Why this chapter is worth the time it takes
Two facts make Rotational Motion different from the chapters before it. First, its questions are long - a typical one involves setting up a moment of inertia, applying a torque equation, and then bringing in energy conservation or the rolling condition. Second, it combines with almost everything: energy conservation for rolling bodies, momentum conservation for collisions involving rotation, and circular motion throughout.
The consequence for preparation is that this chapter cannot be crammed in the last week the way Modern Physics can. It needs enough practice for the setup to become automatic, because in the exam the time cost is in deciding how to start, not in the algebra afterwards.
How JEE actually asks Rotational Motion
As elsewhere, NTA publishes no chapter-wise weightage - figures circulating online are coaching estimates from past papers, and the syllabus and pattern for your session are in the official NTA information bulletin.
What past papers do show is that rotational questions are usually hybrid. A body rolls down an incline and you are asked for its speed at the bottom, which needs both a moment of inertia and energy conservation. A rod is struck at one end and you are asked for the angular velocity afterwards, which needs angular momentum conservation about the pivot. Preparing moments of inertia in isolation therefore leaves you short of what the paper asks.
Key concepts, compressed
- ✓Moment of inertia is rotational inertia: it measures how hard it is to change a body's rotation, and it depends on how the mass is distributed relative to the axis.
- ✓Torque is the rotational analogue of force, τ = r × F, and its magnitude is rF sinθ where θ is the angle between the position vector and the force.
- ✓The rotational form of Newton's second law is τ = Iα.
- ✓Angular momentum L = Iω is conserved when the net external torque is zero, which is why a spinning skater speeds up on pulling their arms in.
- ✓A rolling body's total kinetic energy is ½mv² + ½Iω², and with v = ωR this can be written entirely in terms of v.
- ✓The parallel-axis theorem, I = I(cm) + Md², converts a known moment of inertia about the centre of mass to any parallel axis.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Torque | τ = rF sinθ = Iα | |
| Angular momentum | L = Iω | = mvr for a particle |
| Rotational KE | KE = ½Iω² | |
| Rod about its centre | I = ML²/12 | |
| Rod about its end | I = ML²/3 | from the parallel-axis theorem |
| Ring about its axis | I = MR² | |
| Disc about its axis | I = MR²/2 | |
| Solid sphere about a diameter | I = 2MR²/5 | |
| Hollow sphere about a diameter | I = 2MR²/3 | |
| Parallel-axis theorem | I = I(cm) + Md² | |
| Rolling condition | v = ωR | |
| Rolling KE | KE = ½mv² + ½Iω² | |
| Rolling down an incline | a = g sinθ / (1 + I/MR²) | |
| Centre of mass | x(cm) = Σmᵢxᵢ / Σmᵢ |
The five mistakes that cost the most marks
- ✓Using a moment of inertia about the wrong axis. A rod about its centre is ML²/12 but about its end is ML²/3 - four times larger. Always name the axis before writing the value.
- ✓Assuming angular momentum is conserved. It is conserved only when the net external torque about the chosen axis is zero, which is often not the case.
- ✓Forgetting the rolling condition v = ωR. Without it, rolling problems have two unknowns and one equation.
- ✓Counting only translational kinetic energy for a rolling body. Its energy is split between ½mv² and ½Iω², and the split depends on shape.
- ✓Assuming a heavier or larger body rolls down faster. The acceleration depends only on I/MR², which is a pure shape factor - mass and radius cancel out entirely.
Practice set 1: moment of inertia
1. A uniform rod of mass 2 kg and length 3 m rotates about an axis through its centre, perpendicular to its length. Find its moment of inertia.
1.5 kg·m². I = ML²/12 = 2 × 9/12 = 1.5 kg·m². Note that the moment of inertia grows with the square of the length, so doubling the rod's length while keeping its mass would quadruple this value.
2. The same rod now rotates about an axis through one end. Find its moment of inertia.
6 kg·m². Either quote I = ML²/3 = 2 × 9/3 = 6 kg·m², or derive it from the parallel-axis theorem: I = I(cm) + Md² = 1.5 + 2 × (1.5)² = 1.5 + 4.5 = 6 kg·m². The end value is four times the centre value, which is why naming the axis matters more here than anywhere else in the chapter.
