Equations of Motion for Rotational Kinematics Revision Notes | A Level AQA Physics

1. From Linear to Rotational: The Core Analogy

You already know the SUVAT equations for linear motion. Here is the key insight for this entire topic: rotational kinematics under uniform angular acceleration follows exactly the same algebraic structure. Every symbol swaps for its rotational counterpart, and every equation retains its form.

Picture a flywheel spinning up from rest. Its rim traces an arc, but the physics governing how fast it spins and how far it turns is described by a set of four equations — the rotational equivalents of SUVAT. Mastering the analogy means you never need to memorise two separate frameworks; you only need to understand one and translate it.

Time is the one quantity that carries across unchanged — it plays the same role in both frameworks. Every other kinematic quantity has a direct rotational partner. This one-to-one mapping is what AQA means when it asks you to demonstrate awareness of the analogy between rotational and translational dynamics.

2. The Four Equations of Motion for Uniform Angular Acceleration

These equations apply only when angular acceleration α is constant — that is the condition of uniform angular acceleration. If α varies with time, these equations do not hold and calculus-based methods are required instead.

Notice how each equation eliminates one of the five quantities (ω₁, ω₂, α, θ, t). Your strategy in any problem is to identify the three quantities you know and the one you want — then select the equation that contains exactly those four, leaving out the one you neither know nor need.

A common source of confusion is the subscript notation. AQA uses ω₁ for initial angular velocity and ω₂ for final angular velocity — this mirrors the use of u and v in linear SUVAT. Always check which end of the time interval you are working with before substituting values.

3. Deriving the Equations — Where They Come From

Understanding the derivation helps you reconstruct any equation you forget under exam pressure. Start from the definition of uniform angular acceleration: α is constant, so angular velocity changes linearly with time.

If angular velocity increases at a constant rate α from an initial value ω₁, then after time t the angular velocity is $\omega_2 = \omega_1 + \alpha t$. This is Equation 1 — it follows directly from the definition of α as the rate of change of ω.

Because ω changes linearly, the average angular velocity over the interval is $\frac{\omega_1 + \omega_2}{2}$. Multiplying the average angular velocity by time gives the angular displacement: $\theta = \frac{(\omega_1 + \omega_2)}{2} \cdot t$. That is Equation 2.

Substituting Equation 1 into Equation 2 eliminates ω₂ and produces Equation 3: $\theta = \omega_1 t + \frac{1}{2}\alpha t^2$. Eliminating t between Equations 1 and 2 instead produces Equation 4: $\omega_2^2 = \omega_1^2 + 2\alpha\theta$. The four equations are therefore not independent facts to memorise in isolation — they are four rearrangements of the same underlying relationship.

4. Worked Examples — Applying the Equations

The selection strategy is always the same: list your knowns, identify your unknown, choose the equation that links them, then substitute carefully with consistent units. Work through the following example step by step before checking each stage.

Worked example: Finding Angular Displacement During Spin-Up

A turbine rotor starts from rest and accelerates uniformly at $\alpha = 4.0 \text{ rad s}^{-2}$ for $t = 5.0 \text{ s}$. Calculate the angular displacement θ during this time. [3 marks]

1. $\omega_1 = 0 \text{ rad s}^{-1}$ (starts from rest) $\alpha = 4.0 \text{ rad s}^{-2}$ $t = 5.0 \text{ s}$ Unknown: $\theta$ — Identifying knowns and unknowns first prevents substituting into the wrong equation. [M1] for correct identification of starting conditions.
2. $\theta = \omega_1 t + \dfrac{1}{2}\alpha t^2$ — This equation contains ω₁, α, t, and θ — exactly the four quantities involved. ω₂ is not needed and not given, so this is the correct choice. [M1] for selecting this equation.
3. $\theta = (0)(5.0) + \dfrac{1}{2}(4.0)(5.0)^2$ — ω₁ = 0 eliminates the first term. Now expand 5.0² carefully: 5.0 × 5.0 = 25.
4. $\theta = 0 + \dfrac{1}{2} \times 4.0 \times 25$ $\theta = 0.5 \times 4.0 \times 25 = 0.5 \times 100 = 50$ — 0.5 × 4.0 = 2.0; then 2.0 × 25 = 50. Each multiplication step is kept separate to avoid errors.
5. $\theta = 50 \text{ rad}$ — Angular displacement is measured in radians. Always attach the unit — omitting it loses the accuracy mark. [A1] for 50 rad.

