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JEE Main · Physics · Class 11
Rotational motion carries one to three of the 25 JEE Main Physics questions in our reading of past papers, and almost every one of them turns on two decisions: which axis the moment of inertia is about, and whether the body is rolling without slipping. Get those right and the arithmetic is short.
Quick answer
Questions per paper
1–3
Our analysis of past papers
Marks at stake
4–12
At +4 / −1 per question
Difficulty
Moderate–hard
Long questions, short arithmetic
Class
11
Builds on work, energy and power
BrainStride analysis of past papers
I = Σmr² measures how mass is spread about one particular axis. A uniform rod is ML²/12 about its centre and ML²/3 about an end — same rod, different number. Always name the axis before you quote a value.
Parallel axis shifts a known I a distance d away from the centre of mass: I = I_cm + Md². Perpendicular axis applies to plane laminae only: I_z = I_x + I_y. Together they cover most exam variations.
τ = Iα is the rotational F = ma. Torque about a point is τ = rF sin θ, where θ is the angle between the position vector and the force — a force pointing at the axis produces no torque at all.
L = Iω for a rigid body about a fixed axis. When no external torque acts about that axis, L is constant: pull mass inwards and ω rises. Kinetic energy is not conserved in such a change — muscles or internal forces do work.
Rolling means the contact point is instantaneously at rest, so v_cm = ωR and a_cm = αR. Total kinetic energy splits as ½Mv² + ½Iω², which factorises to ½Mv²(1 + I/MR²). The bracket decides every race down an incline.
On an incline, static friction takes exactly the value the rolling constraint needs — f = Mg sin θ · (I/MR²)/(1 + I/MR²) — and only reaches μN at the point of slipping. Setting f = μN too early is the classic way to lose the question.
τ = Iα
Both measured about the same axis.
I = Σ m r² · k = √(I/M)
k is the radius of gyration.
I = I_cm + M d²
d is the distance between the two parallel axes.
I_z = I_x + I_y
Plane laminae only, z perpendicular to the lamina.
Rod (centre) ML²/12 · Rod (end) ML²/3 · Ring MR² · Disc MR²/2 · Solid sphere 2MR²/5 · Hollow sphere 2MR²/3
Each about the axis named; shift with the parallel-axis theorem.
L = Iω · τ_ext = dL/dt
τ_ext = 0 ⇒ I₁ω₁ = I₂ω₂.
KE = ½Mv² (1 + I/MR²)
Rotational share is (I/MR²)/(1 + I/MR²).
a = g sin θ / (1 + I/MR²)
Solid sphere (5/7)g sin θ, disc (2/3)g sin θ, ring (1/2)g sin θ.
f = Mg sin θ · (I/MR²) / (1 + I/MR²)
Solid sphere: f = (2/7)Mg sin θ, so μ_min = (2/7)tan θ.
ω = ω₀ + αt · θ = ω₀t + ½αt² · ω² = ω₀² + 2αθ
Same shape as linear kinematics — reuse the habit.
Why it happens
Standard results are memorised as bare fractions, and under time pressure the axis they belong to is dropped.The fix
Write the axis next to the value every time. If it is not the centre of mass, apply I = I_cm + Md² before anything else.Why it happens
Friction is taught as μN in the block-on-incline chapter, and the habit carries over.The fix
For rolling without slipping, treat f as an unknown, solve the torque and force equations together, and use μN only to test whether the body slips.Why it happens
The problem has two unknowns, v and ω, so it looks unsolvable without the extra equation.The fix
Write v = ωR (and a = αR) on the diagram as soon as you read “rolls without slipping”. It is the equation that closes the system.Why it happens
Energy conservation is practised on sliding blocks, where there is no rotational term.The fix
Use KE = ½Mv²(1 + I/MR²). For a solid sphere that bracket is 7/5, which changes the answer by a fifth.Why it happens
Angular momentum conservation and energy conservation are applied together out of habit.The fix
With L fixed, KE = L²/2I. Halving I doubles the kinetic energy — the extra energy comes from the work done pulling the mass in.5 questions
Question 1
A uniform rod of mass M and length L rotates about an axis through one end, perpendicular to the rod. Its moment of inertia is:
Question 2
A solid sphere, a disc and a ring of equal mass and radius are released together from rest at the top of the same incline and roll without slipping. Which reaches the bottom first?
