Routinely, in order to write down the kinetic ${\displaystyle T={\tfrac {1}{2}}mv^{2}}$  and potential ${\displaystyle V}$  parts of the Lagrangian ${\displaystyle L=T-V}$  in arbitrary generalized coordinates the position of the mass is expressed along Cartesian axes. Here, following the conventions shown in the diagram,

${\displaystyle x=l\sin \theta \cos \phi }$ 

${\displaystyle y=l\sin \theta \sin \phi }$ 

${\displaystyle z=l(1-\cos \theta )}$ .

Next, time derivatives of these coordinates are taken, to obtain velocities along the axes

${\displaystyle {\dot {x}}=l\cos \theta \cos \phi \,{\dot {\theta }}-l\sin \theta \sin \phi \,{\dot {\phi }}}$ 

${\displaystyle {\dot {y}}=l\cos \theta \sin \phi \,{\dot {\theta }}+l\sin \theta \cos \phi \,{\dot {\phi }}}$ 

${\displaystyle {\dot {z}}=l\sin \theta \,{\dot {\theta }}}$ .

Thus,

${\displaystyle v^{2}={\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}=l^{2}\left({\dot {\theta }}^{2}+\sin ^{2}\theta \,{\dot {\phi }}^{2}\right)}$ 

and

${\displaystyle T={\tfrac {1}{2}}mv^{2}={\tfrac {1}{2}}ml^{2}\left({\dot {\theta }}^{2}+\sin ^{2}\theta \,{\dot {\phi }}^{2}\right)}$ 

${\displaystyle V=mg\,z=mg\,l(1-\cos \theta )}$ 

The Lagrangian, with constant parts removed, is^[1]

${\displaystyle L={\frac {1}{2}}ml^{2}\left({\dot {\theta }}^{2}+\sin ^{2}\theta \ {\dot {\phi }}^{2}\right)+mgl\cos \theta .}$ 

The Euler–Lagrange equation involving the polar angle ${\displaystyle \theta }$ 

${\displaystyle {\frac {d}{dt}}{\frac {\partial }{\partial {\dot {\theta }}}}L-{\frac {\partial }{\partial \theta }}L=0}$ 

gives

${\displaystyle {\frac {d}{dt}}\left(ml^{2}{\dot {\theta }}\right)-ml^{2}\sin \theta \cdot \cos \theta \,{