## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We first convert 70,000 rpm to units of rad/s; $\omega = (70,000~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 7330.4~rad/s$ We then assume that the masses of the two samples are $m$ and $(m+10\times 10^{-6}~kg)$. We can find the magnitude of the net force on the rotor. Let $a_c$ be the centripetal acceleration. So; $\sum F = (m+10\times 10^{-6}~kg)~a_c-m~a_c$ $\sum F = (10\times 10^{-6}~kg)~a_c$ $\sum F = (10\times 10^{-6}~kg)(\omega^2~r)$ $\sum F = (10\times 10^{-6}~kg)(7330.4~rad/s)^2(0.12~m)$ $\sum F = 64.5~N$ The net force on the rotor is 64.5 N