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Rotational Equilibrium (HL) (HL IB Physics)

Revision Note

Katie M

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Katie M

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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 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, tau space equals space F r (as sin space 90 degree space equals space 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

Examiner 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

SozIF7Xh_1-4-2-beam-with-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 FL (reaction force from the left support) and FR (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 FL 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 space cross times space g space cross times space 1.5 space equals space 50 space cross times space g space cross times space r

r space equals space fraction numerator 30 up diagonal strike g cross times 1.5 over denominator 50 up diagonal strike g end fraction space equals space fraction numerator 30 space cross times space 1.5 over denominator 50 end fraction

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.

net-torque-on-a-wheel-we

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

tau space equals space F r space sin space theta

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

Torque of the 10 N force:  tau space equals space 10 space cross times space 0.25 space cross times space sin space 90 degree space equals space 2.5 space straight N space straight m

Torque of the 9 N force:  tau space equals space 9 space cross times space 0.25 space cross times space sin space 90 degree space equals space 2.25 space straight N space straight m

Total clockwise torque = 2.5 + 2.25 = 4.75 N m

Step 3: Calculate the torque in the anti-clockwise direction

net-torque-on-a-wheel-we-solution

Torque of the 12 N force:  tau space equals space 12 space cross times space 0.1 space cross times space sin space 30 degree space equals space 0.6 space straight N space straight m

Total anti-clockwise torque = 0.6 N m

Step 4: Determine the net resultant torque

  • Resultant torque = sum of clockwise torques − sum of anti-clockwise torques
  • Resultant torque:  sum tau = 4.75 − 0.6 = 4.15 N m, clockwise
  • Direction of angular acceleration: clockwise

Examiner 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

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.