Short on time? This page is the recap for the calculus-based course (separate from Physics 1). For the derivation, graphs and worked examples, use the full study guide.
Recap
- A rigid system’s ω changes only when the net external torque is not zero.
- α = Στ/I. α is proportional to the net torque, has the same sense, and is inversely proportional to I about that axis.
- Derivation: F_t = m rα on each piece; multiply by r; add: Στ = (Σmr²)α = Iα. Internal torques cancel in third-law pairs.
- Two equations, one object: a_cm = ΣF/M for the center of mass, α = Στ/I for the rotation. Pivot forces appear only in the first.
- State the positive sense of rotation before you start, and keep it.
Key relationships
| Relationship | Units | Notes |
|---|---|---|
| Στ = Iα | N·m = kg·m²·rad/s² | about a fixed axle or the center of mass |
| I = Σmr² = ∫r² dm; I = I_cm + Md² | kg·m² | depends on the axis (Topic 5.4) |
| α(t) = Στ(t)/I | rad/s² | holds at every instant |
| ω(t) = ω₀ + (1/I)∫₀ᵗ Στ dt | rad/s | area under Στ–t graph ÷ I gives Δω |
| θ(t) = θ₀ + ∫₀ᵗ ω dt | rad | |
| a_t = rα (no slipping string or rim) | m/s² | links a hanging mass to the wheel |
| T = m(g − a) for a mass accelerating downward | N | not mg |
Assumptions behind the numbers
- Rigid system, one plane of rotation, fixed axis unless stated.
- g = 9.8 m/s². Light strings that do not slip unless stated.
- Constant-α equations only when the net torque is constant.
Mistakes to avoid
- α in the sense of motion. α follows the net torque.
- Wrong axis for I. Use the parallel axis theorem for an off-center axle.
- T = mg for an accelerating hanging mass.
- Leaving out a friction torque in the net torque.
- Ignoring the axle force in ΣF. It has no torque about the axle, but it is a force.
- Greatest ω at greatest torque. ω peaks when Στ = 0.
- Constant-α formulas with Στ(t). Integrate instead.
Quick self-check
- A net torque of 0.60 N·m acts on a wheel with I = 0.15 kg·m². Find α. (0.60 ÷ 0.15 = 4.0 rad/s²)
- The net torque doubles and I halves. By what factor does α change? (2 ÷ ½ = 4)
- A uniform disk (2.0 kg, radius 0.10 m) lies on frictionless ice. A string wound round its rim is pulled with a horizontal 4.0 N force. Find α and a_cm. (I = 0.010 kg·m², α = 0.40 ÷ 0.010 = 40 rad/s²; a_cm = 4.0 ÷ 2.0 = 2.0 m/s². Two separate equations.)
Next: practice questions.