Calculate new RPM after change in g-force in a centrifuge

[EFB]

Homework Statement

A laboratory centrifuge on earth makes n rpm (rev/min) and produces an acceleration of 6.50 g at its outer end.

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

Relevant Equations

Maybe: A=RPM∗pi∗2/60∗r
where A = "acceleration from RPM", r = "radius"

Supposed answer = 0.482n

 

[berkeman]

Welcome to PF.

Relevant Equations:: RCF = (RPM)^2 * 1,1118*10^(-5)*r

What is RCF? And where did you get this formula? And what are the units of each quantity in that formula?

EDIT -- It appears the OP has edited his post to change his Relevant Equation. Whatever.

@EFB -- Instead of editing your original post (OP) after it has been replied to, please just add a new reply that updates what you were wanting to post. That can eliminate confusion by your first responding users. Thanks.

 

[haruspex]

Relevant Equations:: Maybe: A=RPM∗pi∗2/60∗r

Certainly RPM∗pi∗2/60 is the rotation rate converted to radians per second, which is useful. But multiplying that by a distance, r, gives something with the dimension length/time. That is not an acceleration.
What sort of acceleration does a centrifuge produce (the clue is in the name)?
What is the formula for that sort of acceleration?
What direction does it act in?

 

[berkeman]

And why does the local value of g matter for centrifuge operation?

 

[hutchphd]

Is that the exact statement of the problem? The term g is used to represent two different things. Good grief.

 

[haruspex]

Is that the exact statement of the problem? The term g is used to represent two different things. Good grief.

Good point. So which one is meant in "How many rpm … should it make to produce 4 g at its outer end?"?

 

[jbriggs444]

And why does the local value of g matter for centrifuge operation?

A centrifuge rotating at 0 RPM on earth produces a g force of 1 gee.

A centrifuge producing a resultant of 6.5 gee on earth must be generating some lesser amount of centripetal acceleration.

 

[EFB]

Homework Statement:: A laboratory centrifuge on earth makes n rpm (rev/min) and produces an acceleration of 6.50 g at its outer end.

This centrifuge is now used in a space capsule on the planet Mercury, where g is 0.378 what it is on earth. How many rpm (in terms of n ) should it make to produce 4 g at its outer end?
Relevant Equations:: Maybe: A=RPM∗pi∗2/60∗r
where A = "acceleration from RPM", r = "radius"

Supposed answer = 0.482n

Edit: g_Mercury = 0.378 * g_Earth
Question: How many rpm(in terms of n) should it make to produce 4 g_Mercury at its outer end?

 

[haruspex]

Edit: g_Mercury = 0.378 * g_Earth
Question: How many rpm(in terms of n) should it make to produce 4 g_Mercury at its outer end?

That's clearer. So please post an attempt. You could start by trying to answer my questions in post #3.

 

[EFB]

Certainly RPM∗pi∗2/60 is the rotation rate converted to radians per second, which is useful. But multiplying that by a distance, r, gives something with the dimension length/time. That is not an acceleration.
What sort of acceleration does a centrifuge produce (the clue is in the name)?
What is the formula for that sort of acceleration?
What direction does it act in?

a_centrifugal = v²/r
The direction of a_centrifugal is straight outwards from the edge of the circular motion.

 

[EFB]

That's clearer. So please post an attempt. You could start by trying to answer my questions in post #3.

Since a_c = v² and r is constant, a_c changes according with v.
v is represented by n rpm.
n rpm => a_c,Earth/0.378 g_Mercury
Wanted: k, where k is constant such that k * n => a_c = 4 g_Mercury:
k * n => 4/(a_c,Earth/0.378)
Since k * n rpm is the velocity and a is altered according with v:
k = √(4/(a_c,Earth/0.378))
Ans: 0.482*n rpm
I get the right answer, but do not fully understand what I have done. I have tried to write down the solution as I best understand it.

 

[haruspex]

Since a_c = v² and r is constant, a_c changes according with v.
v is represented by n rpm.
n rpm => a_c,Earth/0.378 g_Mercury
Wanted: k, where k is constant such that k * n => a_c = 4 g_Mercury:
k * n => 4/(a_c,Earth/0.378)
Since k * n rpm is the velocity and a is altered according with v:
k = √(4/(a_c,Earth/0.378))
Ans: 0.482*n rpm
I get the right answer, but do not fully understand what I have done. I have tried to write down the solution as I best understand it.

Your working is very hard to follow. Mistakes like leaving out an "r" and omitting a power of 2 in:
"Since a_c = v²/r and r is constant, a_c changes according with v²"
don't help. And you keep using"=>" in ways that are not mathematically defined. Not sure how to read them.
If you can write it out in a way that others can follow, you'll probably understand it better yourself.

 

[Lnewqban]

...
I get the right answer, but do not fully understand what I have done. I have tried to write down the solution as I best understand it.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/6-3-centripetal-force/