#### Answer

(a) $\omega_{min} = \sqrt{\frac{g}{L}}$
(b) $\omega_{min} = 29.9~rpm$

#### Work Step by Step

(a) We can find the angular velocity $\omega_{min}$ such that the force of gravity is equal to the centripetal force at the top.
$m\omega_{min}^2~L = mg$
$\omega_{min} = \sqrt{\frac{g}{L}}$
(b) We can find $\omega_{min}$ in units of rad/s
$\omega_{min} = \sqrt{\frac{g}{L}}$
$\omega_{min} = \sqrt{\frac{9.80~m/s^2}{1.0~m}}$
$\omega_{min} = 3.13~rad/s$
We can convert $\omega_{min}$ to units of rpm.
$\omega_{min} = (3.13~rad/s)(\frac{1~rev}{2\pi~rad})(\frac{60~s}{1~min})$
$\omega_{min} = 29.9~rpm$