System of Particles and Rotational Motion
'Give me a lever long enough and a place to stand, and I will move the earth.' — Archimedes
1. Chapter Overview
So far, we treated objects as POINT masses. But real objects have SIZE and SHAPE. This chapter extends mechanics to EXTENDED bodies and ROTATIONAL motion. You will learn about the CENTRE OF MASS, TORQUE, ANGULAR MOMENTUM, and the MOMENT OF INERTIA — the rotational analogue of mass.
2. Centre of Mass
- Definition: The POINT where the ENTIRE mass of a system is assumed to be concentrated (for translational motion)
- For two particles: X_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
- For n particles: r_cm = (Σmᵢrᵢ)/(Σmᵢ) = (Σmᵢrᵢ)/M
- In vector form: R_cm = (m₁r₁ + m₂r₂ + ... + mₙrₙ)/(m₁ + m₂ + ... + mₙ)
Properties of Centre of Mass
- The CM moves as if ALL external forces act at the CM
- If NO external force acts, velocity of CM is CONSTANT
- The CM of a symmetric body lies at its geometric centre
- The CM need NOT lie within the body (e.g., a ring, a horseshoe)
Motion of Centre of Mass
- V_cm = (Σmᵢvᵢ)/M
- A_cm = (Σmᵢaᵢ)/M = F_ext/M
- Law: The CM accelerates only due to NET external force
3. Torque (Moment of Force)
- Torque τ = r × F (cross product)
- Magnitude: τ = rFsinθ = r⊥F (where r⊥ is the PERPENDICULAR distance from axis to line of force)
- SI unit: N·m
- Direction: Along the axis of rotation (right-hand rule)
Couple
- Two EQUAL and OPPOSITE forces that produce rotation
- Torque of a couple = Force × perpendicular distance BETWEEN the forces
4. Angular Momentum
- Angular Momentum L = r × p = r × mv
- Magnitude: L = mvr sinθ = r⊥p
- SI unit: kg·m²/s
- Relation with torque: τ = dL/dt (Net torque = rate of change of angular momentum)
- Conservation of Angular Momentum: If net external torque = 0, L = CONSTANT
Examples of Conservation
- Diver tucking in during somersault (reduces I, increases ω)
- Ice skater spinning — pulling arms in increases spin speed
- Planet orbiting the Sun (area swept per unit time is constant — Kepler's second law)
5. Moment of Inertia (I)
- Definition: Rotational analogue of MASS
- I = Σmᵢrᵢ² (sum of mass × distance² from axis)
- SI unit: kg·m²
- Depends on: Mass distribution AND axis of rotation
Key Moments of Inertia
| Body | Axis | I |
|---|---|---|
| Thin ring | Through centre, perpendicular to plane | MR² |
| Disc | Through centre, perpendicular to plane | MR²/2 |
| Solid sphere | Through centre | 2MR²/5 |
| Thin rod | Through centre, perpendicular to rod | ML²/12 |
| Thin rod | Through end, perpendicular to rod | ML²/3 |
Parallel Axis Theorem
- I = I_cm + Md² (d = distance between axes)
- Used when axis is PARALLEL to one through CM
Perpendicular Axis Theorem
- I_z = I_x + I_y (for a LAMINA in x-y plane)
- Only for PLANAR objects
Worked Problem
Q: Find MI of a thin ring of mass 2 kg and radius 0.5 m about its diameter. A: For a ring, I_diameter = ½MR² = ½×2×(0.5)² = 0.25 kg·m²
6. Rotational Kinetic Energy
- K_rot = ½Iω² (analogous to ½mv²)
- Total KE of rolling body = ½MV² + ½Iω²
- For a rolling body: K_rot / K_trans = I/(MR²)
Worked Problem
Q: A solid sphere rolls down an incline of height 10 m. Find speed at bottom. A: Energy conservation: Mgh = ½MV² + ½Iω² For sphere: I = 2/5 MR², ω = V/R Mgh = ½MV² + ½×(2/5 MR²)×(V²/R²) = ½MV² + (1/5)MV² = (7/10)MV² V = √(10gh/7) = √(10×10×10/7) = √(1000/7) = 11.95 m/s
7. Equations of Rotational Motion
Analogy between LINEAR and ANGULAR quantities:
| Linear | Angular |
|---|---|
| Mass m | Moment of Inertia I |
| Force F | Torque τ |
| Linear momentum p = mv | Angular momentum L = Iω |
| F = ma | τ = Iα |
| W = F·s | W = τ·θ |
| K = ½mv² | K = ½Iω² |
Rotational Kinematics (constant α)
- ω = ω₀ + αt
- θ = ω₀t + ½αt²
- ω² = ω₀² + 2αθ
8. Common Mistakes
- Torque is NOT the same as force: A large force applied at the axis produces ZERO torque
- MI depends on the axis: The SAME body has DIFFERENT I for different axes
- Angular momentum ≠ momentum × distance: L = r × p (cross product), not rp
- Conservation of L requires net external TORQUE = 0, not force = 0
- Rolling vs. sliding: Pure rolling implies v_cm = ωR. If this condition breaks, sliding occurs
9. CBSE Exam Focus
- Centre of mass derivation for two-particle and n-particle systems
- Torque and angular momentum relation τ = dL/dt
- MI of various bodies using theorems (5-mark)
- Rolling without slipping — energy method (5-mark)
- Angular momentum conservation problems (diver, skater)
- Parallel and perpendicular axis theorems applications
10. Key Formulas
- r_cm = Σmᵢrᵢ/Σmᵢ
- τ = r × F, |τ| = rFsinθ
- L = r × p, |L| = mvr sinθ
- τ = dL/dt = Iα
- I = Σmᵢrᵢ²
- K_rot = ½Iω²
- Rolling KE = ½MV² + ½Iω²
11. Self-Test (5+ Q&A)
Q1: Two masses 2 kg and 3 kg are at (1,0) and (4,0). Find CM. A: X_cm = (2×1 + 3×4)/(5) = 14/5 = 2.8 m from origin.
Q2: The MI of a disc about a perpendicular axis through centre is ½MR². Find MI about a parallel axis at the rim. A: Using parallel axis theorem: I = I_cm + Md² = ½MR² + MR² = (3/2)MR².
Q3: A torque of 10 N·m acts on a body with I = 5 kg·m² for 4 s from rest. Find ω achieved. A: α = τ/I = 10/5 = 2 rad/s². ω = αt = 2×4 = 8 rad/s.
Q4: A diver with I = 15 kg·m² in tucked position rotates at 3 rad/s. On straightening, I becomes 25 kg·m². Find new ω. A: Conservation of L: I₁ω₁ = I₂ω₂ → 15×3 = 25×ω₂ → ω₂ = 1.8 rad/s.
Q5: State the perpendicular axis theorem. Can it be applied to a sphere? A: Perpendicular axis theorem: I_z = I_x + I_y for a lamina in x-y plane. It CANNOT be applied to a sphere because a sphere is NOT a planar object (lamina).
12. Conclusion
This chapter bridges ROTATIONAL and TRANSLATIONAL mechanics. The centre of mass concept simplifies extended-body problems, while torque and angular momentum provide the rotational analogues of force and linear momentum. The moment of inertia is a CRITICAL concept that depends on BOTH mass distribution and axis choice. Rolling motion beautifully combines translation and rotation. These concepts are essential for engineering entrance exams.
