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, that's 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, that's 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 . 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

 

[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.