Concept Core
Every linear mechanics concept has a rotational twin — learn them together.
Linear ↔ Rotational Analogy — The Master Map
| Linear | Symbol | Rotational | Symbol |
|---|
| 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 |
| Newton 2nd: F=ma | — | τ = Iα | — |
| KE = ½mv² | — | KE = ½Iω² | — |
| Momentum p=mv | — | Angular momentum L=Iω | — |
Torque
Torque = turning effect of a force:
τ = r × F = rF sinθ
r = perpendicular distance from axis to line of action of F. Unit: N·m.
τ = 0 when F passes through the axis (r=0) or F is parallel to r (sinθ=0).
Moment of Inertia (I)
The rotational equivalent of mass — resistance to rotation:
I = Σmᵢrᵢ² = ∫r² dm
Key values: Solid sphere: 2MR²/5 | Hollow sphere: 2MR²/3 | Disc/cylinder (axis): MR²/2 | Ring (axis): MR² | Rod (centre): ML²/12 | Rod (end): ML²/3
Parallel & Perpendicular Axis Theorems
Parallel Axis: I = I_cm + Md² (shift axis by distance d from centre of mass)
Perpendicular Axis (for laminas only): Iz = Ix + Iy (z-axis ⊥ to plane containing x,y axes)
Angular Momentum & Conservation
L = Iω = mvr (for a point mass)
Conservation of L: When net torque = 0, L is constant: I₁ω₁ = I₂ω₂
Example: Ice-skater pulls arms in → I decreases → ω increases → spins faster. L is conserved.
Rolling Motion
A rolling body has both translational and rotational KE:
KE_total = ½mv² + ½Iω²
v = Rω (rolling without slipping)
KE = ½mv²(1 + k²/R²)
k = radius of gyration. Disc: k²=R²/2. Ring: k²=R². Solid sphere: k²=2R²/5.
Formula Vault
All rotational mechanics formulas — paired with their linear counterparts.
Torque
τ = rF sinθ = Iα
r = moment arm; θ = angle F makes with r
Moment of Inertia
I = Σmᵢrᵢ²
Mass × (distance from axis)²
Rotational KE
KE = ½Iω²
ω in rad/s; I in kg·m²
Angular Momentum
L = Iω = mvr
Conserved when τ_net = 0
Newton's 2nd (rotation)
τ = Iα
Analogue of F = ma
Parallel Axis Theorem
I = I_cm + Md²
d = distance from CM to new axis
Perp. Axis Theorem
Iz = Ix + Iy
Laminas only; z ⊥ plane
Ring (central axis)
I = MR²
All mass at radius R
Disc/Solid Cylinder
I = MR²/2
About central axis
Solid Sphere
I = 2MR²/5
About diameter
Rolling KE
KE = ½mv²(1 + k²/R²)
v = Rω for rolling without slip
Conservation of L
I₁ω₁ = I₂ω₂
When net torque = 0
Worked Examples
5 problems — torque, MOI, rolling, angular momentum, and a classic trap.
EasyFind torque: F = 10 N at 30° to a rod of length 2 m▾
A force of 10 N acts at 30° to a rod of length 2 m from the pivot. Find the torque.
1
τ = rF sinθ = 2 × 10 × sin30° = 2 × 10 × 0.5 = 10 N·m
✓ τ = 10 N·m
EasyFind I of a rod (mass M, length L) about one end using parallel axis theorem▾
Moment of inertia of a rod about its centre is ML²/12. Find I about one end.
1
Parallel axis theorem: I = I_cm + Md²
2
d = L/2 (distance from centre to end)
3
I = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/12 + 3ML²/12 = ML²/3
✓ I (about one end) = ML²/3
MediumIce skater reduces I from 4 kg·m² to 1 kg·m² — find new ω▾
An ice skater spins at 2 rad/s with I = 4 kg·m². She pulls her arms in, reducing I to 1 kg·m². Find her new angular velocity.
1
No external torque → angular momentum conserved: I₁ω₁ = I₂ω₂
2
4 × 2 = 1 × ω₂ → ω₂ = 8 rad/s
✓ ω₂ = 8 rad/s (spins 4× faster)
EAPCET LevelA disc and a ring roll down the same incline — which reaches the bottom first?▾
A solid disc (I=MR²/2) and a ring (I=MR²) of same mass and radius roll down the same frictionless incline from rest. Which has greater speed at the bottom?
