IB Physics HLRigid Body MechanicsHL onlyPaper 1 & 2~11 min read
Rotational Equilibrium
A see-saw that hangs perfectly level, a shelf bracket that holds firm, a plank a child can walk along without it tipping — these are all in rotational equilibrium. It’s the rotational twin of the balanced-forces idea you already know: instead of the forces cancelling, it’s the turning effects that cancel. Master this and you can solve a huge family of “balanced beam” problems using one simple principle — the clockwise turns must equal the anticlockwise turns.
📘 What you need to know
A body is in rotational equilibrium if the resultant torque acting on it is zero
Such a body stays at rest or rotates at constant angular velocity
This is the rotational version of Newton’s first law (translational equilibrium)
Principle of torques: total clockwise torque = total anticlockwise torque
A resultant torque causes angular acceleration — the body speeds up or slows its spin
The direction of the angular acceleration matches the direction of the net torque
You can take torques about any point — choose one that removes unknown forces
What rotational equilibrium means
If the resultant (net) torque on a body is zero, it is in rotational equilibrium. Just as an object with balanced forces won’t start moving or change speed, an object with balanced torques won’t start spinning or change its rate of spin — it stays at rest or rotates at a constant angular velocity. This is exactly analogous to Newton’s first law, just for rotation.
Principle of torques
Total clockwise torque = Total anticlockwise torque
This is often called the principle of torques (or principle of moments). Any point where the clockwise turning effects exactly balance the anticlockwise ones is in rotational equilibrium. A beam balanced on a pivot is the classic example of a rigid, extended body in this state.
A balanced beam. When the anticlockwise torque (F₁r₁ + F₃r₃) equals the clockwise torque (F₂r₂), the resultant torque is zero and the beam is in rotational equilibrium.
WE 1
A light beam is pivoted at its centre. A 20 N force acts downward 0.30 m to the left of the pivot. What downward force F acting 0.40 m to the right of the pivot keeps the beam in rotational equilibrium?
Step 1 — apply the principle of torques
clockwise torque = anticlockwise torque
Step 2 — write each torque as force × distance
F × 0.40 = 20 × 0.30
Step 3 — rearrange for FF = (20 × 0.30) ÷ 0.40 = 6.0 ÷ 0.40F = 15 NThe smaller distance needs the larger force to give the same turning effect.
Balancing a beam: worked comparison
To test whether a beam is balanced, take torques about a convenient point (often the centre) and compare the clockwise and anticlockwise totals. If they’re equal, the beam is in rotational equilibrium; if not, it will start to turn. Here’s how four beams compare:
Beam
Anticlockwise torque
Clockwise torque
Balanced?
A
10 × 50 = 500
27 × 30 = 810
No
B
50 × 50 = 2500
71 × 30 = 2130
No
C
15 × 50 = 750
25 × 30 = 750
Yes
D
43 × 50 = 2150
12 × 30 = 360
No
Only beam C has equal clockwise and anticlockwise torques (750 N cm each), so only C is in rotational equilibrium. Notice the units cancel consistently — as long as both sides use the same units (here N cm), you can compare directly.
A powerful trick: you can take torques about any point, not just the pivot. Choosing a point that a troublesome unknown force passes through makes that force’s torque zero (its distance is zero), so it drops out of your equation entirely. Pick your point to make the algebra easy — it’s a genuine time-saver in exams.
Unbalanced torque and angular acceleration
When the torques don’t balance, there’s a resultant torque — and just as a resultant force causes linear acceleration, a resultant torque causes angular acceleration. The body speeds up or slows its rotation, and the direction of that angular acceleration matches the direction of the net torque (clockwise or anticlockwise).
A downward force one side of the pivot turns the beam anticlockwise; the other side turns it clockwise. If the two torques are unequal, the net torque spins the beam that way with angular acceleration.
WE 2
A child of mass 25 kg sits 1.6 m from the pivot of a see-saw. At what distance from the pivot must an adult of mass 60 kg sit on the other side to balance it?
Step 1 — for balance, clockwise torque = anticlockwise torque
(60g) × d = (25g) × 1.6
Step 2 — the g cancels from both sides
60 × d = 25 × 1.6
Step 3 — rearrange for dd = (25 × 1.6) ÷ 60 = 40 ÷ 60d = 0.67 m (2 s.f.)The heavier adult sits closer to the pivot — less distance balances more weight.
WE 3
A pivoted beam has an 8 N force producing an anticlockwise torque 0.50 m from the pivot, plus a 5 N force 0.60 m and a 3 N force 0.50 m both producing clockwise torques. Find the resultant torque and its direction.
Step 1 — total the anticlockwise torqueACW = 8 × 0.50 = 4.0 N mStep 2 — total the clockwise torqueCW = (5 × 0.60) + (3 × 0.50) = 3.0 + 1.5 = 4.5 N mStep 3 — resultant = larger − smallerτ = 4.5 − 4.0 = 0.5 N m0.5 N m, clockwiseThe clockwise torque wins, so the beam has an angular acceleration in the clockwise direction.
🛠️ Solving a rotational equilibrium problem
Choose a point to take torques about — pick one that removes an unknown force if you can.
Identify each force and its perpendicular distance from that point.
Sort into clockwise and anticlockwise turning effects.
For equilibrium, set total clockwise torque = total anticlockwise torque and solve.
For a resultant, subtract the smaller total from the larger and state the direction.
💡 Top tips
Take torques cleverly. Choosing a point on the line of an unknown force makes its torque zero — one fewer unknown.
Mass cancels for balance. When both sides are weights, g appears on both sides and drops out.
Keep units consistent. Both sides in N cm or both in N m — then you can compare directly.
Tipping point. A plank tips when the support it’s about to lift off carries zero reaction force — set that force to zero.
Quick recap: A body is in rotational equilibrium when the resultant torque is zero, so it stays at rest or spins at constant angular velocity. The principle of torques says total clockwise torque = total anticlockwise torque. An unbalanced torque produces angular acceleration in its own direction, and you can take torques about any convenient point to simplify the maths.
⚠ Common mistakes
Forgetting that a body can be in rotational equilibrium while rotating at constant speed, not just at rest
Using the wrong distance — it must be the perpendicular distance to the chosen point
Mixing units (N cm on one side, N m on the other) when comparing torques
Forgetting that a body needs both zero net force and zero net torque to be in full equilibrium
Not setting the lifting support’s reaction to zero at the tipping point
You’ve now seen that a resultant torque produces angular acceleration — but exactly how much? That depends on how the mass is spread out and on the equation linking them. First we need the rotational versions of displacement, velocity and acceleration, which is what we build next: angular displacement, velocity and acceleration.
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