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!
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].
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 7.77x10^{5}. If you square root that value then divide by [itex]2\pi[/itex] you sould obtain the correct answer.
thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P
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