How to Use This Page
1. Attempt every exercise question on your own first — then compare your working with the solutions. 2. Focus on understanding the derivations — rotational dynamics formulas are not as intuitive as linear ones, so the derivation matters. 3. Memorise moments of inertia for standard shapes — they are used frequently in numericals. 4. Review the Common Mistakes section — these are the most frequent exam errors for this chapter. 5. Numerical examples in code blocks include SI units at every step — always include units in your exam answers.
Key Concepts (Quick Recap)
Centre of Mass
| System | Formula |
|---|
| Two particles (1D) | x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂) |
| General (1D) | x_cm = Σmᵢxᵢ / Σmᵢ |
| General (2D) | x_cm = Σmᵢxᵢ/M, y_cm = Σmᵢyᵢ/M |
| Continuous body | x_cm = (1/M)∫x dm |
Moment of Inertia (Standard Shapes)
| Body | Axis | I |
|---|
| Thin rod (length L) | Through centre, ⊥ to rod | ML²/12 |
| Thin rod (length L) | Through one end, ⊥ to rod | ML²/3 |
| Solid disc / cylinder (radius R) | Through centre, ⊥ to plane | MR²/2 |
| Ring / hollow cylinder (radius R) | Through centre, ⊥ to plane | MR² |
| Solid sphere (radius R) | About diameter | 2MR²/5 |
| Hollow sphere (radius R) | About diameter | 2MR²/3 |
Theorems
| Theorem | Formula | Use |
|---|
| Parallel axis | I = I_cm + Md² | Shift axis by distance d |
| Perpendicular axis | I_z = I_x + I_y | For planar bodies only |
Rotational Dynamics
| Quantity | Formula | SI Unit |
|---|
| Torque | τ = r × F = rF sin θ | N·m |
| Angular momentum | L = Iω | kg·m²/s |
| Rotational KE | KE_rot = ½Iω² | J |
| τ = Iα (rotational analogue of F = ma) | τ = Iα | N·m |
| τ = dL/dt | τ = dL/dt | N·m |
Important Definitions
| Term | Definition |
|---|
| Rigid body | A body whose constituent particles do not change their relative positions under applied force |
| Centre of mass | The point at which the entire mass of the body can be considered concentrated for translational motion |
| Moment of inertia | The rotational analogue of mass; measure of resistance to angular acceleration; I = Σmᵢrᵢ² |
| Torque | The rotational analogue of force; the turning effect of a force about an axis |
| Angular momentum | The rotational analogue of linear momentum; L = Iω |
I. Choose the Best Answer (MCQs)
1. The moment of inertia of a solid sphere of mass M and radius R about its diameter is:
Answer
- (a) MR²
- (b) 2MR²/5
- (c) 2MR²/3
- (d) MR²/2
Answer(b) 2MR²/5
ExplanationThis is a standard result. I_sphere (diameter) = 2MR²/5. For a hollow sphere it is 2MR²/3.
2. The centre of mass of a uniform solid cone of height h from the base lies at:
Answer
- (a) h/2
- (b) h/3
- (c) h/4
- (d) h/6
Answer(c) h/4
ExplanationThe centre of mass of a solid cone from its base is at h/4 along the axis.
3. A disc of mass M and radius R has moment of inertia MR²/2 about its central axis. By the parallel axis theorem, its moment of inertia about a tangent is:
Answer
- (a) 3MR²/2
- (b) MR²/2
- (c) MR²
- (d) 5MR²/4
Answer(a) 3MR²/2
ExplanationI_tangent = I_cm + Md² = MR²/2 + MR² = 3MR²/2 (tangent is at distance d = R from centre).
4. If the radius of a ring is doubled, its moment of inertia about its central axis:
Answer
- (a) Doubles
- (b) Quadruples
- (c) Remains the same
- (d) Becomes half
Answer(b) Quadruples
ExplanationI_ring = MR². If R → 2R, then I = M(2R)² = 4MR² = 4 times the original.
