Question

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod of mass 150.0 kg and length 6.6 m. What is the new rotation rate of the satellite (in rev/s)

Answer 1

6.3 rev/s

Step-by-step explanation:

The new rotation rate of the satellite can be found by conservation of the angular momentum (L):

`L_(i) = L_(f)`

`I_(i)*\omega_(i) = I_(f)*\omega_(f)`

The initial moment of inertia of the satellite (a solid sphere) is given by:

`I_(i) = (2)/(5)m_(s)r^(2)`

Where
`m_(s)`: is the satellite mass and r: is the satellite's radium

`I_(i) = (2)/(5)m_(s)r^(2) = (2)/(5)1900 kg*(4.6 m)^(2) = 1.61 \cdot 10^(4) kg*m^(2)`

Now, the final moment of inertia is given by the satellite and the antennas (rod):

`I_(f) = I_(i) + 2*I_(a) = 1.61 \cdot 10^(4) kg*m^(2) + 2*(1)/(3)m_(a)l^(2)`

Where
`m_(a)`: is the antenna's mass and l: is the lenght of the antenna

`I_(f) = 1.61 \cdot 10^(4) kg*m^(2) + 2*(1)/(3)150.0 kg*(6.6 m)^(2) = 2.05 \cdot 10^(4) kg*m^(2)`

So, the new rotation rate of the satellite is:

`I_(i)*\omega_(i) = I_(f)*\omega_(f)`

`\omega_(f) = (I_(i)*\omega_(i))/(I_(f)) = (1.61 \cdot 10^(4) kg*m^(2)*8.0 (rev)/(s))/(2.05 \cdot 10^(4) kg*m^(2)) = 6.3 rev/s`

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

I hope it helps you!