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System of Particles and Rotational Motion – Class 11 Physics Chapter 6 Notes & NCERT PDF

Class 11PhysicsChapter 6NCERT book: Physics Part I - Class 11

Notes and a simple summary of Chapter 6 of the NCERT Class 11 Physics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 6: System of Particles and Rotational Motion

Introduction

So far bodies were treated as point particles. Real bodies have size, and they can rotate. This chapter introduces the centre of mass, extends Newton's laws to systems of particles, and develops rotational analogues of linear motion — angular velocity, torque, angular momentum and moment of inertia — for rigid bodies (bodies whose shape does not change).

Key Concepts

6.1 Motion of a Rigid Body

A rigid body can have pure translation (every particle has the same velocity), rotation about a fixed axis (every particle moves in a circle centred on the axis), or a combination of both — like a cylinder rolling down an incline. A spinning top shows rotation about an axis that itself turns (precession).

6.2 Centre of Mass

For two particles on a line, X = (m₁x₁ + m₂x₂)/(m₁ + m₂). In general R = Σmᵢrᵢ / M. For a uniform symmetric body (ring, disc, sphere, rod) the centre of mass is at the geometric centre. Example 6.1: for three equal particles at the corners of a triangle, the CM is the centroid.

6.3 Motion of the Centre of Mass

M A = F(ext)

The centre of mass moves as if all the mass were concentrated there and all external forces acted on it; internal forces do not affect its motion. A projectile that explodes in mid-air: the fragments fly apart but the CM continues on the original parabola.

6.4 Linear Momentum of a System

P = M V; dP/dt = F(ext). If F(ext) = 0, total linear momentum is conserved and V is constant (radioactive decay of a nucleus at rest: fragments move so that the CM stays at rest).

6.5 Vector Product

A × B = A B sin θ n̂, direction by the right-hand rule. Not commutative: A × B = −B × A. i × j = k, j × k = i, k × i = j; A × A = 0.

6.6 Angular Velocity

v = ω × r; for circular motion v = ω r. Every particle of a rigid body rotating about a fixed axis has the same ω. Angular acceleration α = dω/dt.

6.7 Torque and Angular Momentum

  • Torque (moment of force): τ = r × F, magnitude r F sin θ; unit N m.
  • Angular momentum of a particle: l = r × p.
  • dl/dt = τ — the rotational analogue of F = dp/dt. For a system, dL/dt = τ(ext).
  • If τ(ext) = 0, L is conserved (Example 6.6: a particle moving with constant velocity has constant angular momentum about any point).

    • 6.8 Equilibrium of a Rigid Body

    Two conditions: ΣF = 0 (translational) and Στ = 0 (rotational). A couple — equal and opposite forces with different lines of action — produces rotation without translation (turning a bottle cap, a compass needle in a magnetic field).

  • Principle of moments (lever): load × load arm = effort × effort arm; mechanical advantage = F₁/F₂ = d₂/d₁.
  • Centre of gravity: the point where the total gravitational torque is zero; for uniform g it coincides with the centre of mass.

    • 6.9 Moment of Inertia

    I = Σ mᵢ rᵢ² — rotational inertia; it depends on mass and its distribution about the axis. Rotational KE = ½ I ω². Radius of gyration: I = M k². Standard results given by the book:

  • Thin ring about its axis: M R²
  • Ring about a diameter: M R²/2
  • Thin rod about a perpendicular axis through centre: M L²/12
  • Disc about its axis: M R²/2; about a diameter: M R²/4
  • Hollow cylinder about its axis: M R²; solid cylinder: M R²/2
  • Solid sphere about a diameter: 2 M R²/5

    • A flywheel has a large I so that a vehicle's motion stays smooth.

    6.10 Kinematics of Rotational Motion (fixed axis)

    Same form as linear kinematics with constant α: ω = ω₀ + α t; θ = ω₀ t + ½ α t²; ω² = ω₀² + 2 α θ.

    6.11 Dynamics of Rotational Motion

    τ = I α (analogue of F = m a); work done by torque W = τ θ; power P = τ ω.

    6.12 Angular Momentum about a Fixed Axis

    L = I ω, and dL/dt = τ. With no external torque, I ω = constant: a skater or a person on a rotating stool pulling in the arms reduces I and spins faster; divers tuck in to rotate faster.

