# Problem 3.79: Centrifuge

by kser
Tags: acceleration, centrifuge, centripetal, circular motion, mercury
 P: 6 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 3.40 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? 2. Relevant equations a = v2/R 3. 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
HW Helper
P: 6,189
 Quote by kser 2. Relevant 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"
 P: 6 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?
HW Helper
P: 6,189

## Problem 3.79: Centrifuge

 Quote by kser 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
 P: 6 yup. i'm sorry i just dont 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....
 P: 6 nevermind! got it. thanks so much!

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