IB Physics HLRigid Body MechanicsHL onlyPaper 1 & 2~11 min read
Newton’s Second Law for Rotation
Newton’s second law, F = ma, tells you how a force accelerates a mass in a straight line. Rotation has an exact mirror of this law — and now that you’ve met torque and moment of inertia, you have all the pieces. Torque takes the place of force, moment of inertia takes the place of mass, and angular acceleration takes the place of linear acceleration. The result, τ = Iα, is the single most useful equation for any problem where a torque spins something up.
📘 What you need to know
Newton’s second law for rotation: τ = Iα
It’s the rotational mirror of F = ma
Torqueτ replaces force, moment of inertiaI replaces mass, angular accelerationα replaces linear acceleration
τ is in N m, I is in kg m2, α is in rad s−2
A resultant torque produces an angular acceleration in the same direction
For a point mass, τ = Iα follows directly from F = ma, τ = Fr and a = rα
The rotational form of Newton’s second law
In linear motion, the force needed to give an object a certain acceleration depends on its mass:
Newton’s second law (linear)F = ma
In rotational motion, the torque needed to give a body a certain angular acceleration depends on its moment of inertia:
Newton’s second law (rotational)τ = Iα
where τ is the resultant torque in newton metres (N m), I is the moment of inertia in kg m2, and α is the angular acceleration in rad s−2. The bigger the moment of inertia, the more torque you need for the same angular acceleration — exactly as more mass needs more force.
A tangential force F on a point mass produces a torque Fr. Combining this with F = ma and a = rα gives τ = Iα.
Where the equation comes from
For a single point mass, the rotational law drops straight out of the linear one. Start with torque as force times distance, substitute F = ma, then swap the linear acceleration for the angular one using a = rα:
Deriving τ = Iα for a point massτ = Fr = (ma)r = m(rα)rτ = (mr2)α = Iα
The mr2 that appears is exactly the moment of inertia of a point mass — so the equation assembles itself. The same relationship holds for any rigid body, using its own moment of inertia.
τ = Fr
sub F = ma →
(ma)r
sub a = rα →
(mr²)α = Iα
Comparing linear and rotational variables
Every quantity in the linear law has a rotational partner. Once you see the pattern, rotational dynamics is just linear dynamics wearing different symbols:
Linear
Rotational
Force, F
Torque, τ
Mass, m
Moment of inertia, I
Acceleration, a
Angular acceleration, α
Newton’s 2nd law, F = ma
Newton’s 2nd law, τ = Iα
WE 1
A resultant torque of 12 N m acts on a wheel of moment of inertia 3.0 kg m2. Calculate its angular acceleration.
Step 1 — rearrange τ = Iα for α
α = τ ÷ I
Step 2 — substituteα = 12 ÷ 3.0α = 4.0 rad s⁻²Just like a = F ÷ m — the same structure, rotational symbols.
WE 2
A wheel of radius 0.20 m has a moment of inertia of 0.60 kg m2. A single tangential force is applied at its rim to give it an angular acceleration of 5.0 rad s−2. Calculate the force required.
Step 1 — find the torque needed with τ = Iατ = 0.60 × 5.0 = 3.0 N mStep 2 — the torque comes from a rim force: τ = Fr, so F = τ ÷ rF = 3.0 ÷ 0.20F = 15 NTwo steps: find the torque from τ = Iα, then the force from τ = Fr.
Linked linear and rotational motion
Many exam questions combine both laws — a hanging block (linear) turning a pulley (rotational). The trick is to write F = ma for the block and τ = Iα for the pulley, then link them with a = Rα (the string doesn’t slip) and solve the pair together.
A block on a pulley links the two laws. The tension turns the pulley (TR = Iα) and slows the block’s fall (mg − T = ma); solve the pair with a = Rα.
WE 3
A block of mass m hangs from a string wrapped around a cylindrical pulley of mass M and radius R (moment of inertia ½MR2). When released, the block falls and the pulley turns. Show that the block’s acceleration is a = mg ÷ (m + M/2).
Step 1 — linear law for the block
mg − T = ma (eq. 1)
Step 2 — rotational law for the pulley (τ = TR, α = a/R)
TR = Iα = (½MR²)(a/R)
T = ½Ma (eq. 2)Step 3 — substitute eq. 2 into eq. 1 and solve for amg − ½Ma = mamg = a(m + M/2)a = mg ÷ (m + M/2)The pulley’s mass slows the fall — a heavier pulley (bigger M) gives a smaller acceleration.
🛠️ Solving a τ = Iα problem
Find the resultant torque on the rotating body (sum clockwise minus anticlockwise).
Identify the moment of inertiaI (given, or from I = mr2 / Σmr2).
Apply τ = Iα, rearranging for whatever you need.
If linear motion is linked, write F = ma too and connect with a = Rα.
Solve the equations together and check the units.
The combined block-and-pulley problem is a classic. The one insight that unlocks it: because the string doesn’t slip over the pulley, the block’s linear acceleration and the pulley’s rim acceleration are the same, so a = Rα. That single link lets you eliminate the tension between the two equations and solve for the acceleration. Write both laws, connect them, solve.
💡 Top tips
Same law, new symbols.τ = Iα works exactly like F = ma — rearrange the same way.
Use the resultant torque. Add clockwise and subtract anticlockwise before applying the law.
Link with a = Rα. For a string over a pulley, the block and rim share the same acceleration.
Watch the tension. The tension on a real (massive) pulley differs above and below — it’s what produces the torque.
Quick recap: Newton’s second law for rotation is τ = Iα — the rotational mirror of F = ma, with torque for force, moment of inertia for mass, and angular acceleration for linear acceleration. For a point mass it follows from τ = Fr, F = ma and a = rα. Linked block-and-pulley problems combine both laws through a = Rα.
⚠ Common mistakes
Using mass instead of moment of inertia in the rotational law
Forgetting to take the resultant torque when several torques act
Not linking linear and rotational motion with a = Rα in pulley problems
Assuming the tension equals the weight — it’s less, because the block accelerates
Mixing up which quantity is rotational and which is linear in a combined problem
You now have the rotational versions of force and mass, and the law linking them. Next we complete the rotational toolkit with the spinning equivalent of momentum — angular momentum — and its own powerful conservation law that explains everything from spinning skaters to collapsing stars.
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