JEE Main Physics Rotational Motion 2027: Moment of Inertia, Torque & Rolling
Moment of inertia values, torque, angular momentum and rolling on an incline, with worked examples and the shortcuts that save time in JEE Main.
Edurack
September 28, 2026
A solid sphere, a disc and a ring roll down the same incline. Who wins? You can answer without a single calculation once you know one idea: the smaller the moment of inertia per unit mass, the faster it rolls. Rotational Motion looks intimidating, but it mirrors linear motion almost line for line.
Every linear formula has a rotational twin: force becomes torque, mass becomes moment of inertia.
Chapter at a Glance
| Snapshot | Detail |
|---|---|
| NTA unit | Unit 5 of 20 — Rotational Motion |
| Priority (trend-based) | High |
| Typical question style | Formula-driven numericals on MOI, torque and rolling bodies |
| Best first step | Memorise the moment of inertia table, then rolling on an incline |
Priority reflects past-paper trends, not an official NTA weightage.
What the NTA Syllabus Covers
- Centre of mass of a two-particle system and of a rigid body
- Moment of a force, torque, angular momentum and its conservation
- Moment of inertia, radius of gyration, parallel and perpendicular axes theorems
- Equilibrium of rigid bodies, equations of rotational motion, comparison of linear and rotational motion
Master These Topics
1. Moment of Inertia: The Table Worth Memorising
Moment of inertia I measures resistance to angular acceleration: I = Σmr².
| Body | Axis | I |
|---|---|---|
| Thin ring | Through centre, perpendicular to plane | MR² |
| Solid disc | Through centre, perpendicular to plane | ½MR² |
| Solid sphere | Through diameter | ⅖MR² |
| Hollow sphere | Through diameter | ⅔MR² |
| Thin rod | Through centre, perpendicular to length | ML²/12 |
| Thin rod | Through one end | ML²/3 |
Two theorems extend the table. Parallel axes: I = I_cm + Md². Perpendicular axes (for flat bodies only): I_z = I_x + I_y.
Worked example: A 1 kg rod of length 1.2 m about an end has I = ML²/3 = 1 × 1.44 / 3 = 0.48 kg·m².
Trap: The perpendicular axes theorem works only for planar (flat) bodies. Do not use it for a sphere or a cylinder.
2. Torque and Angular Momentum
Torque is τ = r × F, and the rotational form of Newton's second law is τ = Iα.
Worked example: A disc (M = 2 kg, R = 0.5 m) has I = ½ × 2 × 0.25 = 0.25 kg·m². A torque of 0.5 N·m gives α = 0.5 / 0.25 = 2 rad/s².
Angular momentum is L = Iω. With no external torque, L stays constant. A skater spinning at 2 rad/s with I = 4 kg·m² pulls in her arms to reduce I to 2 kg·m². New speed: ω = 4 × 2 / 2 = 4 rad/s. Rotational kinetic energy doubles, because she does work pulling her arms in.
3. Rolling Without Slipping
For pure rolling, v = ωR, and total kinetic energy is ½mv² (1 + k²/R²), where k is the radius of gyration. On an incline of angle θ:
a = g sinθ / (1 + k²/R²)
- Solid sphere:
a = 5g sinθ / 7 - Solid disc:
a = 2g sinθ / 3 - Hollow sphere:
a = 3g sinθ / 5 - Ring:
a = g sinθ / 2
So the race order is solid sphere, disc, hollow sphere, ring. Mass and radius do not matter, only the shape.
Common Traps to Avoid
- Using the perpendicular axes theorem on a 3D body.
- Forgetting that I depends on the axis you choose.
- Conserving kinetic energy when angular momentum is conserved. They are different quantities.
- Applying
a = g sinθto a rolling body instead of the reduced rolling acceleration.
60-Second Revision Sheet
- Ring
MR², disc½MR², solid sphere⅖MR², rod endML²/3 τ = Iα,L = Iω,KE_rot = ½Iω²- Rolling:
a = g sinθ / (1 + k²/R²) - Parallel axes
I = I_cm + Md²
Your Study Plan
- Day 1: moment of inertia table plus parallel and perpendicular axes problems.
- Day 2: torque, angular acceleration and pulley-with-mass problems.
- Day 3: angular momentum conservation and collision with rotating bodies.
- Day 4: rolling on inclines and energy methods, timed practice.
Practice Rotational Motion Questions Free → (opens in a new tab)
Continue Your Physics Journey
- Previous chapter: Work, Energy & Power
- Next chapter: Gravitation
- All 20 JEE Main Physics chapters
- Complete JEE Main Syllabus 2027 guide
Frequently Asked Questions
Is Rotational Motion difficult for JEE Main?
It is formula-heavy but predictable. Once the moment of inertia table and rolling formulas are memorised, most problems become substitution and careful setup.
Why does a solid sphere beat a ring on an incline?
A ring has more of its mass far from the axis, so it needs a larger share of the energy for rotation, leaving less for translation.