Answer
$v=1.2\times 10^{-5}m/s$
Work Step by Step
We can find the required tangential speed as follows:
$\omega=\frac{1rev}{12h}$
$\omega=(\frac{1rev}{12h})(\frac{2\pi rad}{rev})(\frac{1h}{3600s})$
$\implies \omega=\frac{\pi}{21600}\frac{rad}{s}$
Now $v=r\omega$
We plug in the known values to obtain:
$v=(8.2cm)(\frac{\pi}{21600}rad/s)$
This simplifies to:
$v=1.2\times 10^{-5}m/s$
