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

Answer:

6.3 rev/s

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:

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$: is the satellite mass and r: is the satellite's radium

$I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2}$

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

$I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2}$

Where $m_{a}$: is the antenna's mass and l: is the lenght of the antenna

$I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2}$

So, the new rotation rate of the satellite is:

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

$\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s$  

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

I hope it helps you!