Chapter 6: System of Particles & Rotational Motion
1. Centre of Mass (COM)
The point where the entire mass of a system is assumed to be concentrated.
COM = (m₁x₁ + m₂x₂) / (m₁ + m₂)
- Depends on mass distribution
- Motion of system can be described using COM
2. Centre of Mass of Rigid Body
- For uniform rod → COM at center
- For symmetrical objects → geometric center
3. Momentum Conservation
Total momentum remains constant if no external force acts.
4. Torque (Moment of Force)
Ï„ = r × F = rF sinθ
- Causes rotational motion
- Unit = N·m
5. Angular Momentum
L = r × p
- Conserved in absence of external torque
6. Conservation of Angular Momentum
If no external torque → angular momentum remains constant.
7. Equilibrium of Rigid Body
- Net force = 0
- Net torque = 0
8. Rotational Motion
θ = angular displacement
ω = angular velocity
α = angular acceleration
ω = angular velocity
α = angular acceleration
Equations of Motion
ω = ω₀ + αt
θ = ω₀t + ½Î±t²
ω² = ω₀² + 2αθ
θ = ω₀t + ½Î±t²
ω² = ω₀² + 2αθ
9. Linear vs Rotational Motion
| Linear | Rotational |
|---|---|
| Displacement (s) | Angular displacement (θ) |
| Velocity (v) | Angular velocity (ω) |
| Acceleration (a) | Angular acceleration (α) |
| Force (F) | Torque (Ï„) |
10. Moment of Inertia
I = Σmr²
- Rotational analog of mass
- Depends on mass distribution
Standard Values
- Rod (center) = (1/12)ML²
- Ring = MR²
- Solid sphere = (2/5)MR²
11. Radius of Gyration
I = Mk²
- k = radius of gyration
12. MCQ One-Liners
- Centre of mass depends on mass distribution.
- COM of uniform rod is at center.
- Torque = r × F.
- Unit of torque is N·m.
- Angular momentum = r × p.
- Angular momentum is conserved.
- Moment of inertia depends on mass and distance.
- Unit of I is kg·m².
- Radius of gyration = √(I/M).
- Equilibrium → net force and torque zero.
- Angular velocity unit = rad/s.
- Angular acceleration unit = rad/s².
- Rotational motion uses torque instead of force
- Ring has higher I than disc.
- Moment of inertia increases with distance.
- ω = dθ/dt
- α = dω/dt
- Rigid body has fixed shape.
- Rotation about fixed axis is pure rotation.
- Torque produces angular acceleration.
13. Numerical Problems with Solutions
1. Centre of Mass
m₁=2kg at x₁=0, m₂=3kg at x₂=4
COM = (2×0 + 3×4)/(5) = 12/5 = 2.4 m
2. Torque
F = 10 N, r = 2 m, θ = 90°
Ï„ = 2×10×sin90 = 20 Nm
3. Angular Momentum
m = 2 kg, v = 3 m/s, r = 2 m
L = r×mv = 2×2×3 = 12 kg·m²/s
4. Angular Velocity
ω₀=0, α=2 rad/s², t=5 s
ω = 0 + 2×5 = 10 rad/s
5. Angular Displacement
ω₀=2, α=2, t=3
θ = 2×3 + ½×2×9 = 6 + 9 = 15 rad
6. Moment of Inertia
m = 2 kg, r = 3 m
I = mr² = 2×9 = 18 kg·m²
7. Radius of Gyration
I = 20, M = 5
k = √(20/5) = √4 = 2 m
8. Conservation of Angular Momentum
I₁ω₁ = I₂ω₂
If I decreases → ω increases
