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Revs per min from metres per second

by Taryn
Tags: metres, revs
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Taryn
#1
Jun6-06, 04:14 AM
P: 66
okay so here is the problem and here is wat I did!
I am just tryin to study up for exams now... and this is one of the problems!

A sample of blood is placed in a centrifuge of radius 17.5 cm. The mass of a red corpuscle is 3.0010-16 kg, and the magnitude of the force required to make it settle out of the plasma is 4.0810-11 N. At how many revolutions per second should the centrifuge be operated?

Basically wat I did is found the velocity in metres per second first which I find is 154m/s which I found by usin F=(mv^2)/r
But here is the simple problem that I am confused about... how do I change this to revs/sec. Is it somethin to do with C=2PIr, thats all I can think of!
Thanks for your help!
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Hootenanny
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Jun6-06, 04:29 AM
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Quote Quote by Taryn
Basically wat I did is found the velocity in metres per second first which I find is 154m/s which I found by usin F=(mv^2)/r
But here is the simple problem that I am confused about... how do I change this to revs/sec. Is it somethin to do with C=2PIr, thats all I can think of!
Thanks for your help!
You could do it that way, however it is easier to remember that if we resolve newton's second law radially we achieve;

[tex]F = m\alpha[/tex]

Where alpha is centripetal acceleration and [itex]\alpha = r\omega^2[/itex] where omega is angular velocity (rads/s). Thus;

[tex]\fbox{F = mr\omega^2}[/tex]

You method is completely valid, but is a bit long winded . You can convert radians per second into revolutions per second by dividing by [itex]2\pi[/itex].
Taryn
#3
Jun6-06, 04:46 AM
P: 66
okay I did that and now I am so far off... I did 123686.127rev/sec.
so wat I did was w^2=f/mr
but then I got that in rads per second =7.77E5 which is bigger then I expected.
So I then divided by 2*PI... then it was wrong... the answer is meant to be 140.3

Hootenanny
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Jun6-06, 04:51 AM
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Revs per min from metres per second

Quote Quote by Taryn
okay I did that and now I am so far off... I did 123686.127rev/sec.
so wat I did was w^2=f/mr
but then I got that in rads per second =7.77E5 which is bigger then I expected.
So I then divided by 2*PI... then it was wrong... the answer is meant to be 140.3
You forgot to square root the 7.77x105. If you square root that value then divide by [itex]2\pi[/itex] you sould obtain the correct answer.
Taryn
#5
Jun6-06, 05:02 AM
P: 66
thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P
Hootenanny
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Jun6-06, 05:09 AM
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Quote Quote by Taryn
thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P
No problem
Andrew Mason
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Jun6-06, 08:31 AM
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Quote Quote by Hootenanny
You could do it that way, however it is easier to remember that if we resolve newton's second law radially we achieve;

[tex]F = m\alpha[/tex]

Where alpha is angular acceleration and [itex]\alpha = r\omega^2[/itex] where omega is angular velocity (rads/s). Thus;

[tex]\fbox{F = mr\omega^2}[/tex]
I am sure you meant to say that the quantity [itex]r\omega^2 = v^2/r[/itex] is the centripetal acceleration not angular acceleration.

AM
Hootenanny
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Jun6-06, 09:52 AM
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Quote Quote by Andrew Mason
I am sure you meant to say that the quantity [itex]r\omega^2 = v^2/r[/itex] is the centripetal acceleration not angular acceleration.

AM
Thank-you andrew, duly corrected.


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