Question

A satellite is spinning at 6.2 rev/s. The satellite consists of a main body in the shape of a solid sphere of radius 2.0 m and mass 10,000 kg and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length 3.0 m each and mass 16 kg. The antennas lie in the plane of rotation. What is the angular momentum of the satellite about its center?

Answer 1

The angular momentum of the satellite about its center is 497,785.6 kg⋅m^2/s.

Step-by-step explanation:

To calculate the angular momentum of the satellite about its center, we need to consider the angular momentum of the main body and the antennas separately.

The angular momentum of the main body can be calculated using the formula: L = I × ω, where I is the moment of inertia and ω is the angular velocity. For a solid sphere, the moment of inertia is given by I = (2/5) × m × r^2, where m is the mass and r is the radius. Plugging in the values, we get I = (2/5) × 10000 kg × (2.0 m)^2 = 80000 kg⋅m^2. The angular momentum of the main body is then L_main = 80000 kg⋅m^2 × 6.2 rev/s = 496000 kg⋅m^2/s.

Each antenna can be considered as a rod rotating around one end, so the moment of inertia is given by I = (1/3) * m * L^2, where L is the length. Plugging in the values, we get I = (1/3) × 16 kg × (3.0 m)^2 = 144 kg⋅m^2. The angular momentum of each antenna is then L_antenna = 144 kg⋅m^2 × 6.2 rev/s = 892.8 kg⋅m^2/s.

The total angular momentum of the satellite is the sum of the angular momenta of the main body and the antennas: L_total = L_main + 2 × L_antenna = 496000 kg⋅m^2/s + 2 × 892.8 kg⋅m^2/s = 496000 kg⋅m^2/s + 1785.6 kg⋅m^2/s = 497785.6 kg⋅m^2/s.

Answer 2

`627031 kgm^2/s`

Step-by-step explanation:

First convert angular speed from 6.2 rev/s to rad/s knowing that each revolution is 2π rad:

`\omega = 6.2 * 2\pi = 38.96 rad/s`

The we can calculate the moments of inertia of the satellite by summing up the 2 moments of inertia of the solid sphere and the 2 rods at their ends:

`I = I_s + 2I_r`

`I = (2)/(5)MR^2 + 2(1)/(3)mL^2`

`I = (2)/(5)10000*2^2 + 2(1)/(3)16*3^2`

`I = 16000 + 96 = 16096 kgm^2`

Then the angular momentum is the product of angular velocity and moment of inertia

`\omega I = 38.96 * 16096 = 627031 kgm^2/s`