1. Chapter at a Glance
Rotational motion extends Newton's laws to bodies that spin. Every linear idea has a rotational twin: displacement ↔ angle, velocity ↔ angular velocity, mass ↔ moment of inertia, force ↔ torque, momentum ↔ angular momentum. The chapter also covers the centre of mass and rolling without slipping. It is one of the most important mechanics chapters in JEE Main, with one or two questions every year.
- Centre of mass of systems and bodies with a part removed.
- Moment of inertia using the parallel and perpendicular axis theorems.
- Torque and angular acceleration (pulleys with mass, rods on hinges).
- Rolling on an incline; conservation of angular momentum.
2. Centre of Mass
- The centre of mass moves as if all mass and all external forces were concentrated there: Macm = Fext. Internal forces (explosions) do not change its motion.
- Standard positions: uniform rod — midpoint; semicircular ring — 2R/π from centre; semicircular disc — 4R/3π; solid hemisphere — 3R/8; hollow hemisphere — R/2; solid cone — h/4 from the base.
- For a body with a part removed, treat the removed part as negative mass.
3. Rotational Kinematics and Torque
- Angular displacement θ, angular velocity ω = dθ/dt, angular acceleration α = dω/dt. For constant α: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.
- Linear–angular links: v = ωr, at = αr, ac = ω²r.
- Torque τ = r × F, magnitude rF sinθ (unit N m). τnet = Iα for rotation about a fixed axis.
- Equilibrium of a rigid body: ΣF = 0 and Στ = 0 about any point.
4. Moment of Inertia
Moment of inertia I = Σm r² measures resistance to change in rotation. It depends on mass, its distribution and the axis. I = MK², where K is the radius of gyration.
| Body (mass M) | Axis | I |
|---|---|---|
| Thin ring, radius R | Through centre, ⊥ plane | MR² |
| Disc, radius R | Through centre, ⊥ plane | ½MR² |
| Solid sphere | Diameter | ⅖MR² |
| Hollow sphere | Diameter | ⅔MR² |
| Solid cylinder | Own axis | ½MR² |
| Rod, length L | Centre, ⊥ rod | ML²/12 |
| Rod, length L | One end, ⊥ rod | ML²/3 |
| Ring | Diameter | ½MR² |
| Disc | Diameter | ¼MR² |
- Parallel axis theorem: I = Icm + Md² (d = distance between the parallel axes).
- Perpendicular axis theorem (plane bodies only): Iz = Ix + Iy.
5. Angular Momentum and Rolling
- Angular momentum L = r × p; for a rigid body about a fixed axis L = Iω. τ = dL/dt.
- Conservation: if τext = 0, Iω stays constant (ice skater pulling in arms spins faster; kinetic energy increases because the skater does work).
- Rotational KE = ½Iω²; work by torque W = τθ; power P = τω.
- Rolling without slipping: vcm = ωR. The contact point is momentarily at rest; the top moves at 2vcm.
- Total KE of rolling = ½mv²(1 + K²/R²). Down an incline: a = g sinθ/(1 + K²/R²), v = √(2gh/(1 + K²/R²)).
| Body | K²/R² | Acceleration on incline |
|---|---|---|
| Ring | 1 | ½g sinθ |
| Disc / solid cylinder | ½ | ⅔g sinθ |
| Hollow sphere | ⅔ | ⅗g sinθ |
| Solid sphere | ⅖ | 5⁄7 g sinθ |
6. Formula Master Table
| Linear | Rotational |
|---|---|
| F = ma | τ = Iα |
| p = mv | L = Iω |
| K = ½mv² | K = ½Iω² |
| W = Fs | W = τθ |
| P = Fv | P = τω |
| Parallel axis | I = Icm + Md² |
| Rolling | v = ωR; KE = ½mv²(1 + K²/R²) |
7. Mind Map
- Σmx/Σm
- Removed parts
- Macm = Fext
- τ = Iα
- Standard I values
- Axis theorems
- L = Iω
- Conservation
- v = ωR
- Incline acceleration
- KE split
8. Tips and Tricks
- The smaller K²/R², the faster the body rolls down: solid sphere wins, ring comes last. Mass and radius do not matter.
- For a ring, rolling KE is split 1 : 1 (translation : rotation); for a disc 2 : 1; for a solid sphere 5 : 2.
