Rotational vs. Translational Equilibrium: Independent Conditions
A system can exhibit rotational equilibrium — a constant angular velocity — without being in translational equilibrium at all, and vice versa. These are two genuinely separate conditions, and satisfying one tells you nothing about the other.
🔑Force diagrams — the rotational cousin of free-body diagrams introduced in Lesson 5.3 — describe both the forces and the torques acting on a rigid system, which is exactly what's needed to evaluate these two conditions independently.
Explore four combinations of the two conditions below.
Both equilibrium
Translational: ✓ Rotational: ✓
A book resting motionless on a table.
Rotational only
Translational: ✗ Rotational: ✓
A spinning frisbee in flight — constant spin rate, but its center of mass accelerates downward under gravity.
Translational only
Translational: ✓ Rotational: ✗
A figure skater gliding at constant velocity across the ice while spinning up into a faster pirouette.
Neither
Translational: ✗ Rotational: ✗
A wrench tumbling and accelerating as it's dropped, both spinning up and falling faster over time.
5.5.A.1.ii5.5.A.1.iiiMath
Rotational Equilibrium: Στ = 0
Rotational equilibrium is a configuration of torques such that the net torque exerted on a system is exactly zero:
Σ τᵢ = 0
This directly gives the rotational analog of Newton's first law: a system's angular velocity remains constant if — and only if — the net torque exerted on it is zero.
🔑A system in rotational equilibrium isn't necessarily motionless — it could be spinning steadily at a fast, constant rate. "Equilibrium" here means no change in angular velocity, exactly as translational equilibrium never required an object to be at rest, only to move at constant velocity.
Explore the classic balance-beam application below — the exact lever system from this unit's header art.
The lever-and-fulcrum system from this unit's header art. Adjust each force and its distance from the pivot — balance requires F₁r₁ = F₂r₂, exactly Στ = 0.
F₁40 N
r₁1.5 m
F₂30 N
r₂2.0 m
✓ Balanced — Στ = 0
τ₁ = F₁r₁ = 60.0 N·mτ₂ = F₂r₂ = 60.0 N·m
ExampleGuided Example — Balancing a Seesaw
A 300 N child sits 1.2 m from the pivot of a seesaw. Where must a 400 N adult sit on the other side for the seesaw to remain in rotational equilibrium?
Step 1 — Set up the rotational equilibrium condition
Στ = 0 → τ(child) = τ(adult), since they act in opposite rotational directions.
5.5.A.2Concept
The Rotational Corollary to Newton's Second Law
Run the equilibrium condition in reverse, and a natural corollary appears: if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.
💡This is exactly the rotational echo of Lesson 2.4's own corollary for linear motion — and it directly sets up Lesson 5.6, which turns this qualitative statement into a full quantitative equation: unbalanced torque produces angular acceleration in direct proportion, τ(net) = Iα.
← Back to Lesson 5.5Ready for 5.6? Newton's Second Law in Rotational Form makes this lesson's corollary fully quantitative: τ(net) = Iα.