# Homework Help: Circular Motion of a space station Problem

1. Jun 12, 2005

### kye6338

Hi,
I was hoping I could get some help on these problems. Here's the first problem:
1. A projected space station consists of a circular tube that is set rotating about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of about D = 1.30km. What must be the rotation speed (in revolutions per day) if an effect equal to gravity at the surface of the Earth (1g) is to be felt?

So here's what I did: I converted the diameter from km to m, then got the radius, which is 650 km. Then, I found the velocity of the station by doing v= sqrt r x g . I got 79.81 m/s. And then I used the formula T= 2PiR/V ....and I got 51.17. My question would be what are the units at this point? I assumed it was revolutions per second, then converted to revolutions per day, but that was not correct. Maybe I'm missing a step, so please help!

2. Jun 12, 2005

It is easier so substitute the values you were given at the very end. Only consider the center of mass of the space station. This allows you to write the following:

$$v=\frac{\Delta x}{\Delta t} = \frac{C}{T} = \frac{2\pi (R + d)}{T}$$

and

$$a=\frac{v^2}{(R+d)}$$

where "R" is the radius of the Earth "d" is the distance from the surface of the Earth. I hope it helps.

3. Jun 13, 2005

### kye6338

Do you mean substituting the values given in the problem into those equations? If I did that, wouldn't I have two unknowns with v and T?

4. Jun 13, 2005

### whozum

You can find V with some energy equations, and T is the time of one revolution of the satelite. Notice those are just the equations for average velocity and centripetal acceleration.

He was pointing out that it is easier (and less likely to induce error) if you solve the problem in terms of its variables first, THEN plug in numbers at the very end.

5. Jun 13, 2005

### Astronuc

Staff Emeritus
One is asked to find a rotational speed $\omega$ such that the force exerted on a mass (m) would be equivalent to mg, where g is the acceleration due to gravity at the earth's surface, 9.81 m/sec2.

One could simply use a force balance mg = mv2/r = m$\omega^2$r.

Also, check units on the radius.

For some insight, see -

angular velocity,
centripetal force, and
centripetal acceleration.

Last edited: Jun 13, 2005