Rotational Kinematics: Study Guide (Physics C: Mechanics 5.1)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 5.1 with this title. This guide is the calculus-based one: it defines ω and α as derivatives, integrates angular accelerations that change with time, and derives the constant-α equations rather than just using them. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Rigid systems and the choice of model

Up to now you have treated most moving things as objects: points with no size. That works when every part of the thing moves the same way. A spinning wheel breaks this model. The rim on one side moves up while the rim on the other side moves down. Different points move in different directions at the same instant, so no single velocity describes the wheel.

A rigid system keeps its shape, but its points can move in different directions as it rotates. You cannot model it as a single object. Instead, you describe how the whole system turns about an axis of rotation.

The choice of model depends on the question. A thrown frisbee spins, but if you only want to know where it lands and its spin does not affect the path, you can track its centre of mass and treat it as an object. If the question is about how fast its edge moves, you need the rigid-system description. Ask: does the rotation matter for what I am asked? If not, the object model is fine.

In this course every rotation is about one fixed axis. That makes rotation a one-dimensional problem, with the angle θ playing the part that x played in Unit 1.

Angular position and angular displacement

Pick a fixed reference line through the axis. The angular position θ of a point is the angle between that line and the line from the axis to the point. The angular displacement over an interval is

Δθ = θ − θ₀

measured in radians. One full turn is 2π rad, so 1 rev = 2π rad = 360°, and 1 rad ≈ 57.3°.

You must also choose a positive sense. Usually counterclockwise is positive and clockwise is negative, but either choice is allowed. What matters is that you say which, and say from where you are looking (for example “counterclockwise as seen from above is positive”). A wheel that turns 3.0 rad clockwise then has Δθ = −3.0 rad.

Angular velocity and angular acceleration as derivatives

The definitions copy Unit 1 exactly, with θ in place of x.

Average angular velocity: ω_avg = Δθ / Δt

Instantaneous angular velocity: ω = lim (Δt → 0) Δθ/Δt = dθ/dt

Average angular acceleration: α_avg = Δω / Δt

Instantaneous angular acceleration: α = dω/dt = d²θ/dt²

Units: ω in rad/s, α in rad/s². The radian is a ratio of two lengths, so it often “disappears” from units; keep it in your working anyway so you know an angle is involved. Speeds in revolutions per minute (rpm) must be converted before you use calculus: multiply by 2π rad/rev and divide by 60 s/min. For example, 60 rpm = 2π rad/s ≈ 6.28 rad/s.

Every point of a rigid system turns through the same angle in the same time. So all points share the same θ, ω and α (Figure 1). Points further out move further, but that is a linear quantity and belongs to Topic 5.2.

Spinning faster or slower. Compare signs, just as in Unit 1: ω and α with the same sign → spinning faster; opposite signs → spinning slower. A negative α on its own does not mean slowing down. It means the angular acceleration is clockwise (if counterclockwise is positive).

The analogy with linear motion

Linear (one dimension)	Rotational (one fixed axis)
position x (m)	angular position θ (rad)
v_x = dx/dt (m/s)	ω = dθ/dt (rad/s)
a_x = dv_x/dt (m/s²)	α = dω/dt (rad/s²)
v_x(t) = v_x0 + ∫₀ᵗ a_x dt	ω(t) = ω₀ + ∫₀ᵗ α dt
x(t) = x₀ + ∫₀ᵗ v_x dt	θ(t) = θ₀ + ∫₀ᵗ ω dt

Because the mathematics is the same, every tool from Topic 1.2 carries over: tangent slopes, signed areas, sign charts and integration with initial conditions.

Graphs. On a θ–t graph, the tangent slope is ω. On an ω–t graph, the tangent slope is α and the signed area under the curve is Δθ. On an α–t graph, the signed area is Δω. Area below the time axis counts as negative (clockwise, if counterclockwise is positive). The total angle turned adds the sizes of all areas, just as distance travelled did in Unit 1.

Deriving the constant-α equations

If α is constant, integrate from t = 0:

1. ω(t) = ω₀ + ∫₀ᵗ α dt = ω₀ + αt
2. θ(t) = θ₀ + ∫₀ᵗ (ω₀ + αt) dt = θ₀ + ω₀t + ½αt²
3. Chain rule: α = dω/dt = (dω/dθ)(dθ/dt) = ω(dω/dθ), so α dθ = ω dω. Integrate from (θ₀, ω₀) to (θ, ω): α(θ − θ₀) = ½(ω² − ω₀²), so ω² = ω₀² + 2α(θ − θ₀).

Each step takes α outside an integral. That is allowed only because α is constant. If α depends on time, go back to the integrals.

Predicting changes. The constant-α equations show how quantities depend on each other. Starting from rest:

• At fixed α, the time to reach ω is t = ω/α, so doubling the target ω doubles the time.
• The angle turned is Δθ = ω²/(2α), so doubling the target ω quadruples the angle, while doubling α (same target ω) halves it.

Worked example 3 uses these.

Worked example 1: a camera mount that pans and returns

Question. Take counterclockwise (seen from above) as positive, with θ = 0 at the start. A motorised camera mount turns about a vertical axis with θ(t) = (0.60 rad/s²)t² − (0.10 rad/s³)t³ for 0 ≤ t ≤ 6.0 s. Find ω(t) and α(t), when the mount is momentarily at rest, when it spins faster or slower, its greatest counterclockwise angular velocity, and its angular displacement and total angle turned.

