Answer:
6.3 rev/s
Explanation:
The new rotation rate of the satellite can be found by conservation of the angular momentum (L):
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]
The initial moment of inertia of the satellite (a solid sphere) is given by:
[tex] I_{i} = \frac{2}{5}m_{s}r^{2} [/tex]
Where [tex]m_{s}[/tex]: is the satellite mass and r: is the satellite's radium
[tex] 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} [/tex]
Now, the final moment of inertia is given by the satellite and the antennas (rod):
[tex] I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2} [/tex]
Where [tex]m_{a}[/tex]: is the antenna's mass and l: is the lenght of the antenna
[tex] 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} [/tex]
So, the new rotation rate of the satellite is:
[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]
[tex]\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[/tex] Â
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
I hope it helps you! Â
