A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Respuesta :

The angular velocity of the disk must be 2.25 rpm

Explanation:

The centripetal acceleration of an object in circular motion is given by

[tex]a=\omega^2 r[/tex]

where

[tex]\omega[/tex] is the angular velocity

r is the distance of the object from the axis of rotation

For the space station in this problem, we have

[tex]a=\frac{g}{2}=\frac{9.8}{2}=4.9 m/s^2[/tex] is the centripetal acceleration

The diameter of the disk is

d = 175 m

So the radius is

[tex]r=\frac{175}{2}=87.5 m[/tex]

So, a point on the rim has a distance of 87.5 m from the axis of rotation. Therefore, we can re-arrange the previous equation to find the angular velocity:

[tex]\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{4.9}{87.5}}=0.237 rad/s[/tex]

And this is the angular velocity of any point along the disk. Converting into rpm,

[tex]\omega=0.236 \frac{rad}{s}\cdot \frac{60 s/min}{2\pi rad/rev}=2.25 rpm[/tex]

Learn more about circular motion:

brainly.com/question/2562955

#LearnwithBrainly