Question

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 113000 g's? if the answer has 4 digits or more, enter it without commas,
e.g. 13500?

Answer 1

The acceleration experienced by the particle is given by
$a=113000 g=113000 \cdot 9.81 m/s^2$
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed $\omega$ of the particle and its distance r from the axis by the relationship
$a= \omega ^2 r$
In our problem, $r=6 cm=0.06 m$, so we can solve for $\omega$:
$\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s$
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to $( \frac{1}{2 \pi} )$ revolutions, while $1 s = \frac{1}{60} min$. So we the conversion is$\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm$