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Problem 3.79: Centrifuge

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kser
#1
Sep18-09, 08:00 PM
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
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rock.freak667
#2
Sep18-09, 08:48 PM
HW Helper
P: 6,202
Quote Quote by kser View Post

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"
kser
#3
Sep19-09, 01:06 PM
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?

rock.freak667
#4
Sep19-09, 01:32 PM
HW Helper
P: 6,202
Problem 3.79: Centrifuge

Quote Quote by kser View Post
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
kser
#5
Sep20-09, 06:21 PM
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....
kser
#6
Sep22-09, 09:41 AM
P: 6
nevermind! got it. thanks so much!


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