0.2.1 State one function of a flywheel - AQA - A-Level Physics - Question 2 - 2018 - Paper 6

To find the frictional torque, we can use the change in gravitational potential energy and equate it to the work done against the torque.

1. Calculate the decrease in potential energy:

   $\Delta PE = mgh = (0.020 \text{ kg})(9.81 \text{ m/s}^2)(0.075 \text{ m}) (1 - \cos(3.00))$ where h is the vertical drop of the magnet. Since the angle moved is 3.00 radians, the height change can be calculated using the radius:

   $h = 0.075(1 - \cos(3.00 \text{ rad}))$

2. The work done is equal to the torque ($\tau$) times the angle moved in radians:

   $W = \tau \cdot \theta$

3. Setting the loss in potential energy equal to the work done:

   $\Delta PE = \tau \cdot \theta$

4. Rearranging gives:

   $\tau = \frac{\Delta PE}{\theta} \approx 7 \times 10^{-4} \text{ N m}$

We can determine the moment of inertia ($I$) using the relationship involving torque and angular deceleration:

1. The angular displacement ($\theta$) is given in rotations, so convert to radians:

   $\theta = 73 \text{ rotations} \times 2\pi \text{ rad/rotation} = 459.7 \text{ rad}$

2. Calculate the average angular speed ($\omega$):

   $\omega_{avg} = \frac{\omega_i + \omega_f}{2} = \frac{25.0 + 0}{2} = 12.5 \text{ rad/s}$

3. Use the formula for angular deceleration ($\alpha$) as follows:

   $\alpha = \frac{\Delta \omega}{\Delta t}$

   Where $\Delta \omega$ is the change in angular speed and $\Delta t$ is the time taken. Since the time is not provided, we need to assume a constant torque and calculate using the average speed.

4. Finally apply the relationship between torque, moment of inertia, and angular acceleration:

   $\tau = I \cdot \alpha$

5. From this expression, rearranging gives:

   $I = \frac{\tau}{\alpha}$