The angular speed of the lander must be 3.92 rpm
Explanation:
The (centripetal) acceleration of the lander is given by:
[tex]a=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the length of the arm
In this problem, we have
r = 7.75 m
[tex]a=1.31 m/s^2[/tex] (we are told that the acceleration of the lander must be the same as the acceleration due to gravity at the surface of Europa, which is [tex]1.31 m/s^2[/tex]
Therefore we can find the angular speed of the lander:
[tex]\omega=\sqrt{\frac{a}{r}}=\sqrt{\frac{1.31}{7.75}}=0.411 rad/s[/tex]
And keeping in mind that
[tex]1 rev = 2\pi rad[/tex]
[tex]1 min = 60 s[/tex]
We can convert this angular speed into rpm:
[tex]\omega = 0.411 rad/s \cdot \frac{60 s/min}{2\pi rad/rev}=3.92 rpm[/tex]
Learn more about rotational motion:
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