Revolutions per sec (centripital acceleration)

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Homework Help Overview

The problem involves determining the rotation rate in revolutions per second required for astronauts to experience a specified centripetal acceleration while in circular motion. The context is set within the framework of centripetal acceleration, specifically relating to a mechanical device used in astronaut training.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the confusion surrounding the units of acceleration, questioning the use of "kg" instead of "m/s²" and the meaning of "3g" in the context of the problem. There is an exploration of how to relate centripetal acceleration to the period of rotation.

Discussion Status

The discussion is ongoing, with participants clarifying the terminology and units involved. Some guidance has been offered regarding the interpretation of "g" as a measure of acceleration, and there is an acknowledgment of the need to eliminate velocity in the equations to find the period.

Contextual Notes

There is a noted confusion regarding the units of acceleration and the implications of using "3g" versus "3.00 kg." Participants are working through these assumptions and the definitions involved in the problem setup.

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[SOLVED] Revolutions per sec (centripetal acceleration)

Homework Statement


As their booster rockets separate, 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 centripetal acceleration. 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 centripetal accelration of 3.00kg while in a circular motion with the radius of 9.45m


Homework Equations


a_c = v^2/r

T= 2pi*r/ v


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 centripetal 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 :smile:
 
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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.
 
thanks Kurdt
 
Last edited:
I don't think so. You will never get an acceleration reported in units of mass unless there is a typo or mistake. :smile:
 
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..
 
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.
 
Oh I get it now... I thought it meant grams..
Thanks Kurdt :smile:
 

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