Short on time? This page is the recap. For the derivation, diagrams and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- A non-zero net torque changes a rigid system’s angular velocity. α = τ_net / I.
- α is directly proportional to the net torque and inversely proportional to the rotational inertia.
- α has the same sense as the net torque (clockwise or counterclockwise), not necessarily the same sense as ω.
- The law comes from F = ma for a point mass: F = m(rα), so rF = (mr²)α.
- A rotating object obeys two separate laws: ΣF = m a_cm for its centre of mass and Στ = Iα for its spin.
- A rope that does not slip links them: a = Rα.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| α = τ_net / I, Στ = Iα | rad/s², N·m, kg·m² | find angular acceleration or net torque |
| I = Σmr² | kg·m² | rotational inertia of point masses (extended objects are given) |
| a_T = rα | m/s² | link a rope or rim to the rotation |
| ω = ω₀ + αt, ω² = ω₀² + 2αΔθ | rad/s, rad | follow up with rotational kinematics (constant α) |
| a = mg / (m + I/R²) | hanging mass on a drum | a typical derived result, not one to memorise |
Assumptions behind the numbers
- The system is rigid, so every part has the same α.
- The rope is light and does not slip, so a = Rα.
- Axles are frictionless unless a friction torque is stated.
- g = 9.8 m/s².
Mistakes to avoid
- T = mg for an accelerating hanging mass. T is smaller when the mass speeds up downward.
- Equal tensions on both sides of a massive pulley. They differ; their difference times R equals Iα.
- α in the direction of ω. It follows the net torque.
- Counting the axle force’s torque. It acts at the axis: zero lever arm.
- Doubling r doubles I. For point masses, I ∝ r², so it quadruples.
- Mixed sign conventions. Choose “down positive” for the hanging mass and the matching rotation sense for the drum.
Quick self-check
- A net torque of 6.0 N·m acts on a wheel with I = 2.0 kg·m². Find α. (3.0 rad/s²)
- A motor exerts 5.0 N·m on a rotor with I = 0.50 kg·m²; friction exerts 1.0 N·m the other way. Find α. (8.0 rad/s²)
- The net torque is tripled and I is doubled. By what factor does α change? (× 1.5)
- A wheel spins clockwise; the net torque is counterclockwise. Is it speeding up? (No: it slows down.)
Next: practice questions.