# Revolutions per sec (centripital acceleration)

1. Oct 23, 2007

### ~christina~

[SOLVED] Revolutions per sec (centripital acceleration)

1. The problem statement, all variables and given/known data
As their booster rockets seperate, space shuttle astronauts typically feel acclerations up to 3g where g= 9.80ms^2. In their training astronauts ride in a device in which they experience such accelrations as centripital accleration. Specifically, the astronaut is fastened securely at the end of a mechanical arm and that turns at constant speed in a horizontal circle.

determine the rotation rate in revolutions per sec required to give a astronaut a centripital accelration of 3.00kg while in a circular motion with the radius of 9.45m

2. Relevant equations
$$a_c = v^2/r$$

T= 2pi*r/ v

3. The attempt at a solution

all I know is that the

$$a_c$$ = 3.00kg
and that

r= 9.45m

I'm not sure what I do with this though since wouldn't the accelration usually be in m/s^2 but in this case it is in kg...how can this be?

well I also don't have the velocity either...

I was thinking of substituting the centripital accelration equation into the period T equation to get rid of v however I do know that speed is constant but velocity magnitude doesn't change since speed doesn't change but the magnitude I know does..

Can someone help me out with this...

Thanks

2. Oct 23, 2007

### Kurdt

Staff Emeritus
I think the question is supposed to say "centripetal acceleration of 3g" not 3.00 kg. Just seems like a typo. Your initial thoughts are quite correct. You can eliminate v to work out what T will be.

3. Oct 23, 2007

### ~christina~

thanks Kurdt

Last edited: Oct 23, 2007
4. Oct 23, 2007

### Kurdt

Staff Emeritus
I don't think so. You will never get an acceleration reported in units of mass unless there is a typo or mistake.

5. Oct 23, 2007

### ~christina~

wait so how can it be in g then ??

you did say that it was in g not kg but now you say it can't be in units of mass...I'm confused..

6. Oct 23, 2007

### Kurdt

Staff Emeritus
g in this context is the acceleration due to gravity. So when somebody says there was an acceleration of 3g, they mean that the acceleration is 3 times the magnitude of the acceleration due to gravity. g was given in your question as g=9.80ms-2.

7. Oct 23, 2007

### ~christina~

Oh I get it now... I thought it meant grams..
Thanks Kurdt