Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
This chapter extends Newton's laws to systems of particles and rotating bodies. A rigid body has both translational motion (centre of mass) and rotational motion (about an axis). The key concepts — torque, angular momentum, moment of inertia, and rotational kinetic energy — are the rotational analogues of force, linear momentum, mass, and kinetic energy. This chapter is heavily tested in NEET and JEE.
Centre of Mass
For a two-particle system:
x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)
General (n particles):
r_cm = Σmᵢrᵢ / Σmᵢ
Centre of mass of uniform bodies:
- Uniform rod: midpoint
- Uniform triangle: centroid (intersection of medians)
- Uniform disc/sphere: geometric centre
Motion of centre of mass: F_ext = M × a_cm (Newton's second law for the system)
Rotational Motion — Analogues
| Translational | Rotational |
|---|---|
| Displacement (s) | Angular displacement (θ) |
| Velocity (v) | Angular velocity (ω) |
| Acceleration (a) | Angular acceleration (α) |
| Mass (m) | Moment of inertia (I) |
| Force (F) | Torque (τ) |
| Linear momentum (p = mv) | Angular momentum (L = Iω) |
| KE = ½mv² | Rotational KE = ½Iω² |
| F = ma | τ = Iα |
| Work = F·d | Work = τ·θ |
Torque
τ = r × F (vector); magnitude τ = rF sinθ
- r = position vector of point of application of force from the axis
- θ = angle between r and F
- Unit: N·m
Net torque on a system: τ_net = Iα
Moment of Inertia (MI)
I = Σmᵢrᵢ² (for discrete masses) I = ∫r² dm (for continuous body)
Standard Results
| Body | Axis | I |
|---|---|---|
| Thin rod (length L) | Through centre, perpendicular | ML²/12 |
| Thin rod (length L) | Through one end, perpendicular | ML²/3 |
| Uniform disc (radius R) | Through centre, perpendicular to plane | MR²/2 |
| Uniform disc (radius R) | Diameter | MR²/4 |
| Solid sphere (radius R) | Diameter | 2MR²/5 |
| Hollow sphere (radius R) | Diameter | 2MR²/3 |
| Thin ring (radius R) | Through centre, perpendicular to plane | MR² |
| Thin ring (radius R) | Diameter | MR²/2 |
Theorems
Parallel Axis Theorem: I = I_cm + Md² (I about any axis = MI about parallel axis through CM + Md²)
Perpendicular Axis Theorem (only for laminas): I_z = I_x + I_y (MI about z-axis = sum of MI about two perpendicular axes in the plane)
Angular Momentum
L = I × ω (scalar form); L = r × p (vector form)
Conservation of angular momentum: If τ_net = 0, then L = Iω = constant.
Example — ice skater: Pulls arms in → I decreases → ω increases (L conserved).
Rolling without slipping: v_cm = Rω Total KE = ½mv² + ½Iω² = ½mv²(1 + I/mR²)
Equations of Rotational Motion
| Equation | Form |
|---|---|
| ω = ω₀ + αt | Analogous to v = u + at |
| θ = ω₀t + ½αt² | Analogous to s = ut + ½at² |
| ω² = ω₀² + 2αθ | Analogous to v² = u² + 2as |
Worked Examples
Example 1: Find the MI of a thin rod of mass M and length L about an axis through one end. Using parallel axis theorem: I_end = I_cm + Md² = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3 ✓
Example 2: A disc of MI 2 kg·m² rotating at 4 rad/s. Find angular momentum. L = Iω = 2 × 4 = 8 kg·m²/s
Example 3: A skater has I = 4 kg·m² at ω = 2 rad/s. Pulls arms → I = 1 kg·m². Find new ω. By conservation of L: I₁ω₁ = I₂ω₂ → 4 × 2 = 1 × ω₂ → ω₂ = 8 rad/s
Example 4: A solid cylinder (M, R) rolls down without slipping. Find linear acceleration on incline θ. a = g sinθ / (1 + I/mR²) = g sinθ / (1 + ½) = 2g sinθ / 3
Important Questions
One Mark
- Define torque. Give its SI unit.
- State the theorem of parallel axes.
- What is the MI of a solid sphere about its diameter?
- Define angular momentum.
Short Answer (3 marks)
- Prove the perpendicular axis theorem for a lamina.
- A wheel of MI 0.5 kg·m² is accelerated from rest to 20 rad/s in 5 seconds. Find the torque applied.
- Explain the conservation of angular momentum with an example from daily life.
Long Answer (5 marks)
- Derive the expression for the kinetic energy of a rolling body. Show that for a solid sphere rolling without slipping, KE_rotation = 2/7 of total KE.
- State and prove the theorem of parallel axes. Using it, find the MI of a disc about a tangential axis in its plane.
Common Mistakes
- Parallel axis theorem: applying it between any two axes, not just a CM-axis and a parallel axis — one axis MUST pass through the CM.
- Perpendicular axis theorem: applying it to 3D bodies — it is valid only for laminas (2D flat objects).
- Rolling condition: v = Rω applies only when rolling WITHOUT slipping — don't use it for sliding.
- Angular momentum conservation: applying it even when there IS an external torque (friction, gravity about a non-CM axis).
Board Exam Tips
- Derivation of MI of a rod and disc (about various axes using theorems) is a standard 5-mark question.
- "Derive the equation for acceleration of a rolling body" appears as a 5-mark board question.
- Angular momentum conservation problems (skater, bullet-disc) appear as 3-mark numericals.
- NEET: MI standard values, rolling motion acceleration, and torque calculations appear in 2–3 questions per paper.