AP Physics C: Mechanics · Unit 5: Torque and Rotational Dynamics · Lesson 5.6

Deep Dive: Newton's Second Law in Rotational Form

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
5.6.A.1Concept

When Does Angular Velocity Change?

A system's angular velocity changes precisely when the net torque exerted on it is not equal to zero. This is the direct fulfillment of Lesson 5.5's corollary — unbalanced torque doesn't just theoretically imply changing angular velocity, it fully determines exactly how much that velocity changes.

τ(net) ≠ 0Unbalanced torque → angular velocity is changing (α ≠ 0)τ₁τ₂ (equal, opposite)τ(net) = 0 → α = 0
5.6.A.2Math

Newton's Second Law in Rotational Form

The rate at which a rigid system's angular velocity changes is directly proportional to the net torque exerted on it, and points in that same direction. Angular acceleration is inversely proportional to the system's rotational inertia:

τ(net) = Iα
🔑This is exactly F=ma from Lesson 2.5, wearing rotational clothing: torque plays the role of force, rotational inertia plays the role of mass, and angular acceleration plays the role of linear acceleration. Every intuition you built around F=ma transfers directly here.

Explore the proportionality directly below.

A single object with adjustable net torque and rotational inertia. Watch angular acceleration respond — directly proportional to torque, inversely proportional to I.

τ(net) (N·m)12
I (kg·m²)3.0
ατ(net) →
α = τ/I = 12/3.0 = 4.00 rad/s²

Double the torque with I fixed, and α doubles. Double I with torque fixed, and α is cut in half — exactly the same proportionality pattern as F=ma back in Lesson 2.5.

ExampleGuided Example — Spinning Up a Disk

A solid disk with rotational inertia 0.4 kg·m² is initially at rest. A constant net torque of 2.4 N·m is applied. Find the disk's angular velocity after 5 seconds.

Step 1Apply Newton's second law in rotational form
τ(net) = Iα  →  α = τ/I = 2.4/0.4
Math

Applying τ = Iα: Pulleys with Mass and Friction

Lesson 2.5 always assumed an ideal, massless pulley — tension was simply the same on both sides of the string. Now that pulleys can have real rotational inertia, that assumption breaks down entirely. Some of the tension's torque goes into spinning the pulley itself, and any friction at the axle contributes its own opposing torque:

Iα = Tr − τ(f)
⚠️With a massive pulley, tension is generally different on either side of a system involving multiple masses — treating tension as uniform throughout, the way Unit 2 always did, is no longer valid once the pulley itself has mass.

Combine this with the tangential relationship from Lesson 5.2 (a = rα) and Newton's second law for the hanging mass, and the whole system can be solved together. Explore this directly below — the full pulley system from this unit's header art.

The pulley system from this unit's header art — but now with a real, massive pulley and friction at the axle, exactly matching Iα = Tr − τ(f). Adjust everything and watch the full system respond.

Pulley I0.08 kg·m²
Pulley r0.10 m
hanging m2.00 kg
friction τf0.50 N·m
m
System accelerates — friction can't hold it back
α = (mgr − τf) / (I + mr²) = 15.00 rad/s²a (hanging mass) = rα = 1.50 m/s²T (tension) = 17.0 N

Notice tension no longer just equals mg minus ma the way it did with Unit 2's massless pulleys — some of gravity's pull now goes into spinning up the pulley itself.

ExampleWorked Example — Full System Solve

A pulley with rotational inertia 0.05 kg·m² and radius 0.08 m has negligible friction. A 3 kg mass hangs from a string wrapped around it. Find the system's acceleration and the string's tension.

← Back to Lesson 5.6That's a wrap on Unit 5! Unit 6 covers Energy and Momentum of Rotating Systems — rotational kinetic energy, angular momentum, and rolling without slipping.