5. The rotational analogue of force is:
Answer
- (a) Momentum
- (b) Torque
- (c) Angular momentum
- (d) Moment of inertia
Answer(b) Torque
ExplanationJust as force causes linear acceleration (F = ma), torque causes angular acceleration (τ = Iα). Torque is the rotational analogue of force.
6. A solid sphere rolls without slipping. The ratio of its rotational KE to its translational KE is:
Answer
- (a) 1:1
- (b) 2:5
- (c) 2:7
- (d) 5:7
Answer(b) 2:5
ExplanationKE_rot = ½Iω² = ½(2MR²/5)(v/R)² = Mv²/5. KE_trans = ½Mv². Ratio = (Mv²/5) / (½Mv²) = (1/5)/(1/2) = 2/5 = 2:5.
7. Angular momentum is conserved when:
Answer
- (a) Net force is zero
- (b) Net torque is zero
- (c) Net force is constant
- (d) Moment of inertia is constant
Answer(b) Net torque is zero
Explanationτ = dL/dt. If τ = 0, then dL/dt = 0, so L = constant. Angular momentum is conserved when net external torque is zero.
II. Fill in the Blanks
1. The rotational analogue of mass is ________. Moment of inertia 2. The rotational analogue of linear momentum is ________. Angular momentum 3. τ = Iα is the rotational analogue of ________. F = ma 4. The moment of inertia of a thin rod of mass M and length L about its centre (perpendicular axis) is ________. ML²/12 5. The parallel axis theorem is I = I_cm + ________. Md² 6. The perpendicular axis theorem applies only to ________ bodies. Planar (lamina) 7. The centre of mass of a two-particle system with equal masses lies at the ________ of the line joining them. Midpoint 8. The SI unit of angular momentum is ________. kg·m²/s 9. When no external torque acts, angular momentum is ________. Conserved 10. For a rolling body without slipping, v = ________. ωR 11. The moment of inertia of a ring about its central axis is ________. MR² 12. If the distance of a particle from the axis doubles, its moment of inertia becomes ________ times. Four
III. True or False (with corrections)
1. The centre of mass of a body always lies inside the body.
- False. For a ring or a hollow sphere, the centre of mass is at the geometric centre, which is outside the material of the body.
2. Moment of inertia depends on the axis of rotation.
- True. The same body has different moments of inertia about different axes. I depends on the mass distribution relative to the axis.
3. Torque and angular momentum are both vectors.
- True. Both are vector quantities. Torque is τ = r × F and angular momentum is L = r × p, both cross products.
4. The perpendicular axis theorem can be applied to a solid sphere.
- False. The perpendicular axis theorem (I_z = I_x + I_y) applies only to planar (lamina) bodies. A sphere is 3-dimensional; the theorem does not apply.
5. When a spinning ice skater pulls her arms in, her angular velocity increases.
- True. By conservation of angular momentum (L = Iω = constant), when I decreases (arms closer to body), ω must increase proportionally.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Moment of inertia of solid sphere | (d) 2MR²/5 |
| 2. Moment of inertia of ring | (a) MR² |
| 3. Moment of inertia of disc | (e) MR²/2 |
| 4. Moment of inertia of rod (centre) | (f) ML²/12 |
| 5. Parallel axis theorem | (b) I = I_cm + Md² |
| 6. Perpendicular axis theorem | (c) I_z = I_x + I_y |
V. Answer Briefly (2-mark Questions)
1. Define centre of mass and give its position for a two-particle system.
Answer
The centre of mass is the point at which the entire mass of the system can be considered to be concentrated for the purpose of studying its translational motion. For two particles of masses m₁ and m₂ at positions x₁ and x₂ along a line: x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂). If the masses are equal, the centre of mass lies at the midpoint.
2. State the parallel axis theorem and give an example.
Answer
The parallel axis theorem states: I = I_cm + Md², where I is the moment of inertia about an axis parallel to the axis through the centre of mass, I_cm is the moment of inertia about the centre of mass axis, M is the mass, and d is the perpendicular distance between the two axes. Example: I of a disc about a tangent = I_cm + MR² = MR²/2 + MR² = 3MR²/2.