3. A uniform disc of mass 4 kg and radius 0.5 m rotates about its central axis. Find its moment of inertia.
0.5 kg·m². I = MR²/2 = 4 × 0.25/2 = 0.5 kg·m². Compare with a ring of the same mass and radius, which has I = MR² = 1 kg·m² - twice as much, because all its mass sits at the maximum distance from the axis.
4. A solid sphere of mass 5 kg and radius 0.2 m rotates about a diameter. Find its moment of inertia.
0.08 kg·m². I = 2MR²/5 = 2 × 5 × 0.04/5 = 0.08 kg·m². The solid sphere has the smallest I/MR² of the standard bodies at 0.4, which is exactly why it wins rolling races - a fact several later questions depend on.
5. A ring of mass 2 kg and radius 0.5 m rotates about one of its diameters. Find its moment of inertia.
0.25 kg·m². Use the perpendicular-axis theorem: for a planar body, I(z) = I(x) + I(y). About the central axis I(z) = MR² = 0.5 kg·m², and by symmetry the two diameters are equivalent, so each is half of that: 0.25 kg·m². This theorem applies only to planar bodies, which is a condition questions test directly.
6. State the parallel-axis theorem and the condition for using it.
I = I(cm) + Md², where d is the perpendicular distance between the two axes. The condition is that the two axes must be parallel and one of them must pass through the centre of mass. Applying it between two arbitrary parallel axes, neither through the centre of mass, is invalid - a restriction that is frequently overlooked.
7. Two bodies have the same mass. Must they have the same moment of inertia?
No. Moment of inertia depends on how the mass is distributed relative to the axis, not just on how much there is. A ring and a disc of identical mass and radius differ by a factor of two, and the same body has different values about different axes. This is the single most important conceptual point in the chapter.
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Practice set 2: torque and angular acceleration
8. A force of 10 N is applied perpendicular to a wrench at a distance of 0.5 m from the axis. What is the torque?
5 N·m. τ = rF sinθ = 0.5 × 10 × sin90° = 5 N·m. Because the force is perpendicular, sinθ = 1 and the torque is simply rF. This is also why a longer spanner loosens a bolt more easily - the same force at a greater r produces more torque.
9. A torque of 5 N·m acts on a body of moment of inertia 2.5 kg·m². What is its angular acceleration?
2 rad/s². τ = Iα, so α = τ/I = 5/2.5 = 2 rad/s². This is the direct rotational analogue of a = F/m, and the correspondence is worth holding in mind throughout the chapter: torque plays the role of force and moment of inertia the role of mass.
10. A force is applied along a line passing through the axis of rotation. What torque does it produce?
Zero. The position vector and the force are then parallel, so sinθ = 0 and τ = rF sinθ = 0. This is why pushing a door at its hinge does not open it, however hard you push - only the component of force perpendicular to the position vector produces turning.
11. A force of 10 N acts at 30° to the position vector, at a distance of 2 m from the axis. What is the torque?
10 N·m. τ = rF sinθ = 2 × 10 × sin30° = 2 × 10 × 0.5 = 10 N·m. Using cos30° instead gives about 17.3 N·m and is the standard error - torque uses the perpendicular component of the force, which is the sine, while work uses the parallel component, which is the cosine.
12. A disc of moment of inertia 0.5 kg·m² starts from rest under a constant torque of 2 N·m. What is its angular velocity after 4 s?
16 rad/s. First α = τ/I = 2/0.5 = 4 rad/s². Then ω = ω₀ + αt = 0 + 4 × 4 = 16 rad/s. The rotational kinematic equations mirror the linear ones exactly, so every technique from the Kinematics chapter transfers directly.
13. What is the rotational kinetic energy of that disc at t = 4 s?
64 J. KE = ½Iω² = ½ × 0.5 × 16² = ½ × 0.5 × 256 = 64 J. Check it by the work-energy theorem for rotation: the angular displacement in 4 s is ½αt² = ½ × 4 × 16 = 32 rad, and the work done is τθ = 2 × 32 = 64 J. The two agreeing is a good habit to build.
