Figure 1 shows a fairground ride - AQA - A-Level Physics - Question 1 - 2022 - Paper 6

Power output (P) is calculated using the formula: $P = T \cdot \omega$

Where T is the torque and $heta$ is the angular velocity. We need to use the value of torque obtained in the previous section, alongside the maximum angular velocity achieved during the first 2 s, to calculate:

[ P = 546 \text{ W} ]

To find the maximum torque T_max, we analyze the relationship: $T = I \cdot \alpha$

Where I is the moment of inertia and $heta$ is the angular acceleration. Considering the highest angular acceleration obtained in the graph can be related with the torque equation.

Thus, applying known values: [ T_{max} = 990 ext{ N m} ]

The angular impulse J is given by the product of torque and the time interval: $J = T \cdot \Delta t$

With $heta$ having the corresponding values between 2.0 s and 7.0 s: [ J = 3.8 \times 10^5 \text{ N m s} ]

Upon reviewing the torque variation over the time period displayed in the graphs, the correct graph demonstrates the expected characteristic changes in torque corresponding to the angular velocity changes in the given time. The correct option is ticked on the provided graph.