Spherical pendulum

Lagrangian mechanics

Main article: Lagrangian mechanics

Routinely, in order to write down the kinetic T=\tfrac{1}{2}mv^2 and potential V parts of the Lagrangian L=T-V in arbitrary generalized coordinates the position of the mass is expressed along Cartesian axes. Here, following the conventions shown in the diagram, :x=l\sin\theta\cos\phi :y=l\sin\theta\sin\phi :z=l(1-\cos\theta). Next, time derivatives of these coordinates are taken, to obtain velocities along the axes :\dot x=l\cos\theta\cos\phi,\dot\theta-l\sin\theta\sin\phi,\dot\phi :\dot y=l\cos\theta\sin\phi,\dot\theta+l\sin\theta\cos\phi,\dot\phi :\dot z=l\sin\theta,\dot\theta. Thus,

: v^2=\dot x ^2+\dot y ^2+\dot z ^2 =l^2\left(\dot\theta ^2+\sin^2\theta,\dot\phi ^2\right)

and

: T=\tfrac{1}{2}mv^2 =\tfrac{1}{2}ml^2\left(\dot\theta ^2+\sin^2\theta,\dot\phi ^2\right) : V=mg,z=mg,l(1-\cos\theta)

The Lagrangian, with constant parts removed, is

: L=\frac{1}{2} ml^2\left( \dot{\theta}^2+\sin^2\theta\ \dot{\phi}^2 \right)

• mgl\cos\theta.

The Euler–Lagrange equation involving the polar angle \theta

: \frac{d}{dt}\frac{\partial}{\partial\dot\theta}L-\frac{\partial}{\partial\theta}L=0 gives : \frac{d}{dt} \left(ml^2\dot{\theta} \right) -ml^2\sin\theta\cdot\cos\theta,\dot{\phi}^2+ mgl\sin\theta =0 and : \ddot\theta=\sin\theta\cos\theta\dot\phi ^2-\frac{g}{l}\sin\theta When \dot\phi=0 the equation reduces to the differential equation for the motion of a simple gravity pendulum.

Similarly, the Euler–Lagrange equation involving the azimuth \phi,

: \frac{d}{dt}\frac{\partial}{\partial\dot\phi}L-\frac{\partial}{\partial\phi}L=0 gives : \frac{d}{dt} \left( ml^2\sin^2\theta \cdot \dot{\phi} \right) =0 . The last equation shows that angular momentum around the vertical axis, |\mathbf L_z| = l\sin\theta \times ml\sin\theta,\dot\phi is conserved. The factor ml^2\sin^2\theta will play a role in the Hamiltonian formulation below.

The second order differential equation determining the evolution of \phi is thus :\ddot\phi,\sin\theta = -2,\dot\theta,\dot{\phi},\cos\theta.

The azimuth \phi, being absent from the Lagrangian, is a cyclic coordinate, which implies that its conjugate momentum is a constant of motion.

The conical pendulum refers to the special solutions where \dot\theta=0 and \dot\phi is a constant not depending on time.

Hamiltonian mechanics

Main article: Hamiltonian mechanics

The Hamiltonian is

:H=P_\theta\dot \theta + P_\phi\dot \phi-L

where conjugate momenta are

:P_\theta=\frac{\partial L}{\partial \dot \theta}=ml^2\cdot \dot \theta

and

:P_\phi=\frac{\partial L}{\partial \dot \phi} = ml^2 \sin^2! \theta \cdot \dot \phi.

In terms of coordinates and momenta it reads

H = \underbrace{\left[\frac{1}{2}ml^2\dot\theta^2 + \frac{1}{2}ml^2\sin^2\theta\dot \phi^2\right]}{T} + \underbrace{ \bigg[-mgl\cos\theta\bigg]}{V}= {P_\theta^2\over 2ml^2}+{P_\phi^2\over 2ml^2\sin^2\theta}-mgl\cos\theta

Hamilton's equations will give time evolution of coordinates and momenta in four first-order differential equations

:\dot {\theta}={P_\theta \over ml^2} :\dot {\phi}={P_\phi \over ml^2\sin^2\theta} :\dot {P_\theta}={P_\phi^2\over ml^2\sin^3\theta}\cos\theta-mgl\sin\theta :\dot {P_\phi}=0

Momentum P_\phi is a constant of motion. That is a consequence of the rotational symmetry of the system around the vertical axis.

Trajectory

Trajectory of the mass on the sphere can be obtained from the expression for the total energy

:E=\underbrace{\left[\frac{1}{2}ml^2\dot\theta^2 + \frac{1}{2}ml^2\sin^2\theta \dot \phi^2\right]}{T}+\underbrace{ \bigg[-mgl\cos\theta\bigg]}{V}

by noting that the horizontal component of angular momentum L_z = ml^2\sin^2!\theta ,\dot\phi is a constant of motion, independent of time. This is true because neither gravity nor the reaction from the sphere act in directions that would affect this component of angular momentum.

Hence

:E=\frac{1}{2}ml^2\dot\theta^2 + \frac{1}{2}\frac{L_z^2}{ml^2\sin^2\theta}-mgl\cos\theta :\left(\frac{d\theta}{dt}\right)^2=\frac{2}{ml^2}\left[E-\frac{1}{2}\frac{L_z^2}{ml^2\sin^2\theta}+mgl\cos\theta\right]

which leads to an elliptic integral of the first kind for \theta

:t(\theta)=\sqrt{\tfrac{1}{2}ml^2}\int\left[E-\frac{1}{2}\frac{L_z^2}{ml^2\sin^2\theta}+mgl\cos\theta\right]^{-\frac{1}{2}},d\theta

and an elliptic integral of the third kind for \phi

:\phi(\theta)=\frac{L_z}{l\sqrt{2m}}\int\sin^{-2}\theta \left[E-\frac{1}{2}\frac{L_z^2}{ml^2\sin^2\theta}+mgl\cos\theta\right]^{-\frac{1}{2}},d\theta.

The angle \theta lies between two circles of latitude, where

:E\frac{1}{2}\frac{L_z^2}{ml^2\sin^2\theta}-mgl\cos\theta.

References

References

1. Landau, Lev Davidovich. (1976). "Course of Theoretical Physics: Volume 1 Mechanics". Butterworth-Heinenann.