Practice set 3: angular momentum
14. A body of moment of inertia 3 kg·m² rotates at 4 rad/s. What is its angular momentum?
12 kg·m²/s. L = Iω = 3 × 4 = 12 kg·m²/s, directed along the axis of rotation as given by the right-hand rule. The units are equivalent to J·s, which is why the same dimensions appear in Planck's constant in Modern Physics.
15. A spinning skater pulls their arms in, halving their moment of inertia. What happens to their angular velocity?
It doubles. There is no external torque about the spin axis, so angular momentum is conserved: I₁ω₁ = I₂ω₂. Halving I therefore doubles ω. The skater does not push against anything external - the change comes entirely from redistributing their own mass closer to the axis.
16. In that situation, what happens to the skater's kinetic energy?
It doubles. Using KE = L²/2I with L constant, halving I doubles the kinetic energy. The extra energy comes from the muscular work the skater does pulling their arms inward against the outward tendency of the rotating mass. This is a favourite exam question because students expect energy to be conserved alongside angular momentum - it is not.
17. Under what condition is angular momentum conserved?
When the net external torque about the chosen axis is zero. Internal forces cannot change a system's total angular momentum, but any external torque can. A body rolling down an incline has a non-zero torque about most axes, so conservation does not apply there - which is why energy methods are used for rolling instead.
18. A particle of mass 2 kg moves in a straight line at 3 m/s, passing a fixed point at a perpendicular distance of 4 m. What is its angular momentum about that point?
24 kg·m²/s. L = mvr = 2 × 3 × 4 = 24 kg·m²/s, where r is the perpendicular distance from the point to the line of motion. A particle moving in a straight line does have angular momentum about a point not on that line, and it stays constant as the particle moves - which surprises most students the first time.
19. In what direction does the angular momentum of a rotating body point?
Along the axis of rotation, in the sense given by the right-hand rule: curl the fingers of your right hand in the direction of rotation and the thumb gives the direction of L. Angular momentum is an axial vector, which is why it points along the axis rather than in the plane of motion.
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Practice set 4: rolling motion
20. What is the relationship between linear and angular velocity for a body rolling without slipping?
v = ωR, where v is the speed of the centre of mass and R the radius. This condition says the contact point is instantaneously at rest relative to the surface, which is what distinguishes rolling from sliding. Without this relation a rolling problem has two unknowns and cannot be solved.
21. For a solid sphere rolling without slipping, what fraction of its total kinetic energy is rotational?
2/7, or about 29%. Rotational KE = ½Iω² = ½(2MR²/5)(v/R)² = Mv²/5, and translational KE = ½Mv². The total is (1/5 + 1/2)Mv² = (7/10)Mv², so the rotational share is (1/5)/(7/10) = 2/7. Notice the result depends only on shape, not on mass, radius or speed.
22. What is the same fraction for a rolling ring?
1/2. For a ring, I = MR², so rotational KE = ½MR²(v/R)² = ½Mv², exactly equal to the translational kinetic energy. The energy splits evenly, which is more than double the sphere's rotational share - and it is precisely why a ring accelerates more slowly down a slope.
23. A solid sphere rolls without slipping down a 30° incline. What is the acceleration of its centre of mass?
About 3.57 m/s². Using a = g sinθ/(1 + I/MR²) with I/MR² = 2/5: a = 10 × 0.5/(1 + 0.4) = 5/1.4 ≈ 3.57 m/s². Compare with a frictionless sliding block on the same incline, which accelerates at the full g sinθ = 5 m/s² - the sphere is slower because some of the available energy goes into spinning it up.
24. A ring, a disc and a solid sphere are released together from the top of an incline. Which reaches the bottom first?
The solid sphere, then the disc, then the ring. Acceleration depends on I/MR², which is 0.4 for the sphere, 0.5 for the disc and 1 for the ring - a smaller value means a larger acceleration. Mass and radius cancel completely, so a small light sphere and a large heavy one arrive together, and both beat any ring.
