Answer
(a) $T = \frac{mgL}{\sqrt{L^2-r^2}}$
(b) $\omega = \sqrt{\frac{T}{m~L}}$
(c) $T = 5.0~N$
$\omega = 30.2~rpm$
Work Step by Step
Let $\theta$ be the angle that the string makes with the vertical. Note that $cos(\theta) = \frac{\sqrt{L^2-r^2}}{L}$
(a) We can find an expression for the tension $T$ in the string. The vertical component of the tension is equal to the ball's weight.
$T~cos(\theta) = mg$
$T~(\frac{\sqrt{L^2-r^2}}{L}) = mg$
$T = \frac{mgL}{\sqrt{L^2-r^2}}$
(b) The horizontal component of the tension provides the centripetal force to keep the ball moving in a circle.
$T~sin(\theta) = m~\omega^2 ~r$
$T~(\frac{r}{L}) = m~\omega^2 ~r$
$\omega^2 = \frac{T}{m~L}$
$\omega = \sqrt{\frac{T}{m~L}}$
(c) We can find the tension in the string.
$T = \frac{mgL}{\sqrt{L^2-r^2}}$
$T = \frac{(0.500~kg)(9.80~m/s^2)(1.0~m)}{\sqrt{(1.0~m)^2-(0.20~m)^2}}$
$T = 5.0~N$
We can find the angular speed.
$\omega = \sqrt{\frac{T}{m~L}}$
$\omega = \sqrt{\frac{5.0~N}{(0.500~kg)(1.0~m)}}$
$\omega = 3.16~rad/s$
We can convert the angular speed to units of rpm.
$\omega = (3.16~rad/s)(\frac{1~rev}{2\pi ~rad})(\frac{60~s}{1~min})$
$\omega = 30.2~rpm$
