Answer
The particle will experience an acceleration of 100,000 g's when the centrifuge rotates at a rate of 34,000 rpm.
Work Step by Step
First we can find the velocity of the particle:
$\frac{v^2}{r} = a$
$v = \sqrt{ra}$
$v = \sqrt{(0.080 ~m)(100,000)(9.8 ~m/s^2)}$
$v = 280~m/s$
We can use the velocity to find the number of revolutions each second. Let $z$ be the number of revolutions each second. Then,
$z = \frac{v}{2\pi r}$
$z = \frac{280 ~m/s}{(2\pi)(0.080 ~m)}$
$z = 560 ~rev/s$
$(560 ~rev/s)(\frac{60 ~s}{1 ~min}) = 34,000 ~rpm$
The particle will experience an acceleration of 100,000 g's when the centrifuge rotates at a rate of 34,000 rpm.