Rotational Motion

Rotational mechanics is arguably the highest-yield, highest-difficulty Physics topic in JEE Advanced. It fuses moment of inertia, torque, angular momentum conservation, and rolling — often in a single multi-part problem. Mastery of the parallel-axis theorem and rolling constraints is essential.

Key results

• Moment of inertia: I = Σmr²; parallel axis I = I_cm + Md²; perpendicular axis (laminae) I_z = I_x + I_y.
• Standard I: disc ½MR², ring MR², solid sphere (2/5)MR², hollow sphere (2/3)MR², rod about centre ML²/12.
• Torque & dynamics: τ = Iα; angular momentum L = Iω, conserved when net external torque = 0.
• Rolling without slipping: v_cm = ωR; a_cm = αR. KE = ½Mv² + ½Iω² = ½Mv²(1 + I/MR²).
• Rolling down incline: a = g sinθ/(1 + I/MR²); friction is static (does no net work).

Body	I (about centre)	a on incline
Ring/hollow cyl	MR²	g sinθ/2
Disc/solid cyl	½MR²	2g sinθ/3
Solid sphere	(2/5)MR²	5g sinθ/7

Exam Tricks & Tips

• 🎯 Conserve angular momentum when a person pulls in arms or a disc drops on a rotating turntable — I changes, ω adjusts so L = Iω is constant.
• 🎯 Solid sphere wins the rolling race: smaller I/MR² means larger acceleration; ordering is sphere > disc > ring.
• 🎯 On rolling, static friction does no net work but is essential to provide angular acceleration — direction can be up or down the incline.
• 🎯 Angular momentum about a point on the ground simplifies problems where the axis is not through the COM (e.g. a rod hinged at one end).
• 🎯 Instantaneous axis of rotation for a rolling body is the contact point; speed of any point = ωr from that axis.
• ❌ Common mistake: using I_cm when the axis is not through the COM — you must add Md² (parallel-axis theorem) first.

Expected exam pattern

2–3 questions guaranteed: a rolling/incline problem, an angular-momentum-conservation collision (rod + particle), and a moment-of-inertia calculation. Among the most-weighted Physics areas in the paper.

Quick recap

Know standard moments of inertia and the parallel/perpendicular axis theorems cold. Conserve L when external torque vanishes. For rolling, v_cm = ωR and a = g sinθ/(1 + I/MR²); the sphere accelerates fastest.

Cover the answer, recall, then check. 12 cards on rotational motion for JEE Advanced.

Q1. Parallel-axis theorem.
A1. I = I_cm + Md², where d is the distance between the parallel axes.

Q2. Moment of inertia of a solid sphere about a diameter.
A2. (2/5)MR².

Q3. Rolling-without-slipping constraint.
A3. v_cm = ωR and a_cm = αR.

Q4. Acceleration of a body rolling down an incline.
A4. a = g sinθ/(1 + I/MR²).

Q5. Which rolls down fastest: ring, disc, or solid sphere?
A5. Solid sphere (smallest I/MR² = 2/5), then disc, then ring.

Q6. When is angular momentum conserved?
A6. When the net external torque about the chosen axis is zero.

Q7. Perpendicular-axis theorem (for a plane lamina).
A7. I_z = I_x + I_y, with z perpendicular to the plane.

Q8. Total KE of a rolling body.
A8. ½Mv²(1 + I/MR²).

Q9. Moment of inertia of a thin rod about its centre and about one end.
A9. ML²/12 about centre; ML²/3 about one end (parallel axis).

Q10. Does static friction do work on a body rolling without slipping?
A10. No net work — the contact point is instantaneously at rest.

Q11. A particle strikes and sticks to the end of a hinged rod. What is conserved?
A11. Angular momentum about the hinge (external hinge force gives no torque about the hinge).

Q12. Location of the instantaneous axis of rotation for a rolling wheel.
A12. The point of contact with the ground.

Rotational dynamics is where JEE Advanced separates strong candidates: it demands the right axis, the right moment of inertia, and a clean choice between torque equations and angular-momentum conservation. Master rolling and you master a third of the mechanics paper.

Core concept: for rotation, torque plays the role of force, moment of inertia the role of mass, and angular momentum the role of momentum — with the crucial twist that I depends on the axis.

