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Ira Lisetskai [31]
3 years ago
13

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 mas

s. 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)
Physics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

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

Explanation:

The expression for conservation of the angular momentum (L) is

L_{i} = L_{f}  I_{i}\times\omega_{i} = I_{f}\times\omega_{f}

Where

I_{i}\ and \ \omega_{i} initial moment of inertia and angular velocity

I_{f}\ and \ \omega_{f} is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

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

Where m_{s} is the satellite mass

r is the  radus of the sphere

Substititute 1900kg for m and 4.6m for r

I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}

The final moment of inertia of the satellite about the centre of mass

I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}

Where m_{x} is the antenna's mass and

I is the length of the antenna

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

So, the Final rotation rate of the satellite is:

I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s

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

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