Concept Core
Every linear mechanics concept has a rotational twin β learn them together.
Linear β Rotational Analogy β The Master Map
| Linear | Symbol | Rotational | Symbol |
|---|
| Displacement | s | Angular displacement | ΞΈ |
| Velocity | v | Angular velocity | Ο |
| Acceleration | a | Angular acceleration | Ξ± |
| Mass | m | Moment of Inertia | I |
| Force | F | Torque | Ο |
| Newton 2nd: F=ma | β | Ο = IΞ± | β |
| KE = Β½mvΒ² | β | KE = Β½IΟΒ² | β |
| Momentum p=mv | β | Angular momentum L=IΟ | β |
Torque
Torque = turning effect of a force:
Ο = r Γ F = rF sinΞΈ
r = perpendicular distance from axis to line of action of F. Unit: NΒ·m.
Ο = 0 when F passes through the axis (r=0) or F is parallel to r (sinΞΈ=0).
Moment of Inertia (I)
The rotational equivalent of mass β resistance to rotation:
I = Ξ£mα΅’rα΅’Β² = β«rΒ² dm
Key values: Solid sphere: 2MRΒ²/5 | Hollow sphere: 2MRΒ²/3 | Disc/cylinder (axis): MRΒ²/2 | Ring (axis): MRΒ² | Rod (centre): MLΒ²/12 | Rod (end): MLΒ²/3
Parallel & Perpendicular Axis Theorems
Parallel Axis: I = I_cm + MdΒ² (shift axis by distance d from centre of mass)
Perpendicular Axis (for laminas only): Iz = Ix + Iy (z-axis β₯ to plane containing x,y axes)
Angular Momentum & Conservation
L = IΟ = mvr (for a point mass)
Conservation of L: When net torque = 0, L is constant: IβΟβ = IβΟβ
Example: Ice-skater pulls arms in β I decreases β Ο increases β spins faster. L is conserved.
Rolling Motion
A rolling body has both translational and rotational KE:
KE_total = Β½mvΒ² + Β½IΟΒ²
v = RΟ (rolling without slipping)
KE = Β½mvΒ²(1 + kΒ²/RΒ²)
k = radius of gyration. Disc: kΒ²=RΒ²/2. Ring: kΒ²=RΒ². Solid sphere: kΒ²=2RΒ²/5.
Formula Vault
All rotational mechanics formulas β paired with their linear counterparts.
Torque
Ο = rF sinΞΈ = IΞ±
r = moment arm; ΞΈ = angle F makes with r
Moment of Inertia
I = Ξ£mα΅’rα΅’Β²
Mass Γ (distance from axis)Β²
Rotational KE
KE = Β½IΟΒ²
Ο in rad/s; I in kgΒ·mΒ²
Angular Momentum
L = IΟ = mvr
Conserved when Ο_net = 0
Newton's 2nd (rotation)
Ο = IΞ±
Analogue of F = ma
Parallel Axis Theorem
I = I_cm + MdΒ²
d = distance from CM to new axis
Perp. Axis Theorem
Iz = Ix + Iy
Laminas only; z β₯ plane
Ring (central axis)
I = MRΒ²
All mass at radius R
Disc/Solid Cylinder
I = MRΒ²/2
About central axis
Solid Sphere
I = 2MRΒ²/5
About diameter
Rolling KE
KE = Β½mvΒ²(1 + kΒ²/RΒ²)
v = RΟ for rolling without slip
Conservation of L
IβΟβ = IβΟβ
When net torque = 0
Worked Examples
5 problems β torque, MOI, rolling, angular momentum, and a classic trap.
EasyFind torque: F = 10 N at 30Β° to a rod of length 2 mβΎ
A force of 10 N acts at 30Β° to a rod of length 2 m from the pivot. Find the torque.
1
Ο = rF sinΞΈ = 2 Γ 10 Γ sin30Β° = 2 Γ 10 Γ 0.5 = 10 NΒ·m
β Ο = 10 NΒ·m
EasyFind I of a rod (mass M, length L) about one end using parallel axis theoremβΎ
Moment of inertia of a rod about its centre is MLΒ²/12. Find I about one end.
1
Parallel axis theorem: I = I_cm + MdΒ²
2
d = L/2 (distance from centre to end)
3
I = MLΒ²/12 + M(L/2)Β² = MLΒ²/12 + MLΒ²/4 = MLΒ²/12 + 3MLΒ²/12 = MLΒ²/3
β I (about one end) = MLΒ²/3
MediumIce skater reduces I from 4 kgΒ·mΒ² to 1 kgΒ·mΒ² β find new ΟβΎ
An ice skater spins at 2 rad/s with I = 4 kgΒ·mΒ². She pulls her arms in, reducing I to 1 kgΒ·mΒ². Find her new angular velocity.
