State the law of conservation of angular momentum - AQA - A-Level Physics - Question 1 - 2018 - Paper 6

To calculate the total moment of inertia ($I_{total}$), we use the formula: $I_{total} = I_{satellite} + 2m r^2$ Where:

• $I_{satellite} = 71$ kg m²
• $m$ is the mass of each instrument pod = 5.0 kg
• $r$ is the radius from the axis of rotation, which is 4.1 m.

Calculating the contribution from the instrument pods: $I_{pods} = 2 imes 5.0 imes (4.1)^2 = 2 imes 5.0 imes 16.81 = 169.6 \text{ kg m}^2$

Thus, $I_{total} = 71 + 169.6 = 240.6 \text{ kg m}^2$

Rounding, we show that the total moment of inertia is approximately 240 kg m².

As the arms of the satellite are retracted, the moment of inertia ($I$) decreases because the distance of the instrument pods from the axis of rotation is reduced. Due to the conservation of angular momentum, if angular momentum ($L = I imes ext{angular speed}$) is conserved, a decrease in $I$ results in an increase in angular speed ($ext{angular speed} ightarrow ext{increases}$).

Using the conservation of angular momentum, initially: $L_{initial} = I_{initial} imes ext{angular speed}_{initial} = 240 \text{ kg m}^2 \times 1.3 \text{ rad s}^{-1} = 312 \text{ kg m}^2 \text{s}^{-1}$

After retraction, with radius at 0.74 m: $I_{final} = I_{satellite} + 2m r^2$ Where: $I_{final} = 71 + 2 imes 5.0 imes (0.74)^2 = 71 + 5.0 imes 1.092 = 76.46 \text{ kg m}^2$

Using $L_{final} = L_{initial}$, we calculate: $312 = 76.46 imes ext{angular speed}_{final}$

Thus,