Rotational Motion — Moment of Inertia, Torque and Rolling
Rotation is where students who have coped with mechanics by memorising formulae come unstuck, and where students who understood the analogy sail through. Every rotational quantity is the translational one with mass replaced by moment of inertia and distance by angle: force becomes torque, momentum becomes angular momentum, ½mv² becomes ½Iω². If you build that table once and trust it, most of the chapter is a translation exercise. What genuinely needs care is that I depends on the axis, not just the body, and that rolling is two motions at once — and both of those are exactly where the paper sets its questions.
Angular Momentum and Its Conservation
The rotational form of Newton's second law is τ⃗_ext = dL⃗/dt. Its immediate consequence is the most useful single result in the chapter: if the net external torque is zero, L is conserved, so Iω stays constant even if I changes.
🎯 Why a spinning skater speeds up — and why her kinetic energy rises
She pulls her arms in, which reduces I, and since Iω is fixed, ω must rise. Halving I doubles ω. Now compute the kinetic energy: KE = ½Iω² = L²/2I, so halving I doubles the kinetic energy. Where did that energy come from? Not from nowhere — she did work with her muscles pulling her arms inward against the outward pull she felt. This is a favourite question because students assume that if L is conserved then energy must be too, and the two conservation laws have entirely different conditions: L needs zero external torque, energy needs no non-conservative work. Writing KE = L²/2I rather than ½Iω² makes the relationship immediately visible.⚠️ A torque can be zero even when the force is not
Because τ = rF sin θ, a torque vanishes in two distinct ways: the force passes through the axis (r = 0), or the force is along the line joining to the axis (sin θ = 0). This is what makes central forces special — gravity on a planet always points at the Sun, so it exerts no torque about the Sun and the planet's angular momentum is conserved, which is Kepler's second law. In the same way the tension in the string of a whirling stone exerts no torque about the centre. So when a question asks whether L is conserved, do not look for the absence of forces; look for the absence of torque, which is a much weaker and more often satisfied condition.Rolling Without Slipping
Rolling is translation and rotation at once, tied together by the constraint v_cm = ωR. The kinetic energy therefore has two parts: KE = ½mv² + ½Iω², and substituting I = kmR² (where k is ⅖ for a solid sphere, ½ for a disc, 1 for a ring) gives KE = ½mv²(1 + k).
| Body | k in I = kmR² | v at the bottom = √(2gh/(1+k)) |
|---|---|---|
| Solid sphere | 2/5 | √(10gh/7) — fastest |
| Solid cylinder / disc | 1/2 | √(4gh/3) |
| Hollow sphere | 2/3 | √(6gh/5) |
| Ring / hollow cylinder | 1 | √(gh) — slowest |
🎯 Why mass and radius drop out, and the ring always loses
Look at v = √(2gh/(1+k)): there is no m and no R in it. A marble and a cannonball, if both are solid spheres, reach the bottom together — and a big ring loses to a small sphere, but also to a big sphere. The only thing that matters is k, the distribution of mass relative to the axis. A ring keeps all its mass at radius R, so it has the largest k, so the largest share of the released potential energy goes into spinning rather than advancing, and it arrives last. This is a genuinely counter-intuitive result and it is asked almost every year, usually as "which reaches the bottom first" with four bodies of different masses and radii offered to tempt you into a calculation you do not need.⚠️ Friction in rolling does no work — but it is essential
Two facts sit uncomfortably together and both are true. Rolling requires friction: without it the body would slide rather than roll, because friction supplies the torque that spins it up. Yet in pure rolling without slipping, friction does no work, because the contact point has zero instantaneous velocity — there is no sliding at the contact for friction to dissipate. That is why you may use energy conservation freely on a rolling body even though a friction force is present, which feels wrong the first time and is the point of the question. The moment slipping begins, this breaks: kinetic friction then acts at a genuinely sliding contact, energy is dissipated, and v_cm = ωR no longer holds.Key takeaways
- Rotation mirrors translation: m→I, F→τ, p→L, ½mv²→½Iω², and all three equations of motion translate.
- I depends on the AXIS, not only the body — a rod is ML²/12 about its centre and ML²/3 about an end.
- Parallel axes: I = I_cm + Md². Perpendicular axes (planar bodies only): I_z = I_x + I_y.
- Ring MR², disc MR²/2, solid sphere 2MR²/5, shell 2MR²/3.
- τ = 0 when the force passes through the axis or acts along the line to it — hence Kepler's second law.
- L conserved does NOT mean energy conserved: KE = L²/2I, so halving I doubles the kinetic energy.
- Rolling: v_cm = ωR and KE = ½mv²(1+k); down an incline v = √(2gh/(1+k)), with no m or R in it.
- Solid sphere wins the race, ring loses, and mass and radius are irrelevant — only k matters.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The moment of inertia of a uniform rod of mass M and length L about an axis through one end, perpendicular to its length, is:
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Answer: B — ML²/3
By the parallel-axes theorem, I = I_cm + Md² = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3. Option 1 is the value about the centre, offered because a question that does not name the axis is exactly how this distinction is tested — the end value is four times the central one.A solid sphere, a disc and a ring of equal mass and radius roll down the same incline from the same height. The order of arrival at the bottom is:
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Answer: B — Sphere, disc, ring
Since v = √(2gh/(1+k)), the smallest k wins: sphere (k = 2/5), then disc (1/2), then ring (1). The ring keeps all its mass at radius R, so the largest share of the released potential energy goes into spinning rather than advancing. Mass and radius do not appear in the formula at all.A skater spinning with arms outstretched pulls them in, halving her moment of inertia. Her angular velocity and kinetic energy respectively:
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Answer: B — Double and double
No external torque acts, so L = Iω is conserved and halving I doubles ω. Writing KE = L²/2I shows that halving I also doubles the kinetic energy — supplied by the work her muscles do pulling the arms inward. Assuming energy must be conserved because L is conserved is the standard error: the two laws have different conditions.The angular momentum of a planet about the Sun is conserved because:
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Answer: B — Gravity is central, so it exerts no torque about the Sun
Gravity certainly acts, but it always points along the line to the Sun, so sin θ = 0 and the torque about the Sun vanishes. Zero torque, not zero force, is what conservation of angular momentum requires — and this is precisely Kepler's second law, which holds for elliptical orbits and does not need a circle.For a body rolling without slipping on a horizontal surface, the work done by friction is:
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Answer: B — Zero, because the contact point has zero instantaneous velocity
In pure rolling the contact point is instantaneously at rest, so there is no sliding for friction to dissipate and the work is zero — which is why energy conservation may be used freely on a rolling body even though friction is present and indeed necessary. Once slipping begins this fails, because the contact genuinely slides.The perpendicular-axes theorem I_z = I_x + I_y is applicable to:
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Answer: B — Planar (laminar) bodies only
The theorem holds only for a plane lamina, with z perpendicular to the plane and x, y lying in it. Applied to a disc, where symmetry gives I_x = I_y, it yields MR²/2 = 2I_x and hence MR²/4 about a diameter — a value that is much quicker to obtain this way than by integration.A marble and a cannonball, both solid spheres, roll down the same incline from the same height. At the bottom:
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Answer: C — Both have the same speed
v = √(2gh/(1+k)) contains neither mass nor radius, and both bodies are solid spheres with k = 2/5, so both arrive at √(10gh/7). The angle changes how long they take but not the final speed, which depends only on the height dropped. Only the mass distribution, through k, can change the answer.The rotational analogue of Newton's second law F = ma is:
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Answer: B — τ = Iα
Mass maps to moment of inertia and linear acceleration to angular acceleration, giving τ = Iα, or more generally τ = dL/dt. The expression L = Iω is the analogue of momentum p = mv, not of the second law, and τ = Iω is dimensionally wrong — a useful check.