1. Jan 25, 2006

Find the required angular velocity of an ultracentrifuge in $\frac{rev}{min}$ for the radial acceleration of a point 1.00 cm from the axis to equal 400,000g (that is, 400,000 times the acceleration of gravity.)
So $$a_{rad} = (\omega)^{2}r$$. $$400,000g = \omega^{2}(0.01 m)$$. Would I just solve for $$\omega$$? $$\omega$$ would be in m/s? Then to convert to rev/min, you use the fact that $$2\pi(0.01 m)$$ equals 1 revolution?
Would this be the correct way to solve this problem?
Thanks
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2. Jan 25, 2006

### andrevdh

when you solve for $\omega$ with your formula the units is $rad/s$ (if you worked in SI units), which needs to be converted to $rev/min$ now
$$2\pi\ rad=\ 1\ rev$$
therefore
$$1\ rad=\ \frac{1}{2\pi}\ rev$$
and since
$$60\ s=\ 1\ min$$
it follows that
$$1\ s=\ \frac{1}{60}\ min$$
therefore
$$\omega\ =\ \frac{1\ rad}{1\ s}\ =\ \frac{60\ rev}{2\pi\ min}\ =\ \frac{30}{\pi}\ rev/min$$