What angular speed (in revolutions per minute) is needed for a centrifuge to produce an acceleration of 944 times the gravitational acceleration 9.8 m/s 2 at a radius of 7.02 cm ? Answer in units of rev/min.

3335.3r/m

Explanations:

For an object in the centrifuge it is assumed to be traveling at a constant circular speed. In this case there is no tangential acceleration on the object. It is only accelerated outwards at an acceleration given by v^2/r where v is the constant speed and r is the radius of the circle in which it is moving.

The required acceleration of the object in the centrifuge is 944 times that of the gravitational acceleration. This is equal to 944*9.8 = 9251.2 m/s^2.

The radius of the centrifuge is 7.02 cm = 7.02*10^-2 m.

To produce the required acceleration the linear velocity is v, where

a =v^2/(7.02*10^-2) = 9251.2m/s

=> v^2 = 9251.2*7.02*10^-2

v^2 = 649.43

v = √(649.43)

=> v = 25.484 m/s

The angular velocity in terms of revolutions per second is 25.484/(2*pi*7.02*10^-2) = 56.7 revolutions/second.

For revolutions/minutes

Since 60s - 1m

1s - 0.017m

So 56.7r/0.017m = 3335.3r/m

The centrifuge has to rotate at 3335.3 revolutions/minutes to achieve an acceleration equal to 944 times that of the gravitational acceleration.

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