Core ideas

Beginner — moment of inertia

I = Σmr² measures rotational inertia about a chosen axis. Memorise the standard results: ring MR², disc ½MR², solid sphere (2/5)MR², hollow sphere (2/3)MR², rod about centre ML²/12. Two theorems extend these: parallel axis I = I_cm + Md², and perpendicular axis (planar bodies) I_z = I_x + I_y.

Intermediate — torque and angular momentum

τ = Iα for a fixed axis; more generally τ = dL/dt. Angular momentum L = Iω for rigid rotation, or L = r × p for a particle. When net external torque = 0, L is conserved — the ice-skater pulling in her arms spins faster because I drops and Iω stays constant. This is the rotational twin of momentum conservation and is tested constantly.

Advanced — rolling without slipping

Rolling combines translation and rotation with the constraint v_cm = ωR (and a_cm = αR). The contact point is instantaneously at rest, so static friction acts (it does no work). Total KE = ½Mv_cm² + ½I_cm ω² = ½Mv_cm²(1 + I_cm/MR²). For a body rolling down an incline, a = g sinθ/(1 + I/MR²): a solid sphere (I/MR² = 2/5) beats a ring (I/MR² = 1) — less rotational inertia rolls faster, independent of mass and radius. The minimum friction to sustain rolling is μ ≥ tanθ/(1 + MR²/I).

Angular impulse: ∫τ dt = ΔL, the tool for sudden strikes (a ball hit off-centre, a rod struck at one end and pivoting about its CM or a hinge).

Worked example

A uniform rod of mass M, length L, hangs vertically from a frictionless hinge at the top. A bullet of mass m, speed v, strikes it horizontally at the bottom and embeds. Find the angular speed just after. Angular momentum about the hinge is conserved (hinge exerts force but no torque about itself): mvL = I_total ω, where I_total = ML²/3 + mL². So ω = mvL/(ML²/3 + mL²) = mv/(ML/3 + mL) = 3mv/((M + 3m)L). Use L conservation, not momentum — the hinge supplies an unknown impulsive force, so linear momentum is not conserved.

How JEE Advanced tests this

Rolling races and rolling-down-incline energy/acceleration; ballistic-pendulum-style angular-momentum problems (bullet–rod, disc dropped on disc); pure vs impure rolling and the friction check; and combined-motion problems where a spool is pulled by a thread at various angles.

Exam tricks & shortcuts

• Rolling KE = translational KE × (1 + I_cm/MR²); the bracket ranks how "slow" a shape rolls.
• Hinge/pivot present? Use angular momentum about the hinge — its impulse has zero torque there.
• Mnemonic: "Sphere beats disc beats ring" downhill.

Key formulas

• Torque: τ = Iα = r F sinθ; angular momentum: L = Iω; τ = dL/dt.
• Moment of inertia: I = Σmr² = Mk²; parallel axis I = I_cm + Md²; perpendicular I_z = I_x + I_y.
• Standard I: disc ½MR², ring MR², solid sphere (2/5)MR², rod ML²/12.
• Rolling: v = ωR; KE = ½Mv² + ½Iω²; a = g sinθ/(1 + k²/R²).
• Conservation: I₁ω₁ = I₂ω₂ (τ_ext = 0).

Rotational quantities mirror linear ones (I↔m, τ↔F, L↔p); angular momentum is conserved without external torque.

Worked Example

Problem: A solid sphere (moment of inertia I = (2/5)MR²) rolls without slipping from rest down an incline of vertical height 1.4 m. Find its speed at the bottom, and compare it with a block that slides down the same frictionless height. (g = 10 m/s².)

Solution:
For rolling without slipping, v = ωR. Conservation of energy (all PE → translational + rotational KE):
Mgh = ½Mv² + ½Iω²
= ½Mv² + ½(2/5 MR²)(v/R)²
= ½Mv² + (1/5)Mv²
= (7/10)Mv².

Solve for v:
v = √(10gh/7) = √(10 × 10 × 1.4 / 7) = √(140/7 × 10)... compute directly: 10gh/7 = (10×10×1.4)/7 = 140/7 × ... = 200/...
10gh = 140; 140/7 = 20; so v² = 20, v = √20 ≈ 4.47 m/s.

For a sliding block (no rotation): ½Mv² = Mgh → v = √(2gh) = √28 ≈ 5.29 m/s.

The rolling sphere is slower because part of its energy goes into rotation.

Answer: Rolling sphere: v ≈ 4.47 m/s; sliding block: v ≈ 5.29 m/s.