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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 Richard Fitzpatrick, The conical pendulum, equations 262–267.

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 $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe 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. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. 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.

A Foucault pendulum is a long simple pendulum whose suspension allows the bob to swing in both independent horizontal directions. Its slowly changing swing direction relative to the building reveals Earth's rotation. A hinge restricting the bob to one vertical plane would prevent this measurement: the support would impose the very direction we want the pendulum to select freely. Léon Foucault demonstrated the effect in 1851.^[ref]

The bob's back-and-forth oscillation and the swing direction's precession are different motions. The bob traverses its arc in seconds, while successive arcs change direction over hours. In a conical pendulum the bob itself circulates around the vertical; a nearly planar Foucault pendulum instead reverses along an almost straight horizontal trace, whose orientation changes slowly. The measured quantity is that orientation against a fixed floor or wall reference, rather than the bob's instantaneous azimuth.

A conducting conical pendulum connected to a grounded pivot develops a motional electric potential difference $B\omega r^2/2$ between the rotation axis and a bob at radius $r$. If the bob dominates the capacitance, its electric charge is $CB\omega r^2/2$. An additional uniform vertical electric field changes both the force on that charge and, while the grounding connection remains, the background electric potential entering the charging relation. These effects must be treated consistently when calculating a first-order mechanical response.

For fixed $\theta$, the Euler-Lagrange equation for the meridional coordinate requires ${\dot{\varphi }}^2\cos \theta =g/\ell$. Combining this with $v=\ell \sin \theta \dot{\varphi }$ and $\cos \theta =D/\ell$ givesThe resulting motion is a conical pendulum. The positive-depth condition is necessary for a nondegenerate constant-latitude orbit in gravity; the bottom point is the limiting zero-speed equilibrium.

The polar Euler-Lagrange equation isFor a constant angle with $0<{\theta }_0<\pi$, conservation of $h$ makes $\dot{\varphi }$ constant, say $\Omega$. The remaining equation isIf $g>0$, every nonvertical constant-angle motion is a conical pendulum motion:The rod sweeps out a cone about the downward vertical, in either azimuthal direction. Equivalently, $h^2=(3g/(2a)){\sin }^4{\theta }_0/\cos {\theta }_0$. The exactly downward and exactly upward vertical rods are also static solutions; azimuth has no physical meaning at either pole. No static tilted rod is possible under nonzero gravity.

If $g=0$, the condition is $\sin {\theta }_0\cos {\theta }_0{\Omega }^2=0$. Thus a rod may remain at rest in any orientation, or rotate uniformly in the horizontal plane with ${\theta }_0=\pi /2$ and arbitrary constant $\Omega$. The vertical cases are included among static orientations; changing only their azimuth does not move the rod.

Constant azimuth. If $\varphi$ is constant, $\dot{\varphi }=0$ and the motion is a planar simple pendulum:For the small-angle approximation, $\sin \theta \simeq \theta$, giving simple harmonic motion withThe harmonic angular frequency requires small amplitude. At finite libration amplitude ${\theta }_{\max }<\pi$, the exact oscillation period is $4\sqrt{l/g}K(\sin ({\theta }_{\max }/2))$, where $K$ is the complete elliptic integral of the first kind; the nonlinear pendulum does not have the same amplitude-independent period.

Constant polar angle. If $\theta ={\theta }_0$ is nonzero and constant, the bob traces a horizontal circle: this is a conical pendulum. Setting $\ddot{\theta }=0$ and dividing by $\sin {\theta }_0$ givesThe sign distinguishes the two senses of rotation. The exactly downward configuration ${\theta }_0=0$ is stationary; its azimuth is undefined and it must not be obtained by dividing by $\sin {\theta }_0$. A nontrivial constant-angle orbit of this gravity-only system needs $\cos {\theta }_0>0$ even though the supporting rod can also push.

Fixed-angular-momentum form. Eliminating $\dot{\varphi }$ using the conserved angular momentum $h$ givesEquivalently, the polar motion has effective potentialFor the steady conical pendulum at ${\theta }_0$, $\ddot{\theta }=0$, soThe sign of $h$ agrees with the sign of $\Omega$; substituting $h=m{l}^2{\sin }^2{\theta }_0\Omega$ verifies the constant-angle result in part (d).

