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

To find the total moment of inertia ($I_{total}$) when the arms are fully extended, we calculate the contributions from both the satellite body and the instrument pods:

1. The moment of inertia of the satellite body is given as $I_{satellite} = 71 ext{ kg m}^2$.
2. The moment of inertia of each pod (considering they are at a distance of $r = 4.1 ext{ m}$ from the axis of rotation) is calculated using the formula: $I_{pod} = m r^2$ where $m$ is the mass of each pod, which is $5.0 ext{ kg}$: $I_{pod} = 5.0 imes (4.1)^2 = 5.0 imes 16.81 = 84.05 ext{ kg m}^2$ Since there are 2 pods, we multiply this by 2: $I_{pods} = 2 imes 84.05 = 168.1 ext{ kg m}^2$
3. Thus, the total moment of inertia is: $I_{total} = I_{satellite} + I_{pods} = 71 + 168.1 = 239.1 ext{ kg m}^2$ This result rounds to 240 kg m², confirming the total moment of inertia.

As the arms are retracted, the distance of the instrument pods from the axis of rotation decreases, leading to a reduction in the total moment of inertia of the system.

According to the conservation of angular momentum, $L = I imes heta$, where $L$ is the angular momentum, $I$ is the moment of inertia, and $heta$ is the angular speed. Since $L$ must remain constant, a decrease in $I$ must lead to an increase in $heta$. Therefore, as the arms are retracted, the angular speed of the satellite increases.

Initially, the satellite has an angular speed of $1.3 ext{ rad s}^{-1}$. The moment of inertia with the arms fully extended is $I_{total} = 240 ext{ kg m}^2$. When the arms are fully retracted, the new moment of inertia can be calculated:

Assuming the new radius when fully retracted is $0.74 ext{ m}$, we need to re-evaluate the moment of inertia:

1. Moment of inertia of the two pods: $I_{retracted} = 2 imes 5.0 imes (0.74)^2 = 2 imes 5.0 imes 0.5476 = 5.476 ext{ kg m}^2$
2. Total moment of inertia: $I_{new} = I_{satellite} + I_{retracted} = 71 + 5.476 = 76.476 ext{ kg m}^2$

Using conservation of angular momentum: $L_{initial} = L_{final}$ $I_{initial} imes heta_{initial} = I_{new} imes heta_{new}$ Plugging in the known values: $240 imes 1.3 = 76.476 imes heta_{new}$ Calculating $heta_{new}$:

ightarrow ext{approximately } 7.55 ext{ rad s}^{-1}$$ Since this exceeds the maximum permitted angular speed of $2.4 ext{ rad s}^{-1}$, it is concluded that the arms cannot be retracted fully without exceeding the maximum permitted angular speed.