Centrifuge Acceleration Calculations

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The discussion revolves around calculating centrifuge acceleration in two scenarios: on Earth and Mercury. For Part A, the acceleration halfway to the end of the centrifuge is questioned, with some confusion about whether it would be double or half of the outer acceleration of 3.40 g. In Part B, participants discuss how to determine the required rpm on Mercury to achieve 4 g, considering Mercury's lower gravitational pull. The conversation highlights the need to use the correct equations for centripetal acceleration and the importance of understanding how radius affects acceleration. Overall, the participants seek clarity on applying the formulas correctly to solve the problems.
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Homework Statement


A laboratory centrifuge on Earth makes n rpm (rev/min) and produces an acceleration of 3.40 g at its outer end.

Part A: What is the acceleration (in g's) at a point halfway out to the end?

Part B: This centrifuge is now used in a space capsule on the planet Mercury, where gMercury is 0.378 what it is on earth. How many rpm (in terms of n) should it make to produce 4 gMercury at its outer end?

Homework Equations



a = v2/R

The Attempt at a Solution



Well using the equation for centripetal acceleration, I figure if the Radius is half then the acceleration in g's would be double for part A. Is that right? or would it be half?

For part B I'm not exactly sure what to use or do. What kind of equation could I use?

Any help would be great! Thanks so much, really thank you
 
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kser said:

Homework Equations



a = v2/R

since we are given rpm or 'ω', let's use a=ω2r instead.


So you know that at the outer end a= 3.40g=ω2r. Our ω in this case is n

so n2r=3.40g


Now halfway to the end is r/2 and ω is the same so we get now:

a1=n2(r/2) → (n2r)/2=a1

try dividing the two equations in red and get a1/3.40g = "something"
 
what do you mean by "dividing the two equations"? combine them? so a/3.40g = ((n^2r)/2))/n^2r ? I'm sorry, am i just looking into this way too much? I don't get what you're saying. because the way i see it. n is constant so when you half the radius, that also halves the acceleration right?
 
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kser said:
what do you mean by "dividing the two equations"? combine them? so a/3.40g = ((n^2r)/2))/n^2r ? I'm sorry, am i just looking into this way too much? I don't get what you're saying. because the way i see it. n is constant so when you half the radius, that also halves the acceleration right?

yes that is what I meant by divide.

For the second part, the 'r' is the same at the outer end. So in terms of 'n' find the r using what happens on Earth. The use amercury=N2r
 
yup. I'm sorry i just don't get it. ugh. this whole mercury part it just not makin sense.

so i find r in terms of n and get: r = 3.4/n^2 <--- is that even right? can't be because then I don't see how that would give me what I'm looking for. i am just not good at this stuf...
 
nevermind! got it. thanks so much!
 
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