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)

$a = \frac{v^2}{r}$

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:

$v = 2\pi r\cdot \frac{rpm}{60} = \omega r$

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

$v = \omega^2 r = (2\pi\cdot\frac{rpm}{60s})^2\cdot r = (2\pi\cdot\frac{4100}{60s})^2\cdot 0.1m = 18434.2 \frac{m}{s^2}$

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