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

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Figure 1 shows a fairground ride. Figure 1 The ride consists of a rotor that rotates in a vertical circle about a horizontal axis. The rotor has two rigid arms. A ... show full transcript

Worked Solution & Example Answer:Figure 1 shows a fairground ride - AQA - A-Level Physics - Question 1 - 2022 - Paper 6

Step 1

Determine the mean angular velocity of the rotor during the 12 s period.

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Answer

To find the mean angular velocity ar{ heta} over the 12 s period, we will calculate the area under the angular velocity-time graph in Figure 2, dividing that area by the total time of 12 s.

The angular displacement in radians can be calculated by summing the areas of the trapezoids created in the graph, and using the formula:

ar{ heta} = \frac{\Delta \theta}{t_{total}}

where ar{ heta} is the mean angular velocity, heta heta is the angular displacement, and ttotalt_{total} is the total time.

Calculating the areas, we find: [\bar{\omega} = 1.75 \text{ rad s}^{-1}]

Step 2

Calculate the power output of the driving mechanism during the first 2 s shown in Figure 2.

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Answer

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

Where T is the torque and heta 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} ]

Step 3

Calculate the maximum torque applied by the driving mechanism to the rotor during the 12 s period.

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Answer

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

Where I is the moment of inertia and heta 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} ]

Step 4

Calculate the magnitude of the angular impulse on the rotor between t = 2.0 s and t = 7.0 s.

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Answer

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

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

Step 5

Which graph best shows the variation of the torque T applied to the rotor for the 12 s period? Tick (✓) one box.

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Answer

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.

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