Short on time? This page is the recap. For explanations, diagrams and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- Rotational equilibrium: the torques on a system add to zero, Στ = 0.
- Rotational first law: a rigid system keeps a constant angular velocity only if the net torque on it is zero. A steadily spinning fan and a still see-saw both qualify.
- Corollary: if the torques are unbalanced, the angular velocity must be changing.
- A force diagram shows each force at the point where it acts. You need this, not a single-dot free-body diagram, to find torques.
- Independence: a couple gives ΣF = 0 but Στ ≠ 0. A spinning ball in free fall has Στ = 0 about its centre of mass but ΣF ≠ 0.
- Only rotation in one plane is analysed in this course.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| τ = r⊥F = rF sin θ | N·m; θ between r and F | find each torque |
| Στ = 0 | sum with signs, e.g. ccw + | test or impose rotational equilibrium |
| ΣF_x = 0, ΣF_y = 0 | N | test or impose translational equilibrium |
| flat ω–t graph | rad/s against s | show Στ = 0 |
| sloping ω–t graph | slope ≠ 0 | show unbalanced torques |
Assumptions behind the numbers
- The system is rigid: it does not bend or change shape.
- The weight of a uniform object acts at its centre.
- When an object is in both kinds of equilibrium, the net torque is zero about any point, so you may choose the pivot.
- g = 9.8 m/s².
Mistakes to avoid
- No sign convention. State “ccw positive” before writing Στ = 0.
- Full force instead of its perpendicular part. Use T sin θ when the force is at an angle θ to the rod.
- Pivot at a known force. Put it where an unknown acts so that force drops out.
- Equilibrium means at rest. It means constant ω.
- ΣF = 0, so no rotation. A couple can still start it turning.
- Forgetting the object’s own weight. A uniform beam’s weight acts at its middle.
Quick self-check
- A 2.0 kg mass hangs 0.60 m left of a see-saw’s pivot. What mass, hung 0.40 m right of the pivot, balances it (light see-saw)? (3.0 kg)
- Two 4.0 N forces act perpendicular to a 0.50 m bar, one at each end, in opposite directions. Find ΣF and the net torque about the centre. (0 N; 2.0 N·m)
- A uniform 6.0 kg plank rests on supports at its two ends. What force does each support exert? (29.4 N ≈ 29 N each)
- A wheel spins at a steady 8 rad/s. Is the net torque zero? (Yes: constant ω.)
Next: practice questions.