IB Physics HL Rigid Body Mechanics HL only Paper 1 & 2 ~10 min read

Angular Acceleration Formula

You already know the four SUVAT equations for objects speeding up in a straight line. Here’s the good news: rotation has its own identical set. A spinning-up flywheel, a turntable slowing to a stop, a wheel accelerating from rest — all of these are solved with the very same equations, just with the angular quantities swapped in for the linear ones. If you can do linear kinematics, you can already do rotational kinematics; you just need to learn which symbol replaces which.

📘 What you need to know

From linear to rotational kinematics

The four kinematic equations for uniform linear acceleration will be familiar:

Linear (SUVAT) equations v = u + at s = ut + ½at2 v2 = u2 + 2as s = ½(u + v)t

Swapping each linear variable for its rotational partner gives the four rotational kinematic equations:

Rotational (angular SUVAT) equations ωf = ωi + αt Δθ = ωit + ½αt2 ωf2 = ωi2 + 2αΔθ Δθ = ½(ωi + ωf)t
LINEAR ROTATIONAL displacement s ang. displacement θ initial velocity u initial ang. vel. ωᵢ final velocity v final ang. vel. ωᶠ acceleration a ang. acceleration α time t time tsame equations, swapped variables
Each linear variable maps to a rotational partner. Only time stays the same — swap the rest and the SUVAT equations become the angular kinematic equations.

The variable swap

The whole method comes down to this correspondence. Learn which symbol replaces which, and you never have to memorise a second set of equations:

VariableLinearRotational
Displacementsθ
Initial velocityuωi
Final velocityvωf
Accelerationaα
Timett
Pick knowns
3 of 5
→ choose eq.
Match variables
to rotational
→ solve
Answer
in rad
WE 1

A flywheel starts at an angular velocity of 2.0 rad s−1 and accelerates uniformly at 3.0 rad s−2 for 4.0 s. Find its final angular velocity.

Step 1 — knowns are ωi, α, t; want ωf, so use ωf = ωi + αt ωf = ωi + αt Step 2 — substitute ωf = 2.0 + (3.0 × 4.0) ωf = 14 rad s⁻¹ Straight swap of the linear v = u + at — nothing new to learn.
WE 2

A wheel starts from rest and accelerates uniformly at 1.5 rad s−2 for 6.0 s. Through what angle does it turn, and how many rotations is that?

Step 1 — knowns are ωi (= 0), α, t; use Δθ = ωit + ½αt² Δθ = ωit + ½αt² Step 2 — substitute (ωi = 0) Δθ = 0 + ½ × 1.5 × 6.0² = 27 rad Step 3 — convert to rotations (÷ 2π) 27 ÷ 2π = 4.3 rotations Δθ = 27 rad ≈ 4.3 rotations Divide the angle in radians by 2π to count whole turns.
WE 3

A drill accelerates from rest to 50 rad s−1 while turning through 20 rad. Assuming constant angular acceleration, find α.

Step 1 — knowns are ωi (= 0), ωf, Δθ; no time, so use ωf² = ωi² + 2αΔθ ωf² = ωi² + 2αΔθ Step 2 — rearrange for α (ωi = 0) α = ωf² ÷ (2Δθ) Step 3 — substitute α = 50² ÷ (2 × 20) = 2500 ÷ 40 α = 62.5 rad s⁻² This is the “no time” equation — just like v² = u² + 2as.
The trick to picking the right equation is identical to linear SUVAT: list your five quantities (θ, ωi, ωf, α, t), mark the three you know and the one you want, then choose the equation that contains those four and leaves out the one you neither know nor need. And always convert spinning rates to rad s−1 first — RPM and revolutions won’t work directly.

🛠️ Solving a rotational kinematics problem

  1. List the five quantities: Δθ, ωi, ωf, α, t.
  2. Convert units — angular velocity to rad s−1 (RPM ÷ 60 × 2π), angles to radians.
  3. Mark 3 knowns + 1 wanted, and note the one you can ignore.
  4. Pick the equation that contains your four quantities.
  5. Substitute and solve; convert the final angle to rotations with ÷ 2π if asked.

💡 Top tips

Quick recap: The four rotational kinematic equations are the linear SUVAT equations with sθ, uωi, vωf, aα. They apply only for constant angular acceleration. Convert rates to rad s−1 and angles to radians first, and divide an angle by 2π to count rotations.

⚠ Common mistakes

You can now describe how a rotating body’s motion changes over time. The missing piece is what causes that angular acceleration in the first place — and just as force is spread over mass in F = ma, torque is spread over a quantity called the moment of inertia. That’s the next building block.

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