Calculating Artificial Gravity Rotation Frequency in RPM

[Momentum09]

"Artificial Gravity"

Homework Statement

A space station in the form of a large wheel, 333m in diameter, rotates to provide an "artificial gravity" of 9m/s^2 for people located on the outer rim. Find the rotational frequency of the wheel that will produce this effect. Answer in units of rpm [revolutions per minute]

Homework Equations

w = 2pi/T , Frequency = 1/T

The Attempt at a Solution

I first equated the normal force with the centripetal force, which is mg = mv^2/r. Cancelling out the m, I got gr = v^2. After I solved for V, I plugged it into the w = 2pi/T equation to find T, then 1/T. To convert the answer to rpm, I multiplied what I got for F by 60 and divided by 2pi. However I kept getting the wrong answer. Can someone please help? Thank you!

 

[hage567]

After I solved for V, I plugged it into the w = 2pi/T equation to find T

How did you relate v to w when finding T in this step?

Be sure to keep track of the units to make sure things make sense.

What answer did you get?

 

[Momentum09]

I got 3.487429162 rpm

 

[Momentum09]

I also know that w = 2 x pi x frequency

 

[hage567]

Yes, but remember that v is linear velocity, and w is angular velocity. They are not interchangeable. Do you know the relation between those two quantities?

 

[Momentum09]

I know that w = v/r

 

[hage567]

So what do you get if you make use of that? You found what v is, and ultimately what you want to find is w (in rev/min).

You're initial approach is OK, it's just the long way around. But you still need to use v=rw.

 

[Momentum09]

oh! I got it now! Thank you so much!

 

[hage567]

You're welcome.