#### Answer

The net force on the rotor is 64.5 N

#### Work Step by Step

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