    Worked Examples

  • Motor wheel (Ex 6.11). 1200 rpm → 3120 rpm in 16 s. ω₀ = 40π rad s⁻¹, ω = 104π rad s⁻¹. α = (104π − 40π)/16 = 4π rad s⁻². θ = 40π × 16 + ½ × 4π × 256 = 1152π rad = 576 revolutions.
  • Flywheel (Ex 6.12). M = 20 kg, R = 0.2 m, pull 25 N. I = MR²/2 = 0.4 kg m²; τ = 25 × 0.2 = 5 N m; α = τ/I = 12.5 rad s⁻². For 2 m of cord, θ = 10 rad; work = 25 × 2 = 50 J; ω² = 2αθ = 250, so KE = ½ × 0.4 × 250 = 50 J — equal, since there is no friction.
  • Bar on knife-edges (Ex 6.8). 70 cm bar, 4 kg, knife-edges 10 cm from each end, 6 kg load 30 cm from one end (g = 9.8). Taking moments about the nearer knife-edge: R₂ = (4g × 0.25 + 6g × 0.20)/0.5 ≈ 43 N, and R₁ = 10g − R₂ ≈ 55 N.
  • Ladder against a frictionless wall (Ex 6.9). 3 m, 20 kg, foot 1 m from the wall; height reached = √8 ≈ 2.83 m. Wall reaction F₃ = W × 0.5/2.83 ≈ 34.6 N; floor: normal 196 N and friction 34.6 N, resultant ≈ 199 N at about 80° to the horizontal.
  • Centre of mass on a line. Masses 2 kg at x = 0 and 3 kg at x = 2 m: X = (0 + 6)/5 = 1.2 m.

    • Formula / Key-Facts Sheet

  • R(CM) = Σmᵢrᵢ/M; M A(CM) = F(ext); P = M V(CM)
  • τ = r × F; l = r × p; dL/dt = τ(ext)
  • Equilibrium: ΣF = 0, Στ = 0
  • I = Σmr² = M k²; KE(rot) = ½ I ω²
  • τ = I α; W = τ θ; P = τ ω; L = I ω
  • ω = ω₀ + αt; θ = ω₀t + ½αt²; ω² = ω₀² + 2αθ
  • Linear ↔ rotational: m ↔ I, v ↔ ω, a ↔ α, F ↔ τ, p ↔ L

    • Exam Focus

    Typical questions: CM of simple systems (Exercise 6.1–6.3 types), vector products, torque and angular momentum of a particle, equilibrium of rods and ladders, moments of inertia from the standard list, rotational kinematics (rpm problems), and conservation of angular momentum.

  • Outline — "Why does a skater spin faster when pulling in her arms?": no external torque ⇒ L = Iω constant; pulling in arms reduces I ⇒ ω increases.
  • Outline — "Show that the CM of an exploding shell follows the original path": explosion forces are internal ⇒ M A(CM) = M g unchanged.
  • Outline — "Derive W = τθ": dW = F ds = F r dθ = τ dθ ⇒ W = ∫τ dθ.

    • Key Terms

  • Rigid body (दृढ़ पिंड)
  • Centre of mass (द्रव्यमान केंद्र)
  • Translational motion (स्थानांतरीय गति)
  • Rotational motion (घूर्णी गति)
  • Vector product (सदिश गुणनफल)
  • Angular velocity (कोणीय वेग)
  • Torque (बल आघूर्ण)
  • Angular momentum (कोणीय संवेग)
  • Couple (बल युग्म)
  • Principle of moments (आघूर्णों का सिद्धांत)
  • Centre of gravity (गुरुत्व केंद्र)
  • Moment of inertia (जड़त्व आघूर्ण)
  • Radius of gyration (घूर्णन त्रिज्या)
  • Flywheel (गतिपालक चक्र)

    • Common Mistakes

  • Assuming the centre of mass must lie inside the body (it lies at the centre of a ring).
  • Writing A × B = B × A; the vector product changes sign.
  • Using the wrong axis for moment of inertia — the value depends on the axis.
  • Forgetting to convert rpm to rad s⁻¹ (multiply by 2π/60).
  • Checking only ΣF = 0 for a rigid body; Στ = 0 is also required.
  • Thinking internal forces can change the motion of the centre of mass.

    • 💡 Key Learning Points

    • ✓Locate the centre of mass of simple systems and state that it moves under external forces only
    • ✓Use the vector product to define torque τ = r × F and angular momentum l = r × p
    • ✓Apply the conditions of equilibrium of a rigid body, including the principle of moments
    • ✓Use standard moments of inertia and the analogy τ = Iα, L = Iω
    • ✓Solve rotational kinematics problems and apply conservation of angular momentum

    👨‍🏫 Teaching Tips

    • →Balance irregular cardboard shapes on a pencil tip to find the centre of gravity
    • →Use a metre scale lever with coins to verify the principle of moments
    • →Spin on a rotating stool with dumbbells to show conservation of angular momentum
    • →Roll a ring and a disc of the same mass down an incline to show the effect of moment of inertia
    • →Build a linear-to-rotational analogy chart with students
    • →Use a door with the handle at different distances to feel how torque depends on r
    • →Work through the flywheel example (Ex 6.12) and compare work done with KE gained

    📋 Assessment Questions

    1. Define centre of mass and find it for two point masses on a line
    2. State the two conditions for equilibrium of a rigid body
    3. Derive the relation τ = dL/dt for a particle
    4. A wheel speeds up from 1200 rpm to 3120 rpm in 16 s — find α and the number of revolutions
    5. Explain with an example the conservation of angular momentum

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    Keep learning

    All chapters in Physics Part I - Class 11

    1. 1. Units and Measurement
    2. 2. Motion in a Straight Line
    3. 3. Motion in a Plane
    4. 4. Laws of Motion
    5. 5. Work, Energy and Power
    6. 6. System of Particles and Rotational Motion
    7. 7. Gravitation