Question

f the arm is 7.75 m long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the lander is the same as the acceleration due to gravity at the surface of Europa?

Answer 1

The angular speed of the lander must be 3.92 rpm

Explanation:

The (centripetal) acceleration of the lander is given by:

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the length of the arm

In this problem, we have

r = 7.75 m

$a=1.31 m/s^2$ (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 $1.31 m/s^2$

Therefore we can find the angular speed of the lander:

$\omega=\sqrt{\frac{a}{r}}=\sqrt{\frac{1.31}{7.75}}=0.411 rad/s$

And keeping in mind that

$1 rev = 2\pi rad$

$1 min = 60 s$

We can convert this angular speed into rpm:

$\omega = 0.411 rad/s \cdot \frac{60 s/min}{2\pi rad/rev}=3.92 rpm$

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