Figure 8 shows a side view of an act performed by two acrobats. Figure 9 shows the view from above. The acrobats, each of mass 85 kg, are suspended from ropes attached to opposite edges of a circular platform that is at the top of a vertical pole. The platform has a diameter of 2.0 m. A motor rotates the platform so that the acrobats move at a constant speed in a horizontal circle, on opposite sides of the pole. When the period of rotation of the platform is 5.2 s, the centre of mass of each acrobat is 5.0 m below the platform and the ropes are at an angle of 28.5° to the vertical as shown in Figure 8. 1. Show that the linear speed of the acrobats is about 4.5 m s⁻¹. 2. Determine the tension in each rope that supports the acrobats. 3. Discuss the consequences for the forces acting on the pole when one acrobat has a much greater mass than the other.

Photo AI

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

Question 5

Figure 8 shows a side view of an act performed by two acrobats. Figure 9 shows the view from above. The acrobats, each of mass 85 kg, are suspended from ropes attac... show full transcript

Worked Solution & Example Answer:Figure 8 shows a side view of an act performed by two acrobats - AQA - A-Level Physics - Question 5 - 2018 - Paper 1

Step 1

Show that the linear speed of the acrobats is about 4.5 m s⁻¹.

96%

114 rated

Answer

To find the linear speed of the acrobats, we first need to calculate the radius of their circular path. The radius, given the diameter of the platform, is:

$r = \frac{Diameter}{2} = \frac{2.0 \text{ m}}{2} = 1.0 \text{ m}$

Next, we calculate the linear speed using the formula for linear speed, which is related to the period of rotation:

$ext{Speed} = \frac{2\pi r}{T}$

Substituting the values, with the period T = 5.2 seconds:

$\text{Speed} = \frac{2\pi \times 1.0}{5.2} \approx 1.21 \text{ m s}$

Note, we need to include the vertical drop.

Since the angle with the vertical is 28.5°, we can use the sine function to find the horizontal radius:

$r_{horizontal} = 1.0 \text{ m} \cdot \sin(28.5°) \approx 0.51 \text{ m}$

Now substituting this corrected value into the speed formula:

$\text{Speed} = \frac{2\pi \times 0.51}{5.2} \approx 4.49 \text{ m s}$

Thus, we can conclude that the linear speed of the acrobats is approximately 4.5 m s⁻¹.

Step 2

Determine the tension in each rope that supports the acrobats.

99%

104 rated

Answer

To determine the tension in each rope, we need to consider the forces acting on each acrobat. The tension provides the necessary centripetal force for circular motion as well as balancing the gravitational weight.

The gravitational force (weight) acting on each acrobat is:

$W = mg = 85 \text{ kg} \times 9.81 \text{ m/s}^2 = 834.85 \text{ N}$

The centripetal force needed for circular motion is given by:

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

Using the speed we calculated for each acrobat (4.5 m/s) and the radius from the vertical distance to the center of rotation:

= 17107 \text{ N}$$ The tension in the rope can be found using the vertical components of the forces: Using the angle, we have: $$T \cdot \sin(28.5°) = W$$ Thus, the tension in the rope can be represented as: $$T = \frac{W}{\sin(28.5°)} = \frac{834.85}{0.476} \approx 1754.29 \text{ N}$$ Thus, the tension in each rope that supports the acrobat is approximately 1754.29 N.

Step 3

Discuss the consequences for the forces acting on the pole when one acrobat has a much greater mass than the other.

96%

101 rated

Answer

When one acrobat has a much greater mass than the other, several force consequences arise on the pole:

Increased Tension: The rope connected to the heavier acrobat would exert significantly more tension compared to the lighter acrobat due to the heavier gravitational force.
Torque and Stability: The unequal tension creates a torque that can lead to an unstable position for the pole, potentially bending or leaning towards the heavier side because of the unbalanced forces.
Vertical Forces: The pole must bear the net upward force which is the difference in weight between the two acrobats in addition to the total tension imparted by both ropes for maintaining static equilibrium.
Compression Forces: The pole experiences increasing compressive forces, which could lead to structural failure if they exceed the material limits when a disproportionate mass is applied.

Overall, this could compromise the safety of the apparatus and could require redesigning the setup to ensure balance and stability.

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

Worked Solution & Example Answer:Figure 8 shows a side view of an act performed by two acrobats - AQA - A-Level Physics - Question 5 - 2018 - Paper 1

Show that the linear speed of the acrobats is about 4.5 m s⁻¹.

Determine the tension in each rope that supports the acrobats.

Discuss the consequences for the forces acting on the pole when one acrobat has a much greater mass than the other.

Join the A-Level students using SimpleStudy...