State the condition necessary so that the law of conservation of angular momentum applies to a rotating system - AQA - A-Level Physics - Question 2 - 2020 - Paper 5

To find the common angular speed when the clutch engages, we use the principle of angular momentum conservation:

$I_A heta_A + I_B heta_B = (I_A + I_B) heta_{common}$

Where:

• $I_A = 7.2 \, \text{kg m}^2$, \quad $\theta_A = 95 \, \text{rad s}^{-1}$ (clockwise)
• $I_B = 11.5 \, \text{kg m}^2$, \quad $\theta_B = -45 \, \text{rad s}^{-1}$ (anticlockwise)

Now substituting these values:

$7.2 \times 95 + 11.5 \times (-45) = (7.2 + 11.5) \theta_{common}$

Calculating the left side:

$7.2 \times 95 = 684 \quad \text{and} \quad 11.5 \times (-45) = -517.5$

Thus:

$684 - 517.5 = 18.7 = 18.7 \theta_{common}$

Now solving for $\theta_{common}$:

$\theta_{common} = \frac{18.7}{18.7} = 9 \, \text{rad s}^{-1}$

The direction of the common angular speed is clockwise when viewed from the left as the total angular momentum is dominated by the clockwise rotation of shaft A.

For clutch C, with a frictional torque of 600 N m: Using the equation for angular impulse: $\tau \Delta t = I \Delta \omega$

We can calculate: $600 \times 1.5 = (7.2 + 11.5) \times (9 - 0)$

This gives: $900 = 18.7 \times 9 \, \text{(valid)}$

For clutch D, with a frictional torque of 320 N m: $320 \times 1.5 = 18.7 \times (9 - 0)$

This gives: $480 = 18.7 \times 9 \, \text{(invalid)}$

Thus, clutch C meets the criteria while clutch D does not.