- For a mass hanging from a pulley of mass M (disc), a = mg/(m + M/2).
- When angular momentum is conserved, KE = L²/2I, so KE rises as I falls.
- Using the perpendicular axis theorem for a 3D body like a sphere.
- Applying the parallel axis theorem between two axes when neither passes through the centre of mass.
- Thinking friction does work in rolling without slipping.
9. Important One-liners
- Moment of inertia depends on the axis of rotation.
- Internal forces cannot change the motion of the centre of mass.
- The contact point of a rolling body is instantaneously at rest.
- Angular momentum is conserved when net external torque is zero.
- Kepler's second law is a result of angular momentum conservation.
- A solid sphere reaches the bottom of an incline before a hollow sphere of the same mass and radius.
10. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take g = 10 m s⁻².
Average level
Find the moment of inertia of a uniform 3 kg rod of length 2 m about an axis through one end, perpendicular to the rod.
Show answer & solution
Concept: Iend = ML²/3 (or Icm + M(L/2)²).
- I = 3 × 2²/3 = 4 kg m².
- Check: ML²/12 + M(L/2)² = 1 + 3 = 4 kg m².
Answer: 4 kg m²
A 10 N force acts at a point 0.5 m from a hinge, at 30° to the line joining the hinge and the point. Find the torque.
Show answer & solution
Concept: τ = rF sinθ.
- τ = 0.5 × 10 × sin30°.
- = 5 × 0.5 = 2.5 N m.
Answer: 2.5 N m
A flywheel speeds up uniformly from rest to 20 rad s⁻¹ in 4 s. Find its angular acceleration, the angle turned and the number of revolutions.
Show answer & solution
Concept: rotational equations for constant α.
- α = 20/4 = 5 rad s⁻².
- θ = ½ × 5 × 4² = 40 rad.
- Revolutions = 40/2π ≈ 6.37.
Answer: 5 rad s⁻²; 40 rad; about 6.4 revolutions
Masses of 1 kg, 2 kg and 3 kg are at (0, 0), (3, 0) and (0, 4) m. Find the centre of mass.
Show answer & solution
Concept: xcm = Σmx/Σm; ycm = Σmy/Σm.
- Total mass = 6 kg.
- xcm = (1 × 0 + 2 × 3 + 3 × 0)/6 = 1 m.
- ycm = (1 × 0 + 2 × 0 + 3 × 4)/6 = 2 m.
Answer: (1 m, 2 m)
A 2 kg disc rolls without slipping at 4 m s⁻¹. Find its total kinetic energy.
Show answer & solution
Concept: KE = ½mv²(1 + K²/R²), with K²/R² = ½ for a disc.
- Translational KE = ½ × 2 × 16 = 16 J.
- Rotational KE = ½ × 16 = 8 J.
- Total = 24 J.
Answer: 24 J
Advanced level
A solid sphere rolls without slipping down a 30° incline from rest. Find its acceleration and its speed after it has descended a vertical height of 7 m.
Show answer & solution
Concept: K²/R² = 2/5 for a solid sphere.
- a = g sinθ/(1 + 2/5) = (5/7) × 10 × 0.5 ≈ 3.57 m s⁻².
- Energy: mgh = ½mv²(1 + 2/5) = (7/10)mv².
- v² = 10gh/7 = 10 × 10 × 7/7 = 100.
- v = 10 m s⁻¹.
Answer: about 3.57 m s⁻²; 10 m s⁻¹
A person on a frictionless turntable has I = 6 kg m² and spins at 3 rad s⁻¹ with arms stretched. Pulling in the arms reduces I to 2 kg m². Find the new angular speed and the change in kinetic energy.
Show answer & solution
Concept: no external torque ⇒ I₁ω₁ = I₂ω₂.
- ω₂ = 6 × 3/2 = 9 rad s⁻¹.
- KE₁ = ½ × 6 × 9 = 27 J.
- KE₂ = ½ × 2 × 81 = 81 J.
- Increase = 54 J — supplied by the muscular work of pulling the arms in.
Answer: 9 rad s⁻¹; KE increases by 54 J
A light string wrapped round a uniform disc pulley (M = 4 kg, R = 0.2 m) supports a 2 kg block. Find the block's acceleration, the string tension and the pulley's angular acceleration.