Question 3
For a solid sphere rolling without slipping, what fraction of its total kinetic energy is rotational?
Question 4
A skater spinning freely pulls her arms in, halving her moment of inertia. Her kinetic energy:
Question 5
A solid sphere rolls without slipping down a 45° incline. The minimum coefficient of static friction required is about:
A test on one chapter alone tells you little — the exam never asks Rotational Motion in isolation. So this builds a full JEE Main Physics paper that leans on it. Weighted like the exam, a 25-question paper gives Rotational Motion 2 questions; marked weak, it gets 4, and the rest of the paper keeps the other chapters honest. Fill the plan from previous-year questions or your module.
Your paper
Press Generate paper plan for a chapter-by-chapter blueprint: questions per chapter, numericals, marks and time — ready to print and fill from your PYQs, modules or the BrainStride question bank.
Chapter weights are BrainStride’s reading of past papers, not official data. For any other chapter mix, use the JEE Main Physics test generator.
Rolling on an incline · 3 marks · moderate
A solid sphere of mass 2 kg and radius R rolls without slipping down an incline of 30°. Take g = 10 m s⁻². Find its acceleration and the friction force acting on it.
Answer: a ≈ 3.6 m s⁻² down the incline; f ≈ 2.9 N up the incline (μ_min ≈ 0.16).
Angular momentum conservation · 4 marks · moderate
A disc of moment of inertia 2 kg m² spins freely at 6 rad s⁻¹ about a vertical axis. An identical disc, initially at rest, is dropped coaxially onto it and the two rotate together. Find the common angular speed and the kinetic energy lost.
Answer: ω = 3 rad s⁻¹; kinetic energy lost = 18 J (half the original).
Rotation about a fixed axis · 3 marks · moderate
A uniform rod of mass M and length L is hinged at one end and held horizontal. It is released from rest. Find its angular speed when it passes through the vertical position.
Answer: ω = √(3g/L).
In BrainStride’s analysis of past JEE Main papers, rotational motion accounts for about one to three of the 25 Physics questions, which is 4 to 12 marks at +4 and −1. NTA publishes no official chapter weightage, so treat this as a range from past papers rather than a guarantee for any single shift.
Rotational motion questions read as long but solve short. Most of them need one moment of inertia, the rolling constraint v = ωR, and a single conservation law. The difficulty lies in setting the problem up correctly, not in the algebra, which is why practising the setup on twenty varied questions beats reading the theory again.
Six cover almost every question: rod about its centre ML²/12 and about an end ML²/3, ring MR², disc MR²/2, solid sphere 2MR²/5 and hollow sphere 2MR²/3. Learn each with the axis it belongs to, then use the parallel-axis theorem I = I_cm + Md² for any other axis.
Acceleration on an incline is g sin θ divided by (1 + I/MR²). A solid sphere has I/MR² = 2/5 and a ring has 1, so the sphere accelerates at (5/7)g sin θ against the ring’s (1/2)g sin θ. Mass and radius cancel out entirely, so only the mass distribution decides the race.
Only at the point of slipping. While a body rolls without slipping, static friction takes exactly the value the constraint demands, which for a solid sphere on an incline is (2/7)Mg sin θ. Use μN as a test of whether rolling is possible, never as the friction value inside the equations.
Revise by question type rather than by reading. Do five questions each on moment of inertia with a shifted axis, rolling on an incline, angular momentum conservation and combined translation with rotation. If your error is in the setup, the fix is more setups, not more theory.
The 20-minute diagnostic tests this chapter alongside the rest of mechanics and tells you where your marks are leaking — chapter by chapter, mistake by mistake.