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

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)

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Sapo 5 years 2021-08-27T07:20:00+00:00 1 Answers 27 views 0

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  1. Ladonna
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    2021-08-27T07:21:50+00:00
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    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!  

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