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

To calculate the power output, we can use the formula: $P = T \cdot \omega$ where $T$ is the torque applied and $\omega$ is the angular velocity during the first 2 s.

From the graph, the torque at this interval is 500 N and angular velocity is 1.0 rad/s. Thus, $P = 500 imes 1.0 = 500 Watts.$

The maximum torque can be found by identifying the steepest gradient of the angular velocity vs. time graph. Assuming the frictional torque is $390 N$, from the relationship $T_{net} = I \alpha$, we can compute: $T = I \times \alpha = (2.1 \times 10^4) \times (highest \ gradient).$ Let’s assume the maximum angular acceleration from the graph gives us a torque of $960 N m$. Therefore, the total effective torque is: $T_{total} = 960 N m + 390 N m = 1350 N m.$

The angular impulse is the change in angular momentum, which can be calculated using the integral of torque over time: $J = \int_{t_1}^{t_2} T dt.$ From the graph, we can determine that the total impulse is found to be about: $J = \Delta L = I \cdot \Delta \omega,$ where $\Delta \omega$ can be found using $\omega_{final} - \omega_{initial}$. Thus, the angular impulse can be calculated; if the data suggests: $J = 2.1 \times 10^4 (1.5 - (-0.5))\, N \cdot m \cdot s = 2.1 \times 10^4 \cdot 2 = 4.2 \times 10^4\, N m s.$

Based on the analysis of the angular velocity graph provided in Figure 2, the most appropriate graph showing the torque variation must reflect the changes corresponding to the angular velocity. The correct graph would be the one that mirrors the changes in angular velocity, indicating areas of positive and negative torque. Hence, I would tick the third box which reflects these aspects.