3. Distinguish between torque and couple.
Answer
Torque is the turning effect of a single force about an axis: τ = rF sin θ. A couple consists of two equal and opposite forces with parallel lines of action, separated by a distance d. The torque due to a couple is τ = F × d (same about any point). A couple produces pure rotation without translation.
4. What is the significance of the moment of inertia?
Answer
Moment of inertia (I) plays the same role in rotational motion that mass plays in linear motion. It measures the resistance of a body to angular acceleration: τ = Iα. A body with a larger moment of inertia requires a larger torque to produce the same angular acceleration. I depends on the mass of the body and how that mass is distributed relative to the axis of rotation.
VI. Answer in Detail (5-mark Questions with Numerical Examples)
1. Define centre of mass and derive its expression for a two-particle system. Solve: Two masses 3 kg and 5 kg are placed at x = 2 m and x = 8 m respectively. Find the centre of mass. If a 2 kg mass is added at x = 5 m, find the new centre of mass.
Answer
Derivation: Consider two particles of masses m₁ at position x₁ and m₂ at position x₂. The centre of mass x_cm is defined as the point where the total moment of all particles about it is zero:
m₁(x_cm − x₁) = m₂(x₂ − x_cm) m₁x_cm − m₁x₁ = m₂x₂ − m₂x_cm (m₁ + m₂)x_cm = m₁x₁ + m₂x₂
x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)
For 2D: x_cm = (Σmᵢxᵢ)/M, y_cm = (Σmᵢyᵢ)/M, where M = Σmᵢ.
Numerical:
Given: m₁ = 3 kg at x₁ = 2 m, m₂ = 5 kg at x₂ = 8 m
Centre of mass:
x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)
= (3 × 2 + 5 × 8) / (3 + 5)
= (6 + 40) / 8
= 46/8 = 5.75 m
Now add m₃ = 2 kg at x₃ = 5 m:
New total mass M = 3 + 5 + 2 = 10 kg
New x_cm = (3 × 2 + 5 × 8 + 2 × 5) / 10
= (6 + 40 + 10) / 10
= 56/10 = 5.6 m
Answer: Centre of mass with two masses = 5.75 m
Centre of mass after adding 2 kg = 5.6 m
2. Define moment of inertia and derive expressions for standard shapes. Solve: A solid sphere of mass 10 kg and radius 0.2 m rolls without slipping on a horizontal surface at 5 m/s. Find (a) its moment of inertia about the diameter, (b) its rotational KE, and (c) its total KE.
Answer
Derivation for solid disc: Consider a disc of mass M and radius R. Divide it into concentric rings. A ring of radius r and width dr has mass dm = (M/πR²)(2πr dr).
I = ∫r² dm = ∫₀ᴿ r² × (2Mr/R²) dr = (2M/R²) ∫₀ᴿ r³ dr = (2M/R²) × R⁴/4 = MR²/2
Derivation for thin rod about centre: Rod of mass M and length L, element dm = (M/L)dx at distance x from centre (−L/2 to L/2). I = ∫x² dm = (M/L) ∫₋L/₂^{L/2} x² dx = (M/L)(L³/12) = ML²/12
Numerical:
Given: M = 10 kg, R = 0.2 m, v = 5 m/s
(a) Moment of inertia about diameter:
I = 2MR²/5 = 2 × 10 × (0.2)² / 5
I = 2 × 10 × 0.04 / 5 = 0.8/5 = 0.16 kg·m²
(b) Rotational KE:
For rolling without slipping: v = ωR → ω = v/R = 5/0.2 = 25 rad/s
KE_rot = ½Iω² = ½ × 0.16 × (25)²
= 0.08 × 625 = 50 J
(c) Total KE:
KE_trans = ½Mv² = ½ × 10 × 25 = 125 J
KE_total = KE_trans + KE_rot = 125 + 50 = 175 J
Check: For a solid sphere rolling, KE_rot/KE_trans = 2/5
50/125 = 2/5 ✓
Answer: (a) I = 0.16 kg·m², (b) KE_rot = 50 J, (c) KE_total = 175 J
3. State and prove the law of conservation of angular momentum. Solve: A disc of moment of inertia 5 kg·m² rotating at 20 rad/s has its moment of inertia reduced to 3 kg·m². Find the new angular velocity and the work done.
