JEE Main · Physics · Class 11
Rotational motion JEE questions: formulas, patterns and the mistakes that cost marks
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
- Rotational motion accounts for roughly one to three questions in a JEE Main Physics paper, worth 4–12 marks (BrainStride analysis of past papers; NTA publishes no official weightage).
- Most questions reduce to τ = Iα, angular momentum conservation, or a rolling body on an incline.
- The rolling constraint v = ωR converts a rotation problem into a one-variable problem — write it down first.
- Standard moments of inertia are quoted about a specific axis; shift them with the parallel-axis theorem I = I_cm + Md² before using them.
- Static friction on a rolling body is whatever the constraint demands, not μN — that assumption is the single most common error in this chapter.
Rotational Motion at a glance
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
Questions per year
- 20192 Q
- 20202 Q
- 20211.8 Q
- 20222 Q
- 20232.2 Q
- 20242 Q
- 20252 Q
- 20262.2 Q
BrainStride analysis of past papers
Key concepts
Moment of inertia is axis-specific
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 and perpendicular-axis theorems
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.
Torque and angular acceleration
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.
Angular momentum and its conservation
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 without slipping
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.
Friction in rolling is static and unknown
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.
Formulas to know cold
Rotational Newton’s second law
τ = Iα
Both measured about the same axis.
Moment of inertia
I = Σ m r² · k = √(I/M)
k is the radius of gyration.
Parallel-axis theorem
I = I_cm + M d²
d is the distance between the two parallel axes.
Perpendicular-axis theorem
I_z = I_x + I_y
Plane laminae only, z perpendicular to the lamina.
Standard moments of inertia
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.
Angular momentum
L = Iω · τ_ext = dL/dt
τ_ext = 0 ⇒ I₁ω₁ = I₂ω₂.
Kinetic energy of a rolling body
KE = ½Mv² (1 + I/MR²)
Rotational share is (I/MR²)/(1 + I/MR²).
Acceleration down an incline
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.
Friction needed to roll
f = Mg sin θ · (I/MR²) / (1 + I/MR²)
Solid sphere: f = (2/7)Mg sin θ, so μ_min = (2/7)tan θ.
Angular kinematics (constant α)
ω = ω₀ + αt · θ = ω₀t + ½αt² · ω² = ω₀² + 2αθ
Same shape as linear kinematics — reuse the habit.
Where marks leak in this chapter
Quoting a moment of inertia about the wrong axis
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.Setting friction to μN on a rolling body
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.Forgetting the rolling constraint
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.Using ½Mv² alone for a rolling body’s energy
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.Assuming kinetic energy is conserved when ω changes
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.Check yourself
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:
Solved examples
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.
- For a solid sphere, I = (2/5)MR², so I/MR² = 2/5 and the rolling bracket is 1 + 2/5 = 7/5.
- a = g sin θ / (1 + I/MR²) = (10)(0.5) / (7/5) = 5 × 5/7 = 25/7 ≈ 3.6 m s⁻².
- Friction follows from f = Mg sin θ · (I/MR²)/(1 + I/MR²) = (2)(10)(0.5) × (2/5)/(7/5).
- f = 10 × 2/7 = 20/7 ≈ 2.9 N, directed up the incline.
- Check the assumption: μ_min = (2/7)tan 30° = (2/7)(0.577) ≈ 0.16, so any surface with μ above 0.16 supports rolling.
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.
- No external torque acts about the vertical axis, so angular momentum is conserved: I₁ω₁ = (I₁ + I₂)ω.
- (2)(6) = (2 + 2)ω ⇒ ω = 12/4 = 3 rad s⁻¹.
- Initial KE = ½(2)(6)² = 36 J.
- Final KE = ½(4)(3)² = 18 J.
- Energy lost = 36 − 18 = 18 J, dissipated by friction between the two surfaces as they reach a common speed.
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.
- The centre of mass falls a height L/2, so the energy released is Mg(L/2).
- About the hinge, I = ML²/3.
- Energy conservation: Mg(L/2) = ½ (ML²/3) ω².
- gL/2 = L²ω²/6 ⇒ ω² = 3g/L.
- Note that M cancels — the answer is independent of the rod’s mass.
Answer: ω = √(3g/L).
Rotational Motion: common questions
How many questions come from rotational motion in JEE Main?
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.
Is rotational motion hard to score in?
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.
Which moment of inertia formulas must I memorise for JEE Main?
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.
Why does a solid sphere beat a ring down an incline?
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.
When is friction on a rolling body equal to μN?
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.
How should I revise rotational motion in the last month?
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.
Find out whether rotational motion is actually costing you marks
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.