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.
What I depends on
- How much mass there is.
- How far that mass sits from the axis.
- Mass far from the axis counts much more: the distance is 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.
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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.
Hoop or disc?: 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.
How I scales: Every moment of inertia is (number) × mass × (size)². Halve the mass and halve the length and I falls to ½ × ¼ = ⅛.
A uniform rod cut in half: each half has half the mass and half the length, so about its own centre it has one eighth of the original I.
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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.
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Finding I from a graph: Plot angular acceleration α against torque τ: since α = τ ÷ I, it is a straight line through the origin with gradient 1/I. So I = 1 ÷ gradient.
A pulley that has mass
Blocks of mass 3.0 kg and 1.0 kg hang on a string over a pulley of radius 0.10 m and moment of inertia 0.010 kg m². The string does not slip. Find the acceleration of the blocks (g = 9.8 m s⁻²).
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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
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