A.4 · IB Physics
Torque and rotational equilibrium
Drag up to three weights along a uniform 4.00 m beam, move the pivot, and see each torque, the net torque and how the beam starts to rotate when the torques do not cancel.
τ A: τ B: τ C: τ beam: Στ: I: α: θ:
τ = F r⊥ = m g r cos θ
Στ = I α (anticlockwise +)
I = ML²/12 + Md² + Σ m r²
Balance: Στ = 0
L = 4.00 m, g = 9.8 m s⁻²
Try this
- A and B start balanced. Switch on C at 2.80 m, predict which way the beam tips, then drag B until Στ = 0 again.
- Switch off all three weights and move the pivot to 1.20 m. The beam's own weight now has a torque: find the mass for A at 0.00 m that balances it (check: about 1.33 kg).
- Make the beam unbalanced and note α. Now double every mass, including the beam. Does α change? Explain using Στ = I α.
Exam tip: take moments about the pivot so the unknown reaction force there has zero torque, and give every torque a sign (clockwise or anticlockwise) before adding them.