What this chapter is about
Until now, you have treated objects as point masses — their entire mass concentrated at a single point. This chapter extends mechanics to extended bodies, objects with finite size and shape. When a cricket bat swings, a fan rotates, or a potter's wheel spins, different parts of the object move differently. Understanding such motion requires new concepts: the centre of mass, torque, angular momentum and moment of inertia.
You will learn how to locate the centre of mass of a system of particles and of continuous bodies, and why this special point moves as if all external forces act there. The chapter then develops rotational kinematics (describing rotation) and rotational dynamics (explaining why rotation changes). You will see striking parallels: just as force produces linear acceleration, torque produces angular acceleration; just as mass resists linear acceleration, moment of inertia resists angular acceleration.
By the end, you should be able to analyse rolling motion, understand why an ice-skater spins faster when she pulls her arms in, and solve problems involving both translation and rotation together.
Key ideas
- Centre of mass (CM): The point where the entire mass of a system can be thought to be concentrated for analysing its translational motion. For two particles of masses m₁ and m₂ at positions x₁ and x₂, the CM is at x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂).
- Motion of the CM: The CM of a system moves as though all external forces act on a point mass equal to the total mass located there. Internal forces do not affect the CM's motion.
- Torque (τ): The rotational analogue of force. For a force F applied at position r from the axis, τ = r × F. Its magnitude is τ = rF sin θ, where θ is the angle between r and F. SI unit: N m.
- Moment of inertia (I): The rotational analogue of mass. It measures how mass is distributed about the axis. For discrete particles, I = Σ mᵢrᵢ². SI unit: kg m².
- Angular momentum (L): The rotational analogue of linear momentum. For a particle, L = r × p. For a rigid body rotating about a fixed axis, L = Iω. SI unit: kg m²/s.
- Rotational Newton's second law: Net external torque equals rate of change of angular momentum. τ_net = dL/dt. For a rigid body with constant I, τ_net = Iα, where α is angular acceleration.
- Conservation of angular momentum: If net external torque on a system is zero, its total angular momentum remains constant.
- Rolling motion: Combines rotation about the CM and translation of the CM. For rolling without slipping, v_cm = Rω, where R is the radius.
Formulas and facts to remember
- Formula: x_cm = Σ mᵢxᵢ / Σ mᵢ · Meaning: x-coordinate of centre of mass
- Formula: τ = r F sin θ · Meaning: Magnitude of torque
- Formula: I = Σ mᵢrᵢ² · Meaning: Moment of inertia of discrete particles
- Formula: I_parallel = I_cm + Md² · Meaning: Parallel-axis theorem: I about any axis equals I about a parallel axis through CM plus Md²
- Formula: L = Iω · Meaning: Angular momentum of a rotating rigid body
- Formula: τ = Iα · Meaning: Newton's second law for rotation (constant I)
- Formula: KE_rot = ½ Iω² · Meaning: Rotational kinetic energy
- Formula: KE_total (rolling) = ½ Mv_cm² + ½ I_cm ω² · Meaning: Total kinetic energy in rolling motion
- Formula: v_cm = Rω (no slipping) · Meaning: Condition for pure rolling
Moment of inertia for common shapes (about axis through CM):
- Thin rod about centre, perpendicular to length: I = (1/12)ML²
- Solid sphere about diameter: I = (2/5)MR²
- Solid disc about axis through centre, perpendicular to plane: I = (1/2)MR²
- Hollow sphere about diameter: I = (2/3)MR²
Worked examples
### Example 1: Finding the centre of mass
Two children sit on a see-saw plank of negligible mass. Child A (mass 30 kg) sits 2.0 m to the left of the pivot, and child B (mass 20 kg) sits to the right. Where is the centre of mass of the two-child system, taking the pivot as origin?
Solution
Let rightward be positive. Position of A: x₁ = −2.0 m. Position of B: x₂ (unknown, but let us place B at +3.0 m for this problem).
x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂) x_cm = (30 × (−2.0) + 20 × 3.0) / (30 + 20) x_cm = (−60 + 60) / 50 = 0 m
The centre of mass lies exactly at the pivot. This explains why the see-saw balances when these positions are chosen.
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### Example 2: Torque on a door
A student pushes a door with a force of 40 N applied at 0.80 m from the hinges, at an angle of 60° to the door surface. Calculate the torque about the hinge axis.
Solution
Torque magnitude τ = r F sin θ Here r = 0.80 m, F = 40 N, θ = 60°.
τ = 0.80 × 40 × sin 60° τ = 0.80 × 40 × 0.866 τ = 27.7 N m (to 3 significant figures)
The door experiences a torque of about 28 N m tending to rotate it about the hinges.
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### Example 3: Angular momentum conservation
A child stands at the edge of a rotating platform (moment of inertia 120 kg m²) spinning at 0.50 rad/s. The child (mass 40 kg) walks to the centre. If the platform's radius is 2.0 m, find the new angular speed.
Solution
Initial moment of inertia of child about axis: I_child = mr² = 40 × (2.0)² = 160 kg m². Total initial I: I₁ = 120 + 160 = 280 kg m².
When at centre, r = 0, so I_child becomes 0. Total final I: I₂ = 120 + 0 = 120 kg m².
No external torque acts, so angular momentum is conserved: I₁ω₁ = I₂ω₂ 280 × 0.50 = 120 × ω₂ ω₂ = 140 / 120 = 1.17 rad/s
The platform speeds up to about 1.2 rad/s — the same principle an ice-skater uses when pulling arms inward.
Common mistakes
- Confusing torque with force → Torque depends on where and at what angle the force is applied; always include r and sin θ.
- Using the wrong axis for moment of inertia → I changes with axis; apply the parallel-axis theorem when the axis is not through the CM.
- Forgetting that rolling has both translational and rotational KE → Include ½ Mv_cm² and ½ I_cm ω² when energy methods are used.
- Applying conservation of angular momentum when external torque exists → First check that net external torque about the chosen axis is zero.
- Treating moment of inertia as a single number for all axes → I is axis-specific; a rod has different I about its centre and about its end.
Quick revision
- Centre of mass is the mass-weighted average position; it moves as if all external forces act there.
- Torque = r × F; it causes angular acceleration, not linear acceleration.
- Moment of inertia depends on mass distribution and the chosen axis.
- τ = Iα is the rotational analogue of F = ma.
- When net external torque is zero, angular momentum L = Iω is conserved.
- For pure rolling without slipping, v_cm = Rω.