Answer: $\theta = 50 \text{ rad}$

Worked example: Finding Final Angular Velocity Without Time

A grinding wheel is spinning at $\omega_1 = 12 \text{ rad s}^{-1}$ when the power is cut. It decelerates uniformly at $\alpha = -3.0 \text{ rad s}^{-2}$ and rotates through $\theta = 24 \text{ rad}$ before stopping. Calculate the final angular velocity ω₂. [3 marks]

1. $\omega_1 = 12 \text{ rad s}^{-1}$ $\alpha = -3.0 \text{ rad s}^{-2}$ (deceleration, so negative) $\theta = 24 \text{ rad}$ Unknown: $\omega_2$ — The negative sign on α is critical — it indicates the wheel is slowing down. Dropping the sign would give a physically impossible result (speeding up).
2. $\omega_2^2 = \omega_1^2 + 2\alpha\theta$ — Time is neither given nor required. This equation links ω₁, ω₂, α, and θ directly. [M1] for selecting this equation.
3. $\omega_2^2 = (12)^2 + 2(-3.0)(24)$ — Expand 12² carefully: 12 × 12 = 144. Then compute 2 × (−3.0) × 24 separately.
4. $12 \times 12 = 144$ $2 \times (-3.0) \times 24 = -6.0 \times 24 = -144$ $\omega_2^2 = 144 + (-144) = 0$ — The two terms cancel exactly, confirming the wheel comes to rest after 24 rad — consistent with the problem statement. [M1] for correct substitution and arithmetic.
5. $\omega_2 = \sqrt{0} = 0 \text{ rad s}^{-1}$ — The wheel has stopped. ω₂ = 0 rad s⁻¹ is physically sensible — the wheel decelerates to rest. [A1] for 0 rad s⁻¹ with correct unit.

Answer: $\omega_2 = 0 \text{ rad s}^{-1}$ (the wheel comes to rest)

5. The Translational–Rotational Analogy in Depth

The analogy between rotational and translational dynamics runs deeper than just swapping symbols. The reason the equations share the same algebraic form is that both describe motion where the relevant acceleration is constant. The physics of constant acceleration — whether linear or angular — always produces the same mathematical structure.

This table is worth studying carefully. For every translational equation, the rotational version is obtained by the substitutions: $s \to \theta$, $u \to \omega_1$, $v \to \omega_2$, $a \to \alpha$. Time t is unchanged. The column on the right shows which quantity each equation eliminates — this is your selection guide in the exam.

AQA exam questions on this analogy often ask you to compare the two frameworks explicitly. A strong answer names the corresponding pairs, states the condition of uniform acceleration, and notes that time is the shared quantity. Avoid vague statements like 'they are similar' — the mark scheme rewards precise mapping.

6. A More Demanding Problem — Multi-Step Reasoning

Exam questions at A-level frequently require two sequential applications of the rotational equations — for example, a spin-up phase followed by a deceleration phase. The key is to treat each phase separately, using the final state of one phase as the initial state of the next.

Worked example: Two-Phase Rotational Motion Problem

A flywheel accelerates uniformly from rest at $\alpha = 6.0 \text{ rad s}^{-2}$ for $t_1 = 4.0 \text{ s}$, then decelerates uniformly to rest over a further $t_2 = 12.0 \text{ s}$. Calculate: (a) the maximum angular velocity reached, and (b) the total angular displacement. [5 marks]

1. $\omega_2 = \omega_1 + \alpha t_1 = 0 + (6.0)(4.0)$ $\omega_2 = 24 \text{ rad s}^{-1}$ — The flywheel starts from rest so ω₁ = 0. Equation 1 is the direct route to ω₂. [M1] for correct equation; [A1] for 24 rad s⁻¹.
2. $\theta_1 = \left(\dfrac{\omega_1 + \omega_2}{2}\right)t_1 = \left(\dfrac{0 + 24}{2}\right)(4.0)$ $\theta_1 = 12 \times 4.0 = 48 \text{ rad}$ — Using Equation 2 with the average angular velocity. 0 + 24 = 24; 24 ÷ 2 = 12; 12 × 4.0 = 48 rad.
3. Phase 2: $\omega_1 = 24 \text{ rad s}^{-1}$, $\omega_2 = 0$, $t_2 = 12.0 \text{ s}$ $\theta_2 = \left(\dfrac{24 + 0}{2}\right)(12.0) = 12 \times 12.0 = 144 \text{ rad}$ — The initial angular velocity for Phase 2 is the final angular velocity from Phase 1. The flywheel decelerates to rest, so ω₂ = 0 for this phase. [M1] for correctly carrying forward ω = 24 rad s⁻¹.
4. $\theta_{\text{total}} = \theta_1 + \theta_2 = 48 + 144 = 192 \text{ rad}$ — Add the displacements from both phases. [A1] for 192 rad.
5. Phase 1 lasts 4.0 s; Phase 2 lasts 12.0 s (three times longer). The deceleration phase covers more angle, which is consistent with θ₂ = 144 rad > θ₁ = 48 rad. — A longer deceleration phase at the same average speed covers more angular displacement. The ratio 144:48 = 3:1 matches the time ratio 12:4 = 3:1, confirming the result is internally consistent.

Answer: (a) Maximum angular velocity = $24 \text{ rad s}^{-1}$; (b) Total angular displacement = $192 \text{ rad}$

7. Exam Technique and Common Errors

The most reliable way to gain full marks on rotational kinematics questions is to follow a consistent method: write down knowns and unknowns explicitly, state the equation before substituting, show every arithmetic step, and attach units to every quantity including intermediate results.

When an exam question asks you to 'compare rotational and translational dynamics', the mark scheme typically awards one mark per correctly identified pair of analogous quantities, plus a mark for stating the condition of uniform acceleration. A response that lists all five pairs (θ↔s, ω₁↔u, ω₂↔v, α↔a, t↔t) and states the uniform-acceleration condition will score full marks on that part.