Show answer & solution
Concept: block: mg − T = ma; pulley: TR = Iα with I = ½MR² and a = αR.
- From the pulley: T = Ia/R² = ½Ma = 2a.
- Block: 20 − 2a = 2a ⇒ a = 5 m s⁻².
- T = 2 × 5 = 10 N.
- α = a/R = 5/0.2 = 25 rad s⁻².
Answer: 5 m s⁻²; 10 N; 25 rad s⁻²
Find the moment of inertia of a uniform disc (M = 2 kg, R = 0.5 m) about (a) a tangent perpendicular to its plane, (b) a tangent lying in its plane.
Show answer & solution
Concept: parallel axis theorem from the matching centroidal axis (d = R).
- MR² = 2 × 0.25 = 0.5 kg m².
- (a) I = ½MR² + MR² = 1.5 × 0.5 = 0.75 kg m².
- (b) The in-plane centroidal axis is a diameter: Id = ¼MR² (perpendicular axis theorem).
- I = ¼MR² + MR² = 1.25 × 0.5 = 0.625 kg m².
Answer: (a) 0.75 kg m²; (b) 0.625 kg m²
A uniform rod of length 1.2 m, hinged at one end, is released from rest in a horizontal position. Find (a) its initial angular acceleration, (b) its angular speed when it becomes vertical, (c) the speed of the free end at that moment.
Show answer & solution
Concept: τ = Iα about the hinge with I = ML²/3; energy conservation for the swing (centre of mass falls L/2).
- (a) τ = Mg(L/2), so α = Mg(L/2)/(ML²/3) = 3g/2L = 30/2.4 = 12.5 rad s⁻².
- (b) Mg(L/2) = ½(ML²/3)ω² ⇒ ω² = 3g/L = 30/1.2 = 25 ⇒ ω = 5 rad s⁻¹.
- (c) vend = ωL = 5 × 1.2 = 6 m s⁻¹.
Answer: 12.5 rad s⁻²; 5 rad s⁻¹; 6 m s⁻¹
11. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
For a ring rolling without slipping, the ratio of translational to rotational kinetic energy is:
- 1 : 1
- 2 : 1
- 5 : 2
- 1 : 2
Show solution
For a ring I = MR², so ½Iω² = ½MR²(v/R)² = ½Mv². Ratio 1 : 1.
Answer: (A)
The radius of gyration of a solid sphere about a diameter is:
- R
- √(2/5) R
- √(2/3) R
- R/2
Show solution
MK² = ⅖MR² ⇒ K = √(2/5) R.
Answer: (B)
A ring, a disc, a hollow sphere and a solid sphere are released together from the top of an incline. Which reaches the bottom first?
- Ring
- Disc
- Hollow sphere
- Solid sphere
Show solution
a = g sinθ/(1 + K²/R²); the solid sphere has the smallest K²/R² (0.4), so the largest acceleration.
Answer: (D)
The moment of inertia of a thin ring about a tangent in its own plane is:
- MR²
- 2MR²
- (3/2)MR²
- ½MR²
Show solution
About a diameter I = ½MR²; parallel axis: ½MR² + MR² = (3/2)MR². (About a tangent ⊥ plane it would be 2MR².)
Answer: (C)
From a uniform disc of radius R, a circular hole of radius R/2 is cut so that the hole touches the rim. The centre of mass of the remaining part is shifted from the original centre by:
- R/6
- R/4
- R/3
- R/2
Show solution
Mass ∝ area: hole = M/4 at distance R/2; remainder = 3M/4. x = −(M/4)(R/2)/(3M/4) = −R/6 (away from the hole).
Answer: (A)
A wheel with moment of inertia 2 kg m² rotates at 5 rad s⁻¹. Find its angular momentum in kg m² s⁻¹.
Show solution
L = Iω = 2 × 5 = 10.
Answer: 10
12. Quick Revision Summary
- xcm = Σmx/Σm; removed parts as negative mass.
- τ = Iα; L = Iω; conserved when τext = 0.
- Ring MR², disc ½MR², solid sphere ⅖MR², hollow sphere ⅔MR², rod ML²/12 (centre), ML²/3 (end).
- I = Icm + Md²; Iz = Ix + Iy (plane bodies).
- Rolling: v = ωR; a = g sinθ/(1 + K²/R²).