# Simulated gravity

1. Jul 9, 2004

### galoshes

So i was presented this question and I would not even like the answer but just the path to walk on. Can you help?

In order to simulate the Earth gravity on a space habitat of 100m diameter, what spin rate of a donut-shape space habitat, in terms of RPM, should be.

conversion of RPM is 2(pi)/60 radians/sec

where do I even begin????

I have read that a station 2km in diameter would spin at 1 RPM to simulate a confortable earth-like gravity. Just in my head I came up with the answer of 20 RPMs for a station 100m in diameter, but I have to show my work and explain the theory behind it, but i'm just going on intuition.

Is there hope for me?

Oh yea, I have to hand this in, in three hours and it's the only question out of thirty that I don't even know where to start.

Last edited: Jul 9, 2004
2. Jul 9, 2004

### #define

Centripetal force due to spinning = mrw^2
m = mass

Force we experience due to gravity = mg
m = mass
g is acceleration

from them: g = rw^2
this will give you the required angular velocity of the spinning station

3. Jul 9, 2004

### e(ho0n3

Consider centripetal acceleration.

There is always hope. You can't just 'come up' with stuff in physics. It doesn't work that way. Anyways, your intuition is wrong.

Last edited: Jul 9, 2004
4. Jul 9, 2004

### galoshes

thanks for your response. I am confused because I wasn't given the mass of the station just the diameter, so how do I get the centripetal acceleration without mass? I am especially confused because I don't know what to do with 2(pi)/60 rad/sec. I could work with the 60 rad/sec but the 2(pi) is killin me

thanks again

5. Jul 9, 2004

### e(ho0n3

The centripetal acceleration is given by $v^2/R$ where v is the tangential velocity and R is the radius. You don't need mass. Then use rotational kinematics to get the rpms.

6. Jul 9, 2004

### Zorodius

If a particle travels along a circle or circular arc with radius r at constant speed v, it is in uniform circular motion and has an acceleration of magnitude

$$a = \frac{v^2}{r}$$

This is directed radially inward. Since $s = \theta r$, differentiating with respect to time with r held constant yields $v = \omega r$. Substituting this into the equation for acceleration above yields:

$$a_r = \frac{\omega^2 r^2}{r} = \omega^2 r$$

where the subscript r is to emphasize that this is a radial acceleration and not a tangential one.

I believe you can use this to find the velocity you are looking for, since you know you want the acceleration to be equal to g and the radius to be equal to 50.0 meters.

The $2\pi / 60$ is a conversion factor for converting between revolutions per minute and radians per second. In general, many of the rules we use to work with angular measures rely on derivations that hold only for radian measure, and as such, it's a good idea to convert to radians before performing any calculations with them.

radians = revolutions X $2\pi$

7. Jul 9, 2004

### galoshes

Ok so maybe I do want ya'll to do this for me because I have no idea what to do with any of the information given because I get .55 RPM and I know that is WAY off

8. Jul 9, 2004

### Zorodius

$$\sqrt{\frac{a_r}{r}} = \omega$$

$$\sqrt{\frac{9.8 m/s^2}{50.0 m}} = .4427 ~rad/s = \omega$$

$$\frac{.4427 ~rad}{s} \cdot \frac{60 ~s}{1 ~min} \cdot \frac{1 ~rev}{2\pi ~rad} = 4.23 ~RPM$$

Hopefully someone will double-check that.

9. Jul 9, 2004

### galoshes

thanks a million

10. Jul 9, 2004

### JohnDubYa

RE: "Centripetal force due to spinning = mrw^2"

Spinning doesn't cause a force. And the centripetal force is not an actual force that acts on the object. It is the vector sum of all forces that act in the radial direction.

RE: "...where the subscript r is to emphasize that this is a radial acceleration and not a tangential one."

Glad that someone else has dropped the term "centripetal." :)