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Centripetal Accleration

  • Thread starter AznBoi
  • Start date
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1. Homework Statement
It has been suggested that rotating cylinders about 10 mi long and 5 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration?


2. Homework Equations
I would say:
W=(delta)theta/(delta)time
a_centripetal=v^2/r or rW^2


3. The Attempt at a Solution
I don't really get what it's asking. Free-fall acceleration is -9.8 m/s^2 right? So that means the centripetal acceleration has to equal -9.8?

I converted the radius of 5mi to meters and got 8047m.

If what I state above is true, then 9.8=8047*W^2 and the angular speed would equal .0349 rad/s?

Despite what I posted above, I think that you need to convert 9.8 (a tangential acceleration) to centripetal acceleration. So would I have to plug 9.8 for in a in this equation to get the angular accel? a=r(alpha) Angular accel is not the same as centripetal accel. right? So would this be wrong?
 

Answers and Replies

PhanthomJay
Science Advisor
Homework Helper
Gold Member
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1. Homework Statement
It has been suggested that rotating cylinders about 10 mi long and 5 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration?


2. Homework Equations
I would say:
W=(delta)theta/(delta)time
a_centripetal=v^2/r or rW^2


3. The Attempt at a Solution
I don't really get what it's asking. Free-fall acceleration is -9.8 m/s^2 right? So that means the centripetal acceleration has to equal -9.8?

I converted the radius of 5mi to meters and got 8047m.

If what I state above is true, then 9.8=8047*W^2 and the angular speed would equal .0349 rad/s?

Despite what I posted above, I think that you need to convert 9.8 (a tangential acceleration) to centripetal acceleration. So would I have to plug 9.8 for in a in this equation to get the angular accel? a=r(alpha) Angular accel is not the same as centripetal accel. right? So would this be wrong?
Except for the fact that the radius is 2.5 miles, not 5 miles, you got it right the first time. The problem means to say that the centripetal acceleration at the cylinders surface...which is pointed inward toward the center of the cylinder...is equal numerically to the free fall acceleration which points in toward the center of the earth. A person on the surface would feel exactly the same as if he or she were standing on earth. It does not mean to imply that the acceleration is tangential...in fact, with the cylinder assumed to be rotating at constant angular speed, there is no tangential acceleration and no angular acceleration, only a centripetal acceleration.
 
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Basically , in rotational mechanics we deal with centripetal and tangential accelerations , and the tangential velocity is related to angular vel. as V=RW , now if we want your cylinder to have the same 'g' as that on earth , we would like it to pull on a body towards the center , which is to be done by this very centripetal force , your calculations are right , just that I think you took diamteter to be the radius , no problem with units.
 

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