Rotational Equilibrium (DP IB Physics: HL): Revision Note

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 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, $\tau =Fr$ (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

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

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.

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

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

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

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

r = 0.90 m

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

The diagram shows three forces acting on a wheel.

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

$\tau =Fr\sin \theta$

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

Torque of the 10 N force:  $\tau =10\times 0.25\times \sin 90°=2.5\mathrm{N}\mathrm{m}$

Torque of the 9 N force:  $\tau =9\times 0.25\times \sin 90°=2.25\mathrm{N}\mathrm{m}$

Total clockwise torque = 2.5 + 2.25 = 4.75 N m

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

Torque of the 12 N force:  $\tau =12\times 0.1\times \sin 30°=0.6\mathrm{N}\mathrm{m}$

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

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