Why does this setup fix a specific angular velocity? 
First let me clarify that this is not a homework type question as I am not seeking any solution rather I am confused about the question itself.

Question:
Find the angular velocity of rotation.

My solution:
Let the angular velocity about the axis of rotation be $\omega$ and the radius of rotation be
$$
R=3+5 \sin 30^{\circ}=3+\frac{5}{2}=\frac{11}{2} m
$$

$$
\begin{array}{l}
\operatorname{Tcos} 30^{\circ}=\mathrm{mg} \\
T \sin 30^{\circ}=m R \omega^{2} \\
\Rightarrow \tan 30^{\circ}=\frac{R \omega^{2}}{g} \\
\Rightarrow \quad \omega=\sqrt{\frac{2 g}{11 \sqrt{3}}}
\end{array}
$$
According to the source of this question, my answer is correct but I am quite perplexed by the answer because the value of angle rotation comes out to be constant which means that the angular velocity of the swing is always constant with is obviously not true.
As per my intuition such type of swing (as shown in figure) can have different angular speeds, then why is $\omega$ constant?
 A: You got a constant value for the angular velocity because you found the one that gives you exactly an inclination angle equal to $30°$.
If you change the angular velocity the inclination angle will change.
Indeed, calling the inclination angle $\theta$, from your equations, you have that the value of $\theta$ as a function of the angular velocity is
\begin{equation}
\theta(\omega) = \arctan\left(\frac{R\omega^2}{g}\right)
\end{equation}
This makes sense because if the angular velocity tends to zero, we have that $\theta(0) = \arctan(0) = 0\,\mbox{rad}$, so the arm is vertical as expected.
On the other hand, if the angular velocity tends to infinity
$$\lim_{\omega\to\infty}\theta(\omega) = \lim_{\omega\to\infty}\arctan\left(\frac{R\omega^2}{g}\right) = \frac{\pi}{2}\,\mbox{rad}$$
and, therefore, the arm is horizontal (once again this makes sense).
A: 
you obtain the solution also from the equation of motion in rotating frame, which is
$$\ddot\vartheta=-{\frac {\cos \left( \vartheta  \right) {\omega}^{2} \left( r+l\sin
 \left( \vartheta  \right)  \right) }{l}}+{\frac {g\sin \left( 
\vartheta  \right) }{l}}
$$
for steady state is $~\ddot\vartheta=0$, solving for $~\omega^2~$
$$\omega^2={\frac {g\sin \left( \vartheta  \right) }{\cos \left( \vartheta 
 \right)  \left( r+l\sin \left( \vartheta  \right)  \right) }}
$$
hence your  solution is the steady state solution of the EOM.
Remarks :
you can't obtain analytical solution for  $~\vartheta= \vartheta(\omega)~$ (answer @Davide Dal Bosco) but for $\vartheta=\frac \pi2 ~,\cos \left( \vartheta  \right)  \left( r+l\sin \left( \vartheta 
 \right)  \right) 
=0~,\omega\mapsto\infty~$ and for $~\vartheta > \frac \pi2~,\omega~$ is imaginary, hence $~0\le \vartheta\ < \pm\frac \pi2$

your solution
$$\vartheta=\frac{\pi}{6}~,l=5~,r=3\\
\omega=\sqrt{\frac{2\,g}{11\,\sqrt{3}}}$$

A: The inclined portion of the chair suspension is not a rigid bar but a chain or cable.  At rest the chain is vertical. This is when people sit in the chairs.  Then the whole thing starts to spin with slowly increasing speed. As this happens the chain moves away from the vertical at slowly increasing angle. So,  for each value if the angular velocity there is an equilibrium value of the angle. When the motion stops the angle decreases again until the chain is again vertical and you can leave the sit. So,  the whole point of this ride  is that at the beginning the seat is close enough to the ground to get in and then you get lifted high above the ground while your feeling of up and down changes direction as well.  If you had rigidly attached rods you have just an ordinary carousel. The difference is that in this case the joints can produce a force perpendicular to the rod,  if necessary. The chain cannot.
A: When I was growing up in Miami, there was a swing like this in a small amusement park run by the PBA. The central shaft was motor driven and supported several seat swings around the perimeter. For a small fee, you (and others) would occupy a seat while the motor was off. Then it was turned on and you were brought up to a constant speed at a reasonable angle.  In the same park there was a disk (maybe 10 feet in diameter) with handles, mounted on a vertical axis at the center.  You could push to get it rotating, and then jump on (a frequent subject of physics problems).
A: It reminds me of another thought experience question that I was once asked in my 1st year: when a kid is swinging on a swing, how does their angular momentum changes, if it is a closed system and the net forces must be zero? The answer is of course that is is not a closed system - the swing is connected to the ground, and a net force is acted upon the hinges that connect the swing to the ground, which is compensated by the earth.
If I understand correctly, in your question the point is that the angle is fixed, yet in real life we know that these devices can accelerate and de-accelerate. The calculation you had is for the case where no net forces need to act on the base of the merry-go-round in the $y$-direction. In that angular velocity, the tension at the rod is exactly compensated by the gravity. For lower angular velocities (let's say when the person just mount the seat), the entire system of rod+swing+person would want to go-towards the earth, but the base of the system will supply the necessary force, that it gets from the earth. At higher angular velocities the entire system would want to lift upwards, but again the bolts that fix the base of the system to the ground will (hopefully!) give the necessary counter-force to keep the people nauseated but safe.
A: First of all the question didn't mention which axis it is talking about so the natural choice is the grounded pole. Now it is clear that the motion of the COM lf system consisting of the girl and seat is horizontally circular since the COM doesn't go up and down (it is assumed that the girl is fixed just like a statue) and of course, the thing is rotating. Hence the angular velocity will be constant. Now the FBD diagram you made is that of the system constant of the girl and the seat. You implicitly assumed that the chair is measles (and that is justified since it is not mentioned in the question as well). Now after that you find $T$ in terms of unknown variable $m$, then you equated the force in the horizontal direction to the centripetal force and found the angular velocity. This is the angular velocity of COM of the girl + seat system and any line perpendicular to the pole. Why? Assume that a line passes through the COM and is perpendicular to the pole, now imagine similar lines parallel to it and now imagine the lines to rotate, isn't it clear that these lines have the same angular velocity?
In these types of questions where the question is not stated clearly, one should make intelligent guesses about what the question is asking. Since your answer is correct you were right in you're guessing.
