revs per min from metres per second

[Taryn]

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 3.00×10-16 kg, and the magnitude of the force required to make it settle out of the plasma is 4.08×10-11 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!

[Hootenanny]

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;

F = m\alpha

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

\fbox{F = mr\omega^2}

You method is completely valid, but is a bit long winded :biggrin:. You can convert radians per second into revolutions per second by dividing by 2\pi.

[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

[Hootenanny]

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 2\pi you sould obtain the correct answer.

[Taryn]

thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P

[Hootenanny]

thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P

No problem :smile:

[Andrew Mason]

You could do it that way, however it is easier to remember that if we resolve newton's second law radially we achieve;

F = m\alpha

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

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

AM

[Hootenanny]

I am sure you meant to say that the quantity r\omega^2 = v^2/r is the centripetal acceleration not angular acceleration.

AM

Thank-you andrew, duly corrected.