Moment of Inertia: Point Masses and Extended Objects Revision Notes | A Level AQA Physics

1. Why Rotation Resists Change — The Core Idea

You already know that a large mass resists changes in linear motion — that is Newton's first law applied to translation. Rotation has an exact parallel: some objects are much harder to spin up or slow down than others, even when they have the same total mass. The quantity that captures this rotational stubbornness is called the moment of inertia.

Picture a figure skater pulling their arms in during a spin. Their mass has not changed, yet they suddenly rotate faster. The reason is that their moment of inertia has decreased — mass has moved closer to the axis. This single idea underpins everything in this lesson.

In linear dynamics, inertia is mass $m$. In rotational dynamics, inertia depends on both mass and how far that mass sits from the axis of rotation. That distance dependence is what makes moment of inertia so interesting — and so important for engineering applications such as flywheels, turbines, and drive shafts.

2. The Point Mass Formula: $I = mr^2$

Start with the simplest possible rotating object: a single small mass $m$ fixed to a light rod of length $r$, rotating about one end. Every part of the mass is at the same perpendicular distance $r$ from the axis, so the moment of inertia is defined as:

Notice the $r^2$ dependence. Doubling the distance from the axis quadruples the moment of inertia, even though the mass is unchanged. This is the key insight that explains why mass distribution matters so much more than total mass alone.

A quick sanity check on units: $[m] = \text{kg}$ and $[r^2] = \text{m}^2$, so $[I] = \text{kg m}^2$. Always carry units through your working — the examiner will expect them in the final answer.

3. Extending to Multiple Masses: $I = \Sigma mr^2$

Real rotating objects contain many mass elements, each at its own distance from the axis. The total moment of inertia is found by adding up the $mr^2$ contribution from every individual mass. This is written using the summation notation:

The summation approach is powerful because it works for any arrangement of masses. You do not need to know the shape of the object — you only need the mass and distance of each element. For a continuous body (like a solid disc), the sum becomes an integral, but AQA will always supply the resulting expression, so you never need to perform the integration yourself.

Worked example: Calculating $I$ for a System of Three Point Masses

Three masses are attached to a light horizontal rod that rotates about a vertical axis through one end. Mass $m_1 = 0.20$ kg is at $r_1 = 0.10$ m, mass $m_2 = 0.30$ kg is at $r_2 = 0.20$ m, and mass $m_3 = 0.50$ kg is at $r_3 = 0.40$ m. Calculate the total moment of inertia of the system.

1. $I = \Sigma mr^2 = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2$ — [M1] Correct formula stated — this earns the method mark.
2. $m_1 r_1^2 = 0.20 \times (0.10)^2 = 0.20 \times 0.10 \times 0.10 = 0.20 \times 0.010 = 0.0020 \text{ kg m}^2$ $m_2 r_2^2 = 0.30 \times (0.20)^2 = 0.30 \times 0.20 \times 0.20 = 0.30 \times 0.040 = 0.012 \text{ kg m}^2$ $m_3 r_3^2 = 0.50 \times (0.40)^2 = 0.50 \times 0.40 \times 0.40 = 0.50 \times 0.16 = 0.080 \text{ kg m}^2$ — [M1] Each term calculated with explicit squaring shown. Notice how $m_3$, despite being only 2.5 times heavier than $m_1$, contributes 40 times more to $I$ because it is 4 times further from the axis.
3. $I = 0.0020 + 0.012 + 0.080 = 0.094 \text{ kg m}^2$ — [A1] Final answer with correct units. The outermost mass dominates — a pattern you will see repeatedly.
4. $[I] = \text{kg} \times \text{m}^2 = \text{kg m}^2$ ✓ — Units are consistent. The answer is of the order $10^{-2}$ kg m², which is physically reasonable for small masses on a short rod.

Answer: $I = 0.094 \text{ kg m}^2$

4. Extended Objects and Given Expressions

When mass is spread continuously through a body — a solid cylinder, a thin rod, a hollow sphere — you cannot list individual point masses. Instead, the $\Sigma mr^2$ sum becomes an integral over the entire volume. AQA will always provide the resulting formula in the question, so your job is to identify the correct expression, substitute correctly, and interpret the result.

The table below shows the standard expressions you are most likely to encounter. Each one was derived by integrating $r^2 \, dm$ over the appropriate geometry, but you need to know how to use them, not derive them.

Look at the rod example: rotating about the end gives $I = \frac{1}{3}ML^2$, which is exactly four times the through-centre value of $\frac{1}{12}ML^2$. This makes sense — when you shift the axis to the end, the far half of the rod is now much further away, dramatically increasing the $r^2$ contributions from those mass elements.

5. Factors Affecting Moment of Inertia — Qualitative Analysis

The AQA specification requires qualitative knowledge of what affects $I$. Three factors are central, and you need to be able to explain each one using the $r^2$ dependence as your reasoning chain.

• Total mass $M$: for a fixed shape and axis, $I$ scales linearly with $M$. Double the mass, double the moment of inertia.
• Distribution of mass relative to the axis: mass concentrated far from the axis contributes far more than the same mass placed close to the axis, because of the $r^2$ factor. This is the dominant effect in most engineering contexts.
• Choice of axis: the same object can have very different moments of inertia depending on which axis you rotate it about. Moving the axis further from the centre of mass always increases $I$ — this result follows from the parallel axis theorem: $I = I_{\text{cm}} + Md^2$, where $I_{\text{cm}}$ is the moment of inertia about the centre-of-mass axis and $d$ is the perpendicular distance between the two axes.

