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

Power is calculated using the formula:

$P = T \cdot \omega$,

where $T$ is the torque and $\omega$ is the angular velocity.

In the first 2 seconds, we assume the rotor reaches the maximum angular velocity of $1.5 \text{ rad/s}$, and the frictional torque (390 N m) must be overcome.

Calculating power:

Using $\omega = 1.5 \text{ rad/s}$,

$P = 390 \text{ N m} \cdot 1.5 \text{ rad/s} = 585 \text{ W}.$

To find maximum torque, we observe from the graph that the steepest ascent in angular velocity typically corresponds to maximum torque output. From $t = 2.0 s$ to $t = 3.0 s$, the gradient is:

$\Delta \omega / \Delta t = (1.5 - 0) / (2.0 - 0) = 0.75 \text{ rad/s²}.$

Using the moment of inertia, we can find the maximum torque:

$T = I \cdot \alpha = (2.1 \times 10^4 \text{ kg m²}) \cdot (0.75 \text{ rad/s²}) = 15750 \text{ N m}.$

Angular impulse can be calculated using:

$\Delta L = \int T \, dt,$

which is equal to the change in angular momentum. The change in angular velocity over this period is from $\omega = 0$ to $\omega = -1.0$ rad/s.

Thus, the change in angular momentum:

$\Delta L = I \cdot (\omega_{final} - \omega_{initial}) = 2.1 \times 10^4 \cdot (-1.0 - 0) = -2.1 \times 10^4 \text{ kg m²/s}.$

Analyzing the torque, it relates directly to the angular velocity changes seen in the graph. The torque would likely show periods of constant value during acceleration and negative torque when the rotor slows down. The correct box to tick is the one that shows stepped changes aligned with the torque variation shown in the angular velocity graph.