Newton’s Second Law for Rotation

Newton’s second law for rotation states that the net torque on a body about a fixed axis equals the product of its moment of inertia and its angular acceleration, Στ = Iα, which is the rotational analog of F = ma. In brief: a given torque produces a smaller angular acceleration in a body with a larger moment of inertia. Combined with v = ωr and a = αr, the law allows problems such as a falling mass turning a pulley, a rolling wheel or a door being pushed to be solved. The work done by a torque is W = τθ, and the power is P = τω.

Accelerating a flywheel, spinning up a washing machine drum or turning a heavy pulley with a falling bucket all follow Στ = Iα. This guide states the law and its derivation, shows how to solve problems with massive pulleys, work and power in rotation, compares the linear and rotational forms in a table, works through numerical examples and corrects common mistakes.

What does the law say?

For a particle of mass m at distance r from the axis, a tangential force F gives a tangential acceleration a = F ÷ m. With a = αr, F = mαr, and the torque τ = rF = mr2α. Adding the contributions of all the particles in a rigid body, Στ = (Σmr2)α, so

Στ = Iα.

The torques are those about the same axis, taking the sign convention for counterclockwise and clockwise. The equation holds for a rigid body rotating about a fixed axis, or about an axis through the center of mass (with the motion of the center of mass described separately by F = ma). If the net torque is zero, α = 0: the body rotates at constant angular velocity or is at rest (the rotational version of Newton’s first law).

How is it used with pulleys?

A rope over a massive pulley with a moment of inertia I and radius R: the tensions on the two sides are different (unlike in the ideal pulley with no mass), and the difference provides the net torque on the pulley: (T1 − T2)R = Iα, with α = a ÷ R if the rope does not slip. For each hanging mass, apply F = ma. For a single mass m hanging from a rope wound on a disc pulley of mass M (I = ½MR2): mg − T = ma and TR = ½MR2(a ÷ R), so T = ½Ma. Hence mg = (m + ½M)a, and a = mg ÷ (m + M/2). Also for a rotating disc, a rolling wheel or a door, the same method applies: find the torque about the axis, the moment of inertia, and calculate α.

Forces on a mass hanging from a rope wound on a disc pulley Disc: 4.0 kg 3.0 kg T = 11.8 N mg = 29.4 N a = 5.88 m/s²
The 3.0 kg mass of the worked example hanging from a rope wound on a 4.0 kg disc pulley. The arrows are drawn to the same scale. The tension (11.8 N) is smaller than the weight (29.4 N), so the net force of 17.6 N gives the mass an acceleration of 5.88 m/s². The tension also provides the torque that spins the pulley.

What are work and power in rotation?

A constant torque τ acting through an angle θ does work W = τθ (θ in radians), which is the analog of Fs. The power is P = τω, the analog of Fv. This is why a car engine produces a given power at a particular combination of torque and rotational speed: P = τω. The work-energy theorem for rotation: τθ = ΔEk = ½Iωf2 − ½Iωi2.

Linear and rotational forms compared

Analogs of Newton’s second law
Concept Linear Rotational
Cause of accelerationForce FTorque τ
InertiaMass mMoment of inertia I
Second lawΣF = maΣτ = Iα
WorkFsτθ
PowerFvτω
Kinetic energy½mv2½Iω2

Worked example: pulley, wheel and motor

  1. Constant torque on a pulley. A disc pulley with M = 2.0 kg and R = 0.20 m has I = ½MR2 = 0.040 kg·m2. A net torque of 0.50 N·m gives α = τ ÷ I = 0.50 ÷ 0.040 = 12.5 rad/s2. After 4.0 s from rest, ω = 50 rad/s.
  2. Falling mass on a disc. A mass m = 3.0 kg hangs from a rope wound around a disc pulley with M = 4.0 kg and R = 0.20 m. a = mg ÷ (m + M/2) = 29.4 ÷ (3.0 + 2.0) = 5.88 m/s2. The tension: T = ½Ma = 0.5 × 4.0 × 5.88 = 11.8 N (which is less than the weight of 29.4 N). The angular acceleration of the pulley = a ÷ R = 29.4 rad/s2. If the pulley had no mass, a would be g and T zero.
    3.0 11.8 N 29.4 N 5.88 m/s² 4.0
  3. Work done. A torque of 20 N·m turns a shaft through 5.0 revolutions (31.4 rad): W = τθ = 20 × 31.4 = 628 J.
  4. Engine power. An engine delivers a torque of 200 N·m at 3,000 rpm (314 rad/s): P = τω = 200 × 314 = 62.8 kW (84 horsepower).
  5. Braking a wheel. A wheel with I = 0.50 kg·m2 rotates at 40 rad/s. A brake applies a constant torque of 8.0 N·m. α = −8.0 ÷ 0.50 = −16 rad/s2, so it stops in t = 40 ÷ 16 = 2.5 s, after turning θ = ω02 ÷ 2|α| = 1,600 ÷ 32 = 50 rad. Check with energy: ½Iω2 = 0.5 × 0.50 × 1,600 = 400 J = τθ = 8.0 × 50 ✓.

How was the law formulated?

Leonhard Euler established the equations of rotational motion of rigid bodies in the 1750s to 1760s, extending Newton’s laws to extended bodies. The equation Στ = Iα is a special case of the law dL/dt = τ for the angular momentum. It underlies the design of engines, gears, turbines and robotics, where motors produce torque to move arms with given moments of inertia.

Which mistakes are common?

Mistake 1: Assuming that the tension is the same on both sides of a massive pulley

The difference in tensions provides the torque that accelerates the pulley.

Mistake 2: Forgetting to use the same axis for torque and moment of inertia

Both must refer to the same axis.

Mistake 3: Mixing linear and angular accelerations

Use a = αR for a rope that does not slip.

Mistake 4: Using degrees

Στ = Iα requires α in rad/s2.