The figure skater analogy captures all three factors at once. When the skater extends their arms, mass moves to larger $r$, so $I$ increases and — because angular momentum is conserved — angular velocity decreases. When they pull their arms in, $r$ decreases, $I$ falls, and they spin faster. The total mass has not changed at all.

In engineering, this principle is exploited deliberately. A flywheel stores rotational kinetic energy ($E_k = \frac{1}{2}I\omega^2$), and designers maximise $I$ by placing most of the flywheel's mass at the rim — a hollow cylinder geometry — rather than distributing it uniformly. The same mass, arranged differently, stores far more energy at a given angular speed.

6. Applying Given Expressions — Worked Calculations

Now that the qualitative picture is clear, the exam will ask you to use a supplied expression to calculate $I$ and then connect it to other rotational quantities. Here is a representative exam-style problem that brings together the formula, substitution, and unit checking in one chain.

Worked example: Using a Given Expression for a Solid Cylinder

A solid cylinder of mass $M = 4.0$ kg and radius $R = 0.25$ m rotates about its central axis. The moment of inertia of a solid cylinder about its central axis is given as $I = \frac{1}{2}MR^2$. Calculate $I$. [2 marks]

1. $I = \frac{1}{2}MR^2$ — [M1] Quoting the supplied formula earns the method mark. Never skip this step — it shows the examiner your reasoning.
2. $I = \frac{1}{2} \times 4.0 \text{ kg} \times (0.25 \text{ m})^2$ — Carry units at every stage. Expand the square explicitly before multiplying.
3. $(0.25)^2 = 0.25 \times 0.25 = 0.0625 \text{ m}^2$ — Always write out the multiplication longhand. A common error is writing $(0.25)^2 = 0.50$, which is wrong — squaring is not the same as doubling.
4. $I = \frac{1}{2} \times 4.0 \times 0.0625 = 0.5 \times 0.25 = 0.125 \text{ kg m}^2$ — [A1] Final answer with correct units. Check: $\frac{1}{2} \times 4.0 = 2.0$, then $2.0 \times 0.0625 = 0.125$. Consistent.

Answer: $I = 0.125 \text{ kg m}^2$

Now consider a more demanding question that asks you to compare two objects and justify a prediction — the kind of higher-order task that appears in the later parts of AQA questions.

Worked example: Comparing Moments of Inertia for Two Rods

A uniform rod of mass $M = 1.2$ kg and length $L = 0.60$ m can rotate about either (a) an axis through its centre or (b) an axis through one end, both perpendicular to the rod. The expressions are $I_{\text{centre}} = \frac{1}{12}ML^2$ and $I_{\text{end}} = \frac{1}{3}ML^2$. Calculate both values and explain the difference. [4 marks]

1. $L^2 = (0.60)^2 = 0.60 \times 0.60 = 0.36 \text{ m}^2$ — Computing $L^2$ once avoids repeating the squaring step and reduces arithmetic errors.
2. $I_{\text{centre}} = \frac{1}{12} \times 1.2 \times 0.36 = \frac{1.2 \times 0.36}{12} = \frac{0.432}{12} = 0.036 \text{ kg m}^2$ — [M1] Correct substitution into the through-centre expression.
3. $I_{\text{end}} = \frac{1}{3} \times 1.2 \times 0.36 = \frac{0.432}{3} = 0.144 \text{ kg m}^2$ — [M1] Correct substitution into the through-end expression. Note: $0.144 = 4 \times 0.036$, confirming the factor-of-four relationship between the two expressions.
4. When the axis is at the end, the far half of the rod is now at distances up to $L$ from the axis rather than up to $L/2$. Because $I \propto r^2$, these distant mass elements contribute disproportionately more, increasing $I$ by a factor of four. — [B1] + [B1] Two marking points: (1) mass is further from the axis when rotating about the end; (2) the $r^2$ dependence means this has a large effect on $I$.

Answer: $I_{\text{centre}} = 0.036 \text{ kg m}^2$; $I_{\text{end}} = 0.144 \text{ kg m}^2$. Rotating about the end gives a moment of inertia four times larger because mass elements in the far half of the rod are now at much greater perpendicular distances from the axis.

7. Exam Technique and Common Errors

Moment of inertia questions reward methodical working. The marks are structured so that a correct method with an arithmetic slip still earns most of the credit — but only if your working is clearly laid out. Here is how to approach these questions to maximise your mark.

When a question supplies an expression for $I$, treat it as a gift. Quote it explicitly, substitute with units, expand any squared or cubed terms longhand, and state the final answer with units. This four-step chain is worth every mark available on a two- or three-mark calculation question.

For qualitative 'explain' questions, the examiner is looking for two linked ideas: (1) a statement about where the mass is relative to the axis, and (2) a connection to the $r^2$ factor in $I = \Sigma mr^2$. A response that says only 'the mass is further away' without invoking $r^2$ will typically earn one mark out of two. Naming the formula and explaining the squared dependence is what separates a good answer from a full-mark answer.

Mastering moment of inertia — both the point mass formula $I = mr^2$, the summation $I = \Sigma mr^2$, and the qualitative factors governing it — gives you the foundation for every subsequent topic in rotational dynamics, from angular momentum to rotational kinetic energy.