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Centrifuge on Mercury

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A laboratory centrifuge on earth makes n rpm (rev/min) and produces an acceleration of 3.70 g at its outer end.

    a.) What is the acceleration g's at a point halfway out to the end?

    2. Relevant equations

    a = (V^2)/R

    3. The attempt at a solution

    I really don't know what to do. I thought that if you divide the radius by 2, the acceleration would double, according to the equation. That doesn't seem to be the case since the answer 7.40 is incorrect. Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 24, 2010 #2
    Do you remember how to calculate linear velocity from angular velocity? If radius changes, but angular velocity is constant, does linear velocity change?
     
  4. Jan 24, 2010 #3
    Uhhh no I don't know how to calculate the linear velocity. I'm guessing the linear velocity changes if the radius changes eh? Only formula I'm familiar with is the a = V^2/R, and the other variation.
     
  5. Jan 24, 2010 #4
    w * r = v

    because w = radians / second

    What is archlength? radians * radius (for example, 2pi radians (360 degrees) * r = circumference , the entire arch length)
    so we get archlength/ second, which is a velocity along the rim of a circle.

    The fact that v changes with r makes sense too: if you were going 1 revolution(360 degrees) a second, and you were right next to the pivot, you'd travel less distance per second. If you were 70 thousand miles away from the pivot going around every second, you'd be traveling VERY fast.
     
  6. Jan 24, 2010 #5
    Ah okay, thanks. So that would mean since the radius is halved, the linear velocity would be halved as well? If that's the case, then I would assume that the acceleration is quadrupled since both velocity and radius are being halved?
     
  7. Jan 24, 2010 #6
    If a = V^2/ r = w^2 r^2 / r = W^2 r
    then changing r by a factor of 1/2 would half the acceleration.
     
  8. Jan 24, 2010 #7
    Haha wow I got confused there for a sec. I see how it works now, thanks a lot !
     
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