= Conical pendulum
{wiki}
A <conical pendulum> is a pendulum whose bob executes uniform horizontal <circular motion> beneath a fixed pivot, with its string at a constant <angle> $\theta$ to the downward vertical. The string sweeps a <right circular conical surface>; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a <point mass> $m>0$, a taut, massless, inextensible string of length $\ell>0$, uniform <gravitational acceleration> of magnitude $g>0$, and negligible <drag force> and pivot <friction>. A nonzero circular motion requires $0<\theta<\pi/2$. This geometry and the corresponding <forces> are described in https://farside.ph.utexas.edu/teaching/301/lectures/node88.html[Richard Fitzpatrick, The conical pendulum, equations 262–267].
\Image[/conical-pendulum-geometry.png]
{title=Conical pendulum geometry, horizontal orbit, and the two forces on the bob}
The <right triangle> formed by the pivot, the orbit centre, and the bob gives $r=\ell\sin\theta$ and $h=\ell\cos\theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for <tension>, $P$ for the <oscillation period> of one revolution, and $\omega=2\pi/P$ for the <angular speed>; then the bob's <speed> is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent <velocity>.
The two actual <forces> on the bob are <tension> toward the pivot and <gravitational weight> vertically downward. The vertical <acceleration> is zero, while the horizontal <centripetal acceleration> is $v^2/r=\omega^2r$. Thus <Newton's second law> gives
$$
T\cos\theta=mg,\qquad T\sin\theta=m\omega^2r=\frac{mv^2}{r}.
$$
The inward net <force> is the horizontal component of <tension>, rather than an extra interaction to add to these two <forces>. Constant <speed> still requires <acceleration> because the <velocity> changes direction. https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force[Ling, Sanny and Moebs, University Physics 1, §6.3: centripetal acceleration and net force]{ref}
Dividing the two component equations and using $r=\ell\sin\theta$ gives the useful collection of predictions
$$
\omega^2\cos\theta=\frac g\ell,\qquad T=\frac{mg}{\cos\theta},\qquad v^2=g\ell\frac{\sin^2\theta}{\cos\theta},\qquad P=2\pi\sqrt{\frac{\ell\cos\theta}{g}}=2\pi\sqrt{\frac hg}.
$$
If $\phi$ is the azimuthal <angle>, then $\dot\phi=\pm\omega$; either direction of rotation satisfies $\dot\phi^2\cos\theta=g/\ell$. The <mass> cancels from the <angular speed> and <oscillation period>, but not from the <tension>. The motion is a <relative equilibrium>: in a <rotating frame> following the bob at <angular speed> $\omega$, its <position> is fixed. This is a concrete way to see how rotational <symmetry> can turn an apparently moving solution into a stationary one.
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