Rotational Equilibrium (HL) | HL IB Physics Revision Notes 2025

Rotational Equilibrium

A system is said to be in rotational equilibrium when
- There is no resultant force acting
- There is no resultant torque acting

An object in rotational equilibrium will therefore remain at rest, or rotate with a constant angular velocity
- This is analogous to Newton's First Law for translational equilibrium

This means a body is in rotational equilibrium if

The sum of the clockwise moments is equal to the sum of the anticlockwise moments

This is also known as the principle of moments and can be applied to a range of scenarios, such as a balanced beam
- A beam is an example of a rigid, extended body

A balanced beam in rotational equilibrium

Balanced Beam in Rotational Equilibrium, for IB HL Physics Revision Notes

When the resultant force and resultant torque are both zero, the beam will be in rotational equilibrium

Worked example

Four beams of the same length each have three forces acting on them.

Which of the beams is in rotational equilibrium?

Answer: C

A beam is in rotational equilibrium when there is zero resultant force and zero resultant torque acting on it
In rotational equilibrium:

Total clockwise torque = Total anticlockwise torque

Consider beam C, taking torques from the centre of the beam (where its weight acts) :

Torque, $τ = F r$ (as $\sin 90 ° = 1$ )

Total clockwise torque = 15 × 50 = 750 N cm

Total anticlockwise torque = 25 × 30 = 750 N cm

The total clockwise torque (750 N cm) = total anticlockwise torque (750 N cm), therefore, beam C is in rotational equilibrium

The other beams are not in rotational equilibrium because...

Beam A has a resultant torque of 310 N cm anticlockwise
Beam B has a resultant torque of 370 N cm clockwise
Beam D has a resultant torque of 1790 N cm clockwise

Exam Tip

When considering an object in rotational equilibrium, choosing certain points can simplify calculations of resultant torque. Remember you can choose any point, not just the axis of rotation.

To simplify your calculation, choose a point where the torque of (most of) the forces are unknown, or when you need to determine where the resultant torque is zero. To do this, choose a point through which the lines of action of the forces pass

Unbalanced Torque

In the same way that a resultant force produces linear acceleration, a resultant torque produces angular acceleration
The direction of the angular acceleration depends on the direction of the net resultant torque

Beam with an unbalanced torque

If there is a net resultant torque in the clockwise or anti-clockwise direction, the beam will also have an angular acceleration in that direction

Worked example

A uniform plank of mass 30 kg and length 10 m is supported at its left end and at a point 1.5 m from the centre.

R_d529FM_1-4-2-net-resultant-torque-worked-example

Calculate the maximum distance r to which a boy of mass 50 kg can walk without tipping the rod over.

Answer:

Step 1: Analyse the scenario and identify the forces

Let the forces at each support be F_L (reaction force from the left support) and F_R (reaction force from the right support)
- These are vertically upwards

Just before the plank tips over, the system is in rotational equilibrium
When the plank begins to tip over, the left support force F_L will become zero since the rod will no longer touch the support

bQbfjKoF_1-4-2-net-resultant-torque-worked-example-solution

Step 2: Take torques about the right support

Torque = Fr sin θ (θ = 90° for all)
Clockwise torque = 50 × g × r
Anti-clockwise torque = 30 × g × 1.5

Step 3: Equate the clockwise and anti-clockwise torques

$30 \times g \times 1.5 = 50 \times g \times r$

$r = \frac{30 g \times 1.5}{50 g} = \frac{30 \times 1.5}{50}$

r = 0.90 m

Therefore, the plank will begin to tip once the boy is 0.90 m from the right support

Worked example

The diagram shows three forces acting on a wheel.

1-4-2-net-resultant-torque-on-a-wheel-worked-example1-4-2-net-resultant-torque-on-a-wheel-worked-example

Determine the net resultant torque about the axis of rotation O. State whether the angular acceleration that is produced is clockwise or anticlockwise.

Answer:

Step 1: Recall the equation for torque

$τ = F r \sin θ$

Step 2: Find the sum of the torques in the anti-clockwise direction

Torque of the 10 N force: $τ = 10 \times 0.25 \times \sin 90 ° = 2.5 N m$

Torque of the 9 N force: $τ = 9 \times 0.25 \times \sin 90 ° = 2.25 N m$

Total anti-clockwise torque = 2.5 + 2.25 = 4.75 N m

Step 3: Calculate the torque in the clockwise direction

1-4-2-net-resultant-torque-on-a-wheel-worked-example-ma-ib-2025-physics-

Torque of the 12 N force: $τ = 12 \times 0.1 \times \sin 30 ° = 0.6 N m$

Step 4: Determine the net resultant torque

Resultant torque = sum of anti-clockwise torques − sum of clockwise torques
Resultant torque: $\sum τ$ = 4.75 − 0.6 = 4.15 N m, anticlockwise
Direction of angular acceleration: anticlockwise

Exam Tip

You should know that torque is a vector quantity, however, at this level, you will only need to consider whether it produces clockwise or anti-clockwise motion

Clockwise or anticlockwise moment, downloadable AS & A Level Physics revision notes

Rotational Equilibrium (HL) (HL IB Physics)

Revision Note