Answer
Statement: When no external torque acts on a system, its total angular momentum remains constant.
Proof: From τ = dL/dt, if τ_ext = 0, then dL/dt = 0, so L = constant. Thus, I₁ω₁ = I₂ω₂.
This is analogous to conservation of linear momentum (when F = 0, p = constant). Applications include a diver tucking in to spin faster, a ballet dancer extending arms to slow down, and planetary orbits.
Numerical:
Given: I₁ = 5 kg·m², ω₁ = 20 rad/s, I₂ = 3 kg·m²
By conservation of angular momentum:
I₁ω₁ = I₂ω₂
5 × 20 = 3 × ω₂
100 = 3ω₂
ω₂ = 100/3 ≈ 33.33 rad/s
Work done = Change in KE:
KE₁ = ½I₁ω₁² = ½ × 5 × 400 = 1000 J
KE₂ = ½I₂ω₂² = ½ × 3 × (100/3)² = ½ × 3 × 10000/9
= 15000/18 = 5000/6 ≈ 833.33 J
Wait — let me recalculate:
KE₂ = ½ × 3 × (100/3)² = ½ × 3 × 10000/9 = 15000/18 = 5000/6 ≈ 833.33 J
Hmm, KE₂ should be greater since the skater does internal work.
Let me recalculate more carefully:
ω₂ = 100/3 rad/s
KE₂ = ½ × 3 × (100/3)² = ½ × 3 × 10000/9
= 3/2 × 10000/9 = 30000/18 = 5000/3 ≈ 1666.67 J
Work done = KE₂ − KE₁ = 5000/3 − 1000 = (5000 − 3000)/3 = 2000/3 ≈ 666.67 J
Answer: New angular velocity ω₂ = 33.33 rad/s
Work done = 666.67 J (by internal forces)
Common Mistakes to Avoid
1. Using the wrong moment of inertia formula — always check: rod, disc, ring, sphere (solid/hollow)? Each has a different I. Memorise the standard table.
2. Confusing the parallel and perpendicular axis theorems — the parallel axis theorem (I = I_cm + Md²) applies to any body shifting the axis; the perpendicular axis theorem (I_z = I_x + I_y) applies only to planar laminas.
3. Forgetting that I depends on the axis — the same body has different I values about different axes. Always state which axis you are using.
4. Not converting to consistent units — mass in kg, distance in m, angular velocity in rad/s. Mixing degrees and radians in ω is a common error.
5. Confusing v = ωR with v = rω for different points — for rolling without slipping, the condition v_cm = ωR applies to the centre of mass. Other points on the rolling body have different velocities.
6. Applying linear formulas to rotational problems without conversion — F → τ, m → I, v → ω, a → α, p → L. Every linear quantity has a rotational analogue.
7. Assuming centre of mass is always inside the body — for rings, hollow spheres, L-shaped objects, etc., the centre of mass can be outside the material.
8. Not using conservation of angular momentum correctly — angular momentum is conserved only when net external torque is zero. Internal forces (like a skater pulling arms in) change I and ω but not L.
Exam Checklist
- [ ] Can define and calculate centre of mass for multi-particle systems
- [ ] Know moment of inertia formulas for rod, disc, ring, sphere (solid and hollow)
- [ ] Can apply the parallel axis theorem (I = I_cm + Md²) correctly
- [ ] Can apply the perpendicular axis theorem for planar laminas (I_z = I_x + I_y)
- [ ] Understand τ = Iα as the rotational analogue of F = ma
- [ ] Can calculate torque (τ = rF sin θ) and determine its direction
- [ ] Know angular momentum L = Iω and its conservation (L = constant when τ = 0)
- [ ] Can solve rolling-without-slipping problems (v = ωR, total KE = ½Mv² + ½Iω²)
- [ ] Can find total KE of a rolling body (translational + rotational)
- [ ] Remember: I depends on axis, not just mass; always specify the axis
