# Centrifugal along a pendulum

1. Feb 1, 2013

### enzi

This may seem like a very simple question but given my lack of knowledge in the area, I'm stumped:

If we have a rotating "space station" a la 2001: A Space Odyssey where the outer ring is at normal Earth gravity, what is the factor to determine how much gravity would be at various points along the "spokes"? So let's say the "spoke" is 100 metres long, what would the gravity be 25 metres from the ring, 50 metres from the ring, 75 metres from the ring, and 99 metres from the ring (1 metre from the hub)?

Thanks in advance for your help on this and I apologize for being so dense that I can't figure this out on my own! :)

2. Feb 1, 2013

### Staff: Mentor

In terms of angular speed ω, the centripetal force needed to keep an object of mass m moving in a circle of radius r is F = mω2r. So the perceived "gravity" along a spoke increases linearly from the center (where it is zero) to the rim.

3. Feb 1, 2013

### enzi

Linearly! That one I wouldn't have guessed. So we can safely say that 25 metres from the ring we have .75g, 50 metres from the ring we have .5g, 75 metres from the ring we have .25g, and 99 metres from the ring we have .01g. Thanks!

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