1
Using energy conservation: mgh = ½mv² + ½Iω² = ½mv²(1 + k²/R²)
2
Disc: k²/R² = 1/2 → v² = 2gh/(1+1/2) = 4gh/3
3
Ring: k²/R² = 1 → v² = 2gh/(1+1) = gh
4
Disc v² = 4gh/3 > Ring v² = gh → disc is faster
✓ Disc reaches bottom first (less rotational inertia → more translational KE)
Trap QuestionDoes a heavier object roll faster down an incline?▾
Two solid spheres of different masses but same radius roll down the same incline. Which reaches the bottom first? ⚠️ Classic intuition trap.
1
The trap: Students assume heavier object rolls faster due to greater force.
2
For rolling: v² = 2gh/(1 + k²/R²). This depends only on k²/R² — the shape, NOT the mass.
3
Both spheres have I = 2MR²/5 → k²/R² = 2/5 → same formula → same speed.
4
Mass cancels out of the energy equation. All objects of the same shape reach the bottom at the same time.
✓ Both arrive simultaneously — rolling speed depends on shape, not mass
Mistake DNA
4 rotational motion errors that appear in EAPCET distractors.
⚖️
Confusing Torque (N·m) with Work (J) — Same Units, Different Concepts
Both torque and work have units of N·m, but torque is a vector (rotational force) while work is a scalar (energy).
❌ Wrong
τ = 10 N·m means
10 J of work done ✗
(units match but
physics doesn't)
✓ Correct
τ = 10 N·m is torque ✓
Work = τ × θ (in J) ✓
Angle θ must be in
radians
Torque × angular displacement = Work. W = τθ (θ in radians). Torque itself is not energy.
🔄
Using I = ½MR² for a Ring (Should Be MR²)
Confusing disc and ring moments of inertia — the most common MOI error in EAPCET.
❌ Wrong
Ring about central axis:
I = MR²/2 ✗
(that's the disc)
✓ Correct
Ring: I = MR² ✓
Disc/Cylinder: I = MR²/2 ✓
All mass at R vs
distributed up to R
Ring: all mass at exactly radius R → I = MR². Disc: mass spread from 0 to R → average r² = R²/2 → I = MR²/2.
📏
Applying Parallel Axis Theorem With Wrong d
d must be the distance from the centre of mass to the new axis, not to any other point.
❌ Wrong
Rod about one end:
d = L (full length) ✗
I = ML²/12 + ML² ✗
✓ Correct
d = L/2 (from CM
to end) ✓
I = ML²/12 + ML²/4
= ML²/3 ✓
In parallel axis theorem I = I_cm + Md², d is specifically the distance from the centre of mass to the new axis. Always identify CM first.
🌪️
Angular Momentum: Using p=mv Instead of L=Iω
For rotating bodies, L = Iω. For a point mass going in a circle, L = mvr. Using just mv gives wrong units.
❌ Wrong
Ball in circle r=2,
v=3 m/s, m=1 kg:
L = mv = 3 kg·m/s ✗
(missing r)
✓ Correct
L = mvr = 1×3×2
= 6 kg·m²/s ✓
Or L = Iω = mr²×(v/r)
= mvr ✓
For a point mass: L = mvr (units kg·m²/s). For rigid bodies: L = Iω. The extra factor r comes from the moment arm.
Chapter Intelligence
Rotational mechanics links to every dynamics chapter through the analogy table.
EAPCET Weightage (2019–2024)
Torque & angular equations~6 High-Yield PYQ Patterns
MOI of disc/ring/sphereTorque given force & angleIce-skater ω conservationRolling disc vs ring speedParallel axis theorem applicationKE of rolling body
Exam Strategy
- Memorise the 6 key MOI values: Ring MR², Disc MR²/2, Solid sphere 2MR²/5, Hollow sphere 2MR²/3, Rod (centre) ML²/12, Rod (end) ML²/3.
- Rolling: use energy method. Total KE = ½mv²(1+k²/R²). Identify shape → look up k²/R² → solve for v.
- Conservation of angular momentum: identify when net torque = 0 (no external torques). Then I₁ω₁ = I₂ω₂.
- The linear-rotational analogy table is your shortcut: if you know F=ma, you know τ=Iα. If you know KE=½mv², you know KE=½Iω². Don't memorise twice — map once.