Rotational motion and tension on strings

Question:
Consider a rod AB with a horizontal arm AC. The length of AC is $a$ and strings are attached from A and C to hang a mass at D. The lengths of the strings are $a$ as shown.
(1) Find the tension on the string AD.
(2) Rotate the rod AB and the mass obtains a constant circular velocity, $v$. There is no slack with string CD. Find the tensions on strings AD and CD.
(3) Find the velocity and angular velocity of the object if string CD does not get slack.

Answer:
(1) The tensions on AD and CD are equal and balanced with the gravitational force, $mg$. Thus we have the tension as follows:
\[
2T\cos 30^o = mg  \\
\rightarrow T = \frac{\sqrt{3}}{3}mg
\]

(2) The radius of rotation is $\frac{a}{2}$, so the centripetal force is $\frac{mv^2}{a/2}$. Then, write out the equations in $x$- and $y$-axes.
\begin{eqnarray*}

\mbox{For x, } \quad T_1\sin 30^o &=& T_2\sin 30^o + \frac{mv^2}{a/2}  \\
\mbox{For y, } \quad T_1\cos 30^o &+& T_2\cos 30^o = mg
\end{eqnarray*}
Solving the simultaneous equations, we have
\[
T_1 = m\left(\frac{\sqrt{3}g}{3}+\frac{2v^2}{a} \right)  \\
T_2 = m\left(\frac{\sqrt{3}g}{3}-\frac{2v^2}{a} \right)
\]

(3) If the string does not get slack, the tension must be equal to or greater than zero. Let $T_2 \geq 0$.
\[
m\left(\frac{\sqrt{3}g}{3}-\frac{2v^2}{a} \right) \geq 0  \\
\rightarrow \frac{\sqrt{3}g}{3}-\frac{2v^2}{a} \geq 0  \\
\rightarrow \frac{2v^2}{a} \leq \frac{\sqrt{3}g}{3}  \\
\rightarrow v \leq \sqrt{\frac{\sqrt{3}ga}{6}}
\]
Since angular velocity, $\omega$, is velocity divided by radius. we obtain
\[
\omega = \frac{v}{a/2} \leq \sqrt{\frac{2\sqrt{3}g}{3a}}
\]

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