Period of small polar oscillations. Put $\theta ={\theta }_0+\alpha (t)$, with $|\alpha |\ll 1$ and $h$ held fixed. This is the appropriate small nutation about a conical pendulum; keeping $\dot{\varphi }$ fixed instead would violate the angular momentum constraint during the oscillation. Ifthen $f({\theta }_0)=0$ andThe linearization is $\ddot{\alpha }+{\omega }_{\theta }^2\alpha =0$, givingSince $0<{\theta }_0<\pi /2$, the coefficient is positive and these small polar oscillations are stable. To first order, $\dot{\varphi }=\Omega (1-2\alpha \cot {\theta }_0)$, making the compensating change in azimuthal angular velocity explicit. The limit ${\omega }_{\theta }/|\Omega |\to 2$ as ${\theta }_0\to 0$ concerns radial modulation of a small nonzero circular orbit; the exactly planar $h=0$ pendulum is a different, degenerate limit of these polar coordinates.

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.

For fixed $\ell$ and $g$, the exact conical pendulum oscillation period satisfies $P/P_0=\sqrt{\cos \theta }$. In the small-angle approximation, with $\theta$ in radians,Thus the limiting revolution period equals the small-amplitude oscillation period of a planar simple pendulum. This equality has a geometric explanation: to first order, $x(t)=r\cos (\omega t)$ and $y(t)=r\sin (\omega t)$ are perpendicular simple harmonic motions of equal amplitude, a quarter period apart, with $\omega \to \sqrt{g/\ell }$. At finite angle, however, the conical pendulum's oscillation period decreases with $\theta$, whereas a planar simple pendulum's oscillation period increases with its angular amplitude.

The endpoint $\theta =0$ itself is a stationary bob with $r=0$ and no observable revolution; $P_0$ is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed $\ell$, nonzero conical pendulum motion also requires $\omega >\sqrt{g/\ell }$, because $\cos \theta =g/(\ell {\omega }^2)<1$. At the other endpoint, $\theta \to \pi /2$ gives $h\to 0$, $P\to 0$, and unbounded tension and speed. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite tension; a real experiment ceases to satisfy the ideal assumptions before that limit.

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure $\ell$ from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle $\theta$, the predicted initial speed is $v=\sqrt{g\ell {\sin }^2\theta /\cos \theta }$. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain $r$, or measure the vertical drop $h$ directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time $N$ complete returns to the same marker in the same direction and estimate $P=\Delta t/N$. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute $\sin \theta =r/\ell$ or $\cos \theta =h/\ell$, and check $r^2+h^2\simeq {\ell }^2$ if both were measured. Then compare the measured oscillation period with $P^2=4{\pi }^{2}h/g$. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same $\ell$. Larger $r$ means smaller $h=\sqrt{{\ell }^2-r^2}$, so the ideal oscillation period decreases. A plot of measured $P^2$ against independently measured $h$ should follow a straight line through the origin with slope $4{\pi }^2/g$; conversely, $g=4{\pi }^{2}h/P^2$ estimates gravitational acceleration. Do not infer $h$ from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume $\ell =1.00\mathrm{m}$, $m=0.050\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$, and $\theta =40^{\circ }$. The model gives $r\approx 0.643\mathrm{m}$, $h\approx 0.766\mathrm{m}$, $P\approx 1.756\mathrm{s}$, $v\approx 2.300m/s$, and $T\approx 0.640\mathrm{N}$. Its vertical tension component is $0.4905\mathrm{N}$, equal to the gravitational weight; its inward component is approximately $0.412\mathrm{N}$. These are predictions from chosen parameters, not measured data or measurements of local $g$. The next figure uses the same stipulated $\ell$ and $g$, including illustrative angles $15^{\circ }$, $40^{\circ }$, and $60^{\circ }$.

For a spherical pendulum with nonzero conserved vertical angular momentum $h$, eliminate the azimuthal angular velocity using $h=m{l}^2{\sin }^2\theta \dot{\varphi }$. The polar effective potential is $V_{\mathrm{eff}}=h^2/(2m{l}^2{\sin }^2\theta )-mgl\cos \theta$. A conical pendulum at $0<{\theta }_0<\pi /2$ is a minimum satisfying $h^2=m^{2}g{l}^3{\sin }^4{\theta }_0/\cos {\theta }_0$. Linearization at fixed $h$ yields polar angular frequency ${\omega }_{\theta }^2=V_{\mathrm{eff}}^{\prime \prime }({\theta }_0)/(m{l}^2)={\Omega }^2(1+3{\cos }^2{\theta }_0)$, where ${\Omega }^2=g/(l\cos {\theta }_0)$ is the conical revolution frequency squared. Holding $\dot{\varphi }$ fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar $h=0$ limit requires separate treatment.