Figure 8 shows a side view of an act performed by two acrobats - AQA - A-Level Physics - Question 5 - 2018 - Paper 1

To find the tension (T) in each rope, we need to apply the principles of circular motion and consider the forces acting on the acrobat. The forces involved include the weight (W) of the acrobat and the centripetal force (F_c).

First, we know the weight is given by:

$W = mg = 85 \times 9.81 \approx 834 \text{ N}$

The centripetal force can be expressed as:

$F_c = \frac{mv^2}{r}$

From our earlier calculation, we substitute the calculated speed and the effective radius.

Assuming the angle of the rope is at 28.5°, we can resolve the forces:

$T \cos(\theta) = W$

$T \sin(\theta) = F_c$

Substituting leads to:

$T = \frac{W}{\cos(28.5)} \approx \frac{834}{\cos(28.5)} \approx 970 \text{ N}$

Thus, the tension in each rope is approximately 970 N.

If one acrobat has a significantly greater mass than the other, the forces acting on the pole would change crucially. The following points summarize the consequences:

1. Unequal Forces: The greater mass will generate a higher gravitational force, increasing the tension in the rope connected to that acrobat. This unequal tension will create an imbalance in the vertical forces.

2. Torque on the Pole: The uneven distribution of weight leads to torque that could affect the stability of the pole, causing it to bend or tilt. The pole will experience additional compressive forces due to the greater downward force from the heavier acrobat.

3. Safety Risks: An increased load on one side could threaten the structural integrity of the pole and the safety of both acrobats, especially during movement. Accordingly, if the pole isn’t sufficiently reinforced, it may risk collapsing or tipping.

In summary, the unequal mass distribution can lead to variations in forces experienced by the pole, potentially resulting in structural failure or accidents.