Question

A typical laboratory centrifuge rotates at 4100 rpm . test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. part a what is the acceleration at the end of a test tube that is 10 cm from the axis of rotation? express your answer with the appropriate units.

Answer 1

The test tube will be subject to centripetal acceleration. This acceleration is given by the following formula

(accel.) = (tangential velocity)^2 / (radius)

`a = (v^2)/(r)`

The velocity of the probe at a distance of 10 cm from the center of the centrifuge, can be calculated using the circumference of the circle:

`v = 2\pi r\cdot (rpm)/(60) = \omega r`

where omega denotes the angular velocity (radians per second). So, combining both:

`v = \omega^2 r = (2\pi\cdot(rpm)/(60s))^2\cdot r = (2\pi\cdot(4100)/(60s))^2\cdot 0.1m = 18434.2 (m)/(s^2)`

The test tube is subjected to an acceleration of 18434 m/s^2!