25. A solid sphere rolls from rest down a vertical drop of 1.4 m. What is the speed of its centre of mass at the bottom?
About 4.47 m/s. Energy conservation gives mgh = ½mv² + ½Iω², and with I = 2mR²/5 and ω = v/R this becomes mgh = (7/10)mv². So v² = 10gh/7 = 10 × 10 × 1.4/7 = 20, giving v = √20 ≈ 4.47 m/s. A body sliding without friction from the same height would reach √(2 × 10 × 1.4) ≈ 5.29 m/s - faster, because none of its energy goes into rotation.
Practice set 5: centre of mass and mixed
26. A 2 kg mass sits at x = 0 and a 3 kg mass at x = 5 m. Where is the centre of mass?
At x = 3 m. x(cm) = Σmᵢxᵢ/Σmᵢ = (2 × 0 + 3 × 5)/(2 + 3) = 15/5 = 3 m. The centre of mass sits closer to the heavier body, which is a useful check - if your answer came out closer to the 2 kg mass, the weighting has been applied the wrong way round.
27. Where is the centre of mass of a uniform rod, and must the centre of mass always lie within the body?
At its geometric centre for a uniform rod - but no, the centre of mass need not lie within the body. A ring has its centre of mass at the centre of the hole, where there is no material at all, and the same is true of a horseshoe or a hollow sphere. Questions use these shapes specifically to test the point.
28. Can internal forces change the position of a system's centre of mass?
No. Internal forces occur in third-law pairs and cancel, so only external forces can accelerate the centre of mass. An exploding shell's fragments fly apart, but the centre of mass continues along the original projectile path - which is a standard exam scenario combining this chapter with Kinematics.
29. A 1 kg particle is at the origin and a 3 kg particle at (4, 0) m. Find the centre of mass.
At (3, 0) m. x(cm) = (1 × 0 + 3 × 4)/(1 + 3) = 12/4 = 3 m, and y(cm) = 0 since both lie on the x-axis. The centre of mass divides the line joining two particles in the inverse ratio of their masses - here 3:1, so it sits three times further from the light particle than from the heavy one.
30. A flywheel of moment of inertia 10 kg·m² spinning at 20 rad/s is brought to rest in 5 s. What constant torque was applied?
40 N·m, opposing the rotation. First α = (ω − ω₀)/t = (0 − 20)/5 = −4 rad/s². Then τ = Iα = 10 × (−4) = −40 N·m, so the magnitude is 40 N·m and the negative sign indicates it opposes the motion. Cross-check by energy: the initial kinetic energy is ½ × 10 × 400 = 2000 J, the angular displacement is 50 rad, and 2000/50 = 40 N·m.
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How to study this chapter efficiently
- ✓Name the axis before writing any moment of inertia. Most wrong answers in this chapter are correct values about the wrong axis.
- ✓Memorise six standard bodies - rod, ring, disc, hollow cylinder, hollow sphere, solid sphere - about their standard axes. Everything else comes from the parallel and perpendicular axis theorems.
- ✓Learn the linear-to-rotational correspondence as a single table rather than as two sets of formulas: m becomes I, v becomes ω, F becomes τ, p becomes L. Half the chapter is then already familiar.
- ✓For any rolling question, write v = ωR immediately, before anything else. It is the equation that makes the problem solvable and the one most often forgotten.
- ✓Remember that I/MR² is a pure shape number. Any rolling answer that depends on mass or radius is wrong, which makes it a fast way to eliminate options.
- ✓Do not leave this chapter to the last week. Its questions are long, and fluency in the setup is what saves time in the exam - that only comes from spaced practice.
Turn this into active practice
The errors in this chapter are systematic rather than random. A student who uses the centre-of-mass moment of inertia where an end-axis value is needed will keep doing it until something makes the pattern visible, and rereading a formula sheet does not.
The JEE Rotational Motion quiz on QUFF generates fresh questions across moments of inertia, torque, angular momentum and rolling, scores them instantly and explains each answer. Do a timed set, and sort your errors into three piles: wrong axis, forgot the rolling condition, or missed that angular momentum was not conserved. Almost every mistake in this chapter falls into one of those three, and each needs a different correction.
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