1. Differentiate: ω = dθ/dt = 1.2t − 0.30t² = 0.30t(4.0 − t) rad/s.
2. Differentiate again: α = dω/dt = 1.2 − 0.60t rad/s². It is zero at t = 2.0 s.
3. Momentarily at rest when ω = 0: t = 0 and t = 4.0 s. At 4.0 s, θ = 9.6 − 6.4 = 3.2 rad (about 183°), the furthest counterclockwise position.
4. Greatest counterclockwise ω: where α = 0, at t = 2.0 s: ω = 2.4 − 1.2 = 1.2 rad/s.
5. Sign chart:

Interval (s)	ω	α	Rotation
0 to 2.0	+	+	counterclockwise, spinning faster
2.0 to 4.0	+	−	counterclockwise, spinning slower
4.0 to 6.0	−	−	clockwise, spinning faster

6. At 6.0 s: θ = 21.6 − 21.6 = 0 and ω = 7.2 − 10.8 = −3.6 rad/s. Angular displacement over 0 to 6.0 s is zero: the mount is back where it started, now turning clockwise at 3.6 rad/s.
7. Total angle turned: 3.2 rad out plus 3.2 rad back = 6.4 rad.

Check. In Figure 2 the area under the ω–t curve from 0 to 4.0 s is +3.2 rad and from 4.0 to 6.0 s is −3.2 rad. Their sum is zero, matching step 6.

Worked example 2: an angular acceleration that dies away

Question. Take counterclockwise as positive. A test flywheel starts from rest at θ₀ = 0. Its drive gives α(t) = (8.0 rad/s²)e^(−t/2.5 s). Find ω(t), θ(t), the values at t = 5.0 s and the angular velocity the flywheel approaches.

1. ω = 0 + ∫₀ᵗ 8.0e^(−t/2.5) dt = 8.0 × 2.5 [1 − e^(−t/2.5)] = 20(1 − e^(−t/2.5)) rad/s.
2. θ = 0 + ∫₀ᵗ 20(1 − e^(−t/2.5)) dt = 20t − 50(1 − e^(−t/2.5)) rad.
3. At t = 5.0 s, e^(−2) ≈ 0.135: ω = 20 × 0.865 ≈ 17 rad/s and θ = 100 − 50 × 0.865 ≈ 57 rad (about 9.0 revolutions).
4. As t → ∞, e^(−t/2.5) → 0, so ω → 20 rad/s. The flywheel never quite reaches it; after 5.0 s it has about 86% of that value.

Check the units. 8.0 rad/s² × 2.5 s = 20 rad/s. The integral of a rad/s² quantity over seconds gives rad/s, as it must.

Why the constant-α equations fail here. Using the starting value α = 8.0 rad/s² for the whole 5.0 s gives ω = 40 rad/s and θ = 100 rad, both far too large. The angular acceleration shrinks as time goes on.

Worked example 3: constant α and scaling

Question. Take counterclockwise as positive. A lab spin-coater platter starts from rest and reaches 300 rad/s in 1.5 s with constant angular acceleration. (a) Find α and the angle turned. (b) A new recipe needs 600 rad/s with the same α. Predict the new time and angle without repeating the full calculation.

1. (a) α = Δω/Δt = 300 ÷ 1.5 = 200 rad/s².
2. Δθ = ½(ω₀ + ω)t = ½(0 + 300)(1.5) = 225 rad, about 36 revolutions.
3. Check with the third equation: ω² = 2αΔθ gives 300² = 90 000 and 2 × 200 × 225 = 90 000. ✓
4. (b) From rest at fixed α, t = ω/α ∝ ω: doubling ω doubles the time to 3.0 s.
5. Δθ = ω²/(2α) ∝ ω²: doubling ω quadruples the angle to 900 rad.

Interpretation. Twice the final speed costs twice the time but four times the turning, because the platter spends the extra time at higher angular velocities. On the ω–t graph, the triangle is twice as tall and twice as wide, so its area is four times larger.

Common misconceptions

• Using constant-α equations for α(t). Integrate instead (Worked example 2).
• Forgetting ω₀ or θ₀. ∫α dt gives only the change in ω.
• Working in rpm or degrees inside the calculus. Convert to rad/s and rad first.
• “Negative α means slowing down.” Compare the signs of ω and α. In Worked example 1, from 4.0 to 6.0 s both are negative and the mount spins faster.
• “ω = 0 means α = 0.” At t = 4.0 s in Worked example 1, ω = 0 but α = −1.2 rad/s².
• “Points further out have a larger ω.” Every point of a rigid system has the same ω and α. Points further out have larger linear speeds (Topic 5.2).
• Treating ∫ω dt as total angle turned. It is angular displacement. Split where ω = 0 and add sizes.
• Not stating the positive sense. “α = −2.0 rad/s²” means nothing until you say which way is positive.

Where this leads

Topic 5.2 connects these angular quantities to the linear motion of points on the rotating system, through s = rθ, v = rω and a_T = rα. Later in Unit 5, torque explains why α is what it is, and Newton’s second law in rotational form links the two. Try the practice questions now, then use the revision notes and the checklist. The next topic is Connecting Linear and Rotational Motion. You can also return to the course roadmap.