1
No external torque β angular momentum conserved: IβΟβ = IβΟβ
2
4 Γ 2 = 1 Γ Οβ β Οβ = 8 rad/s
β Οβ = 8 rad/s (spins 4Γ faster)
EAPCET LevelA disc and a ring roll down the same incline β which reaches the bottom first?βΎ
A solid disc (I=MRΒ²/2) and a ring (I=MRΒ²) 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 = Β½mvΒ² + Β½IΟΒ² = Β½mvΒ²(1 + kΒ²/RΒ²)
2
Disc: kΒ²/RΒ² = 1/2 β vΒ² = 2gh/(1+1/2) = 4gh/3
3
Ring: kΒ²/RΒ² = 1 β vΒ² = 2gh/(1+1) = gh
4
Disc vΒ² = 4gh/3 > Ring vΒ² = gh β disc is faster
β Disc reaches bottom first (less rotational inertia β more translational KE)
Trap QuestionDoes a heavier object roll faster down an incline?βΎ
Two solid spheres of different masses but same radius roll down the same incline. Which reaches the bottom first? β οΈ Classic intuition trap.
1
The trap: Students assume heavier object rolls faster due to greater force.
2
For rolling: vΒ² = 2gh/(1 + kΒ²/RΒ²). This depends only on kΒ²/RΒ² β the shape, NOT the mass.
3
Both spheres have I = 2MRΒ²/5 β kΒ²/RΒ² = 2/5 β same formula β same speed.
4
Mass cancels out of the energy equation. All objects of the same shape reach the bottom at the same time.
β Both arrive simultaneously β rolling speed depends on shape, not mass
Mistake DNA
4 rotational motion errors that appear in EAPCET distractors.
βοΈ
Confusing Torque (NΒ·m) with Work (J) β Same Units, Different Concepts
Both torque and work have units of NΒ·m, but torque is a vector (rotational force) while work is a scalar (energy).
β Wrong
Ο = 10 NΒ·m means
10 J of work done β
(units match but
physics doesn't)
β Correct
Ο = 10 NΒ·m is torque β
Work = Ο Γ ΞΈ (in J) β
Angle ΞΈ must be in
radians
Torque Γ angular displacement = Work. W = ΟΞΈ (ΞΈ in radians). Torque itself is not energy.
π
Using I = Β½MRΒ² for a Ring (Should Be MRΒ²)
Confusing disc and ring moments of inertia β the most common MOI error in EAPCET.
β Wrong
Ring about central axis:
I = MRΒ²/2 β
(that's the disc)
β Correct
Ring: I = MRΒ² β
Disc/Cylinder: I = MRΒ²/2 β
All mass at R vs
distributed up to R
Ring: all mass at exactly radius R β I = MRΒ². Disc: mass spread from 0 to R β average rΒ² = RΒ²/2 β I = MRΒ²/2.
π
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.
β Wrong
Rod about one end:
d = L (full length) β
I = MLΒ²/12 + MLΒ² β
β Correct
d = L/2 (from CM
to end) β
I = MLΒ²/12 + MLΒ²/4
= MLΒ²/3 β
In parallel axis theorem I = I_cm + MdΒ², d is specifically the distance from the centre of mass to the new axis. Always identify CM first.
πͺοΈ
Angular Momentum: Using p=mv Instead of L=IΟ
For rotating bodies, L = IΟ. For a point mass going in a circle, L = mvr. Using just mv gives wrong units.
β Wrong
Ball in circle r=2,
v=3 m/s, m=1 kg:
L = mv = 3 kgΒ·m/s β
(missing r)
β Correct
L = mvr = 1Γ3Γ2
= 6 kgΒ·mΒ²/s β
Or L = IΟ = mrΒ²Γ(v/r)
= mvr β
For a point mass: L = mvr (units kgΒ·mΒ²/s). For rigid bodies: L = IΟ. The extra factor r comes from the moment arm.
Chapter Intelligence
Rotational mechanics links to every dynamics chapter through the analogy table.
EAPCET Weightage (2019β2024)
Torque & angular equations~6 High-Yield PYQ Patterns
MOI of disc/ring/sphereTorque given force & angleIce-skater Ο conservationRolling disc vs ring speedParallel axis theorem applicationKE of rolling body
Exam Strategy
- Memorise the 6 key MOI values: Ring MRΒ², Disc MRΒ²/2, Solid sphere 2MRΒ²/5, Hollow sphere 2MRΒ²/3, Rod (centre) MLΒ²/12, Rod (end) MLΒ²/3.
- Rolling: use energy method. Total KE = Β½mvΒ²(1+kΒ²/RΒ²). Identify shape β look up kΒ²/RΒ² β solve for v.
- Conservation of angular momentum: identify when net torque = 0 (no external torques). Then IβΟβ = IβΟβ.
- The linear-rotational analogy table is your shortcut: if you know F=ma, you know Ο=IΞ±. If you know KE=Β½mvΒ², you know KE=Β½IΟΒ². Don't memorise twice β map once.
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