A rigid system is in rotational equilibrium when its angular velocity remains constant over time. This includes two cases:
Static equilibrium
omega = 0
System is at rest and remains at rest. The most common AP case — seesaws, balanced beams, hinged doors.
Dynamic equilibrium
omega = constant ≠ 0
System spins at constant angular velocity. No angular acceleration. Less common but equally valid.
The condition for rotational equilibrium is that the net torque on the system must equal zero:
Στ = 0
🔑Στ = 0 is the rotational analog of ΣF = 0. Just as zero net force means constant linear velocity, zero net torque means constant angular velocity. These are independent conditions — a system can satisfy one without satisfying the other.
5.5.A.2Concept
Newton's First Law in Rotational Form
Newton's First Law states that an object maintains its state of motion unless acted upon by a net external force. In rotational form:
Linear N1L: ΣF = 0 → v = constant
Rotational N1L: Στ = 0 → ω = constant
A spinning top, a wheel rotating at steady speed, a balanced seesaw at rest — all are in rotational equilibrium because no net torque acts to change their angular velocity. The moment a net torque appears, angular velocity starts changing. That is 5.6.
💡The infographic makes a crucial point: if torques are not balanced (Στ ≠ 0), the system's angular velocity must be changing. This is the Newton's Second Law corollary — and it is what tau_net = I*alpha (Lesson 5.6) quantifies.
5.5.A.3Concept
Rotational vs. Translational Equilibrium
Rotational and translational equilibrium are independentconditions. A system can be in one without being in the other. On the AP exam, problems often involve systems that satisfy one condition but not both — and you must recognize which equations apply.
✗ NOT rotational equilibrium — Στ ≠ 0, omega is changing
✓ Translational equilibrium — pivot provides normal force, CM stays put
Ball in free fall (not spinning)
✓ Rotational equilibrium — omega = 0 throughout
✗ NOT translational equilibrium — gravity produces net downward force
5.5.A.4Math
The Pivot Choice Strategy
Since Στ = 0 holds for any pivot point when a system is in rotational equilibrium, you are free to choose whichever axis makes the algebra simplest. The key rule: a force acting at the chosen pivot produces zero torque (because r = 0 at that point).
🔑Strategy: Identify the unknown force you most want to eliminate. Place your pivot there. That unknown drops out of the torque equation entirely. You then have one equation with one unknown — solve directly.
Adjust forces and distances on a beam pivoted at the center. Find combinations where the net torque equals zero — the beam stops tilting.
Try F₁=30 N at d=2.0 m and F₂=40 N at d=1.5 m — can you find a third force that brings the net torque to zero? Note how the beam tilts away from equilibrium when torques are unbalanced.
ExampleGuided Example — Finding a Support Force
A 4 m uniform beam (mass 10 kg) is supported at its left end (pin) and at a point 3 m from the left. A 25 kg box sits at the right end. Find the upward force F at the 3 m support. (g = 10 m/s²)
Step 1 — Choose the pivot
Place the pivot at the pin (left end, x = 0). This eliminates the unknown pin reaction force — it acts at our pivot so its torque is zero.
5.5.A.5Math
Balance Problems
The classic AP equilibrium problem types all follow the same framework: draw the force diagram, choose a strategic pivot, write Στ = 0, solve. The scenarios vary — seesaws, balanced beams, hinged structures, signs hanging from rods — but the method is always the same.
⚠️Don't forget the beam's own weight. A uniform beam has its weight acting at its geometric center (the center of mass). Students frequently account for object weights but forget the beam itself contributes a torque about any pivot not at its center.
ExampleWorked Example — Seesaw Balance
Two children sit on a 5 m seesaw pivoted at its center. Child A (40 kg) sits 2.0 m left of the pivot. Where must Child B (25 kg) sit to the right to balance the seesaw? Assume the seesaw board is massless.