Minimum angular velocity for circular motion (pendulum) How can I show that there is a minimum angular velocity $\omega_{min}$, different from zero, such that if we chose an $\omega$ smaller than $\omega_{min}$, then it is not possible to have a circular motion as in the picture?

 A: If you solve the problem for the two forces the vertical and the horizontal force (which is required for the circular motion) you obtain the relation
$$\omega^2=\frac{g}{L\cos\theta}$$
Hence the minimum required $\omega$ is $\sqrt{g/l}$, the same as the angular frequency for motion of a bob in a plane.
A: Expanding the correct answer of @SAKhan:
Assume that the conical pendulum is rotating at an angle $\theta$ at an angular velocity $\omega$.
Note also that the radius of the circle is given by $$r=L\sin(\theta)$$
For the point mass to move in a horizontal circle, the total vertical force is zero:$$T\cos(\theta)=mg$$
The net horizontal force must supply the needed centripetal force:$$T\sin(\theta)=m\omega^2r=m\omega^2L\sin(\theta)$$Combining these two equations to eliminate $T$, we get:$$\omega^2=\frac{g}{L\cos(\theta)}$$
The maximum value of $\cos(\theta)$ is $1$, when $\theta=0$, so $$\omega_{minimum}=\sqrt{\frac{g}{L}}$$
Edited for clarity:
So, what does this math mean?
Assume for the sake of simplicity that the length, $L$ of the pendulum is $9.8$ meters,  Then the equation for $\omega$ reduces to$$\omega = \sqrt{\frac{1}{\cos(\theta)}}$$  Now, we repeatedly start the pendulum into circular motion, each time at a some different angle $\theta$. (This could take some fiddling!) For each of these set-ups, once we are sure that the pendulum is moving in a circle, we measure the angle $\theta$ and the angular velocity $\omega$.  This angular velocity can be measured by taking the period of the circular motion, and dividing it into $2\pi$.
If we took all the data and plot them, we would obtain graph that shows that the value of $\omega$ approaches $1$ as $\theta$ approaches $0$.  Strictly speaking, $\omega=1$ is a lower limit (rather than a minimum) that is approached asymptotically as the angle $\theta$ approaches zero.

The vertical axis is $\omega$, in radians, and  the horizontal axis is $\theta$ in degrees
A: There is such minimun agngular speed: You will always find an an angle that results in circular motion for any given angular speed. The angle is given by:
$$\cos \theta=\frac{g}{L \omega^2}$$
Update: I got to this expression by using the equations of motion:
$m\omega^2L \sin \theta=T\sin\theta$ 
and
$T\cos\theta=mg$
What this means is that the minimum speed is reached for $\theta=0$, where \omega_min=\sqrt{L/g}. Any other circular motion will require a larger angular velocity. Thus, if we give the pendulum a spees less that the minimum, it will not be able to undergo a circular motion and starts to oscilate. 
