = dL/dt, and explain two real-life applications.
- Derive the expression for the acceleration of a body rolling without slipping down an inclined plane of angle θ.
Quick Revision Points
- Centre of mass: R = Σmᵢrᵢ/M; moves under net external force only (internal forces cancel)
- Total momentum P = Mv_cm; if F_ext = 0, v_cm is constant
- Vector product: A × B = AB sin θ n̂; anti-commutative; î × ĵ = k̂
- Angular velocity ω = dθ/dt; v = ω × r; angular acceleration α = dω/dt
- Torque τ = r × F = rF sin θ; rotational Newton’s law τ = Iα
- Angular momentum L = r × p = Iω; τ = dL/dt
- Conservation of L: if τ_ext = 0, then I₁ω₁ = I₂ω₂
- Moment of inertia I = Σmᵢrᵢ²; radius of gyration k = √(I/M); KE = ½Iω²
- Perpendicular axes: I_z = I_x + I_y (laminae); Parallel axes: I = I_cm + Md²
- Standard MI: rod ML²/12, ring MR², disc ½MR², solid sphere (2/5)MR², hollow sphere (2/3)MR²
- Rolling: v_cm = ωR; KE = ½Mv²(1 + k²/R²); a = g sin θ/(1 + k²/R²)
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Centre of Mass
One imaginary point where the whole mass of a system seems to act.
For uniform symmetric bodies, COM sits at the geometric centre.
- Heavier parts pull the COM toward themselves
- Two-particle COM divides the line inversely as masses: m1r1 = m2r2
- COM can lie in empty space (ring, L-shape)
Motion of the Centre of Mass
The COM moves as if all external force acts on the total mass there.
Internal forces (explosions, collisions) never shift the COM.
- An exploding shell’s COM keeps the original parabola
- If F_ext = 0, v_cm stays constant (momentum conserved)
- Man walks on a frictionless boat → boat recoils, COM fixed
Torque
The turning effect of a force about an axis.
r sinθ is the moment arm; force through the axis gives zero torque.
- Only the perpendicular force component turns the body
- Direction follows the right-hand rule
- A couple gives pure rotation (zero net force)
Angular Momentum
The ‘amount of spin’ a body carries about a point.
If net external torque = 0, L stays constant.
- Skater pulls arms in → I drops → ω rises (L fixed)
- Rotational form of Newton’s 2nd law: τ = dL/dt
- Straight-line particle still has L about an off-line point
Moment of Inertia
Rotational inertia: resistance to angular acceleration.
Depends on mass AND how it’s distributed about the axis.
- Farther mass from axis → larger I
- Hollow sphere ⅔MR2; rod about centre 1/12 ML2
- Radius of gyration k: I = Mk2, k = √(I/M)
Axis Theorems
Two shortcuts to find I about a shifted or new axis.
Perpendicular-axis theorem works only for flat laminae.
- Parallel axis shifts by distance d from the COM axis
- Rod about end = 1/12 ML2 + M(L/2)2 = ⅓ ML2
- Perpendicular axis: z-axis ⊥ to the plane of the lamina
Rotational Dynamics
Every linear law has a rotational twin.
Dictionary: F→τ, m→I, v→ω, a→α, p→L.
- τ = Iα is Newton’s 2nd law for rotation
- Angular kinematics apply when α is constant
- Work W = τθ, Power P = τω
Rotational Kinetic Energy
Energy stored in a spinning body.
Work–energy theorem holds: net rotational work = ΔKE_rot.
- Skater pulling arms in does work; KE rises though L is fixed
- Identify which of L, KE, ω is conserved before solving
- Total KE adds translational + rotational parts
Rolling Without Slipping
A rolling body translates and rotates together.
Contact point is instantaneously at rest; friction is static, does no work.
- Total KE = ½Mv2(1 + k2/R2)
- k2/R2: ring 1, disc ½, solid sphere ⅖
- The contact point is the instantaneous axis of rotation
Rolling Down an Incline
Smaller k2/R2 means a faster body — order is set by shape, not mass.
Independent of mass and radius.
- Solid sphere beats disc beats ring down the slope
- Larger k2/R2 → smaller acceleration, later arrival
- Energy: mgh splits into translation + rotation
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Frequently Asked Questions
The centre of mass is the mass-weighted average position of all particles in a system, given by x_cm = (sum of m_i x_i) / (sum of m_i). It moves as if the entire mass and all external force were concentrated at that single point, so F_ext = M a_cm.
Torque is the turning effect of a force about an axis, tau = r F sin(theta), and it is what causes angular acceleration. Moment of inertia, I = sum of m_i r_i squared, is the body’s resistance to that angular acceleration and depends on how mass is distributed about the axis. They are linked by the rotational form of Newton’s second law, tau = I alpha.
Angular momentum L = I omega is conserved whenever the net external torque on the system is zero, since tau = dL/dt. This explains why an ice skater spins faster when pulling in their arms: the moment of inertia drops, so the angular velocity rises to keep L constant.
For rolling without slipping the acceleration is a = g sin(theta) / (1 + k squared / R squared), which is independent of mass and radius. A solid sphere has the smallest k squared / R squared (2/5) while a ring has the largest (1), so the sphere has the greatest acceleration and reaches the bottom first.
Yes, System of Particles and Rotational Motion is part of the NEET Physics syllabus and is consistently a high-weightage chapter, typically contributing one to three questions. Questions commonly test moment of inertia, the parallel and perpendicular axis theorems, conservation of angular momentum, and rolling motion.