The big idea: Try spinning a broom held by its middle, then by one end — the far-out end fights you harder. That resistance to a change in spin is the moment of inertia (I), rotation's version of mass.
It depends not just on how much mass there is, but on how far that mass sits from the axis. Mass far from the axis counts much more (it's the distance squared).
- moment of inertia (kg m²)
- mass of each part (kg)
- distance of that part from the axis (m)
Two 3.0 kg masses sit on a light rod, each 0.40 m from the axis through the centre. Find the moment of inertia about that axis.
Model answer plan
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For a solid object the sum becomes a standard result for each shape. You are given the right formula in the exam question — you just pick the value of I and use it. Same mass and radius, but a hoop (all its mass at the rim) is harder to spin than a disc (mass spread inward).
| Shape (mass M, radius R) | Moment of inertia I |
|---|---|
| Thin hoop / ring, about its centre | M R² |
| Solid disc / cylinder, about its centre | ½ M R² |
| Solid sphere, about its centre | ⅖ M R² |
| Thin rod (length L), about its centre | 1/12 M L² |
Don't memorise these: The exam gives you the moment-of-inertia formula for the shape in the question. Your job is to recognise it and put the numbers in.
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A torque makes a body's spin change — it produces angular acceleration α. The bigger the moment of inertia, the smaller the α for the same torque. It's exactly F = ma, but rotational.
- resultant torque (N m)
- moment of inertia (kg m²)
- angular acceleration (rad s⁻²)
| Straight-line motion | Rotational motion |
|---|---|
| mass m | moment of inertia I |
| force F | torque τ |
| F = ma | τ = Iα |
| momentum p = mv | angular momentum L = Iω |
A flywheel has a moment of inertia of 1.2 kg m². A net torque of 6.0 N m acts on it. Find its angular acceleration.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
How this is tested — rotational dynamics is HL only (A.4):
Paper 1A
- A one-step τ = Iα, or comparing the I of two shapes.
Paper 2
- Combine τ = Iα with the rotational suvat (from 1.4.1): find α from the torque, then the final ω or the angle turned.
The classic trap: Using the shape's total mass as I, or forgetting to square r in I = Σmr². Use the moment of inertia the question gives you, and use the net torque.
Three easy marks: (1) Use the I the question gives you. (2) Use the net torque. (3) Once you have α, the rotational suvat equations finish the question.
A disc (moment of inertia 0.50 kg m²) starts from rest. A constant net torque of 2.0 N m acts on it. Determine its angular velocity after 4.0 s.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.