after HOW many REVOLUTIONS per SECOND should the centrifuge be operated?

Question

A sample of blood is placed in a centrifuge of radius 12.0 cm. The mass of a red blood cell is 3.0X10^-16 kg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 X10^-11 N. At how many revolutions per second should the centrifuge be operated?

Answer 1

Your answer is 167.764 revolutions/second.


You want the red blood cells to travel in a circle of radius 12.0 cm. That lets all the lighter plasma travel in a circle of a lesser radius. Thus, you want the magnitude of the centripetal force (see the source below for explanation of this formula) to be m*v*v/r, where m is the mass of the red blood cell, v is the (translational) velocity of the red blood cell, and r is the radius of the centrifuge. Now, you want this force (m*v*v/r) to be equal to 4.0x10^-11 Newtons. That way the blood will separate from the plasma. So your goal is to solve m*v*v/r=4.0x10^-11 Newtons for v. Then you can convert v to revolutions per second.

First, solve for v. You know m=3.0x10^-16kg. You know r=12.0cm=0.120 m. So you want:

(3.0x10^-16 kg)*v*v/(0.12 m) = 4.0x10^-11 N

Multiply both sides by 0.12 m and divide both sides by 3.0x10^-16 kg. You get:

v*v = 16000 (m/s)^2

Take the square root of both sides to get:

v = 126.491 m/s

Next, you want to convert this into revolutions per second. You know that the radius is 12 cm=0.120 m. That means that the circumference of the centrifuge (or the length of one revolution) is 2*pi*r, which is 2*pi*0.120m=0.754m. So the revolutions per second that you need is:

v/(2*pi*r) = (126.491 m/s)/(0.754 m) = 167.764 revolutions/second.

And that's your answer. Notice that:

m*((167.764 rev/sec)*2*pi*r)^2/r = 4.0x10^-11 Newtons

That's what you want.


So, again, your answer is 167.764 revolutions/second.