Swing a weight on a string so that it traces a horizontal circle. The string sweeps out the surface of a cone, which is why this is called a conical pendulum.
Only two forces act on the weight: the tension S along the string and gravity mg straight down. Nothing is added to this on account of the turning.
The weight circles at a constant height and does not move up or down, so the vertical direction is in balance. Only the tension can hold up gravity, and to do so it must have a vertical component. This is S\cos\theta = mg.
From here we can see that the string can never lie horizontal. At \theta = 90^\circ, \cos\theta is 0, the vertical component disappears, and there is nothing left to hold up gravity. However fast the weight is swung, the string stays at least slightly tilted.
The tension points at an angle, so no equation can be written with it as it stands. Split it into the direction that is in balance and the direction that is accelerating: vertical and horizontal.
The vertical component S\cos\theta balances gravity, as we have just seen. The resultant is 0, and that is why the weight does not move up or down.
The horizontal component S\sin\theta has nothing to cancel it, because gravity points straight down and has no horizontal component. So the whole of it survives as the resultant pointing at the centre: S\sin\theta = mr\omega^2.
Take care that what plays the part of the centripetal force is not the tension itself but its horizontal component. Writing S = mr\omega^2 would put the vertical component to work toward the centre as well.
There are now two equations, but r is still in them, so we rewrite it with the length of the string and the angle. The horizontal dimension in the figure is that r, the radius of the base of the cone, so r = L\sin\theta.
Divide S\sin\theta = mL\sin\theta\,\omega^2 by S\cos\theta = mg. The S goes on the left and the m on the right, leaving \tan\theta = \dfrac{L\sin\theta\,\omega^2}{g}. Since \tan\theta = \dfrac{\sin\theta}{\cos\theta}, the \sin\theta cancels too, and \omega^2 = \dfrac{g}{L\cos\theta}.
As a period this is T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}. The mass has gone, for the same reason as in the simple pendulum. The force pulling the weight, mg, and the reluctance of the weight to be moved, m, are both proportional to the same m, so they cancel.
What is left, L\cos\theta, is the depth from the pivot down to the plane of the orbit. It is marked on the axis in the figure. Call it h and the period is T = 2\pi\sqrt{\dfrac{h}{g}}, exactly the form of the simple pendulum's T = 2\pi\sqrt{\dfrac{L}{g}}. What fixes the period of a conical pendulum is not the length of the string but this depth.
Vary \theta in T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}. The flatter the string lies, the smaller \cos\theta becomes and the shorter the period, which is to say the faster the weight turns.
The tilt in the figure travels between a steep angle and a shallow one. The string has not changed length, yet the shallow one visibly turns faster. Read it as turning fast makes it lie flat rather than lying it flat to turn fast. Fix \theta and the rate of turning is fixed; fix the rate of turning and \theta is fixed.
The depth of the orbital plane, h = L\cos\theta, says the same thing. The faster the weight turns the shallower it circles, and the slower it turns the deeper. The string has not changed length; only the height of the plane has.
Taking \theta toward 90^\circ makes the period look as though it could be shortened without limit. But the tension is S = \dfrac{mg}{\cos\theta}, and that grows without limit instead. The string never goes horizontal partly because nothing would hold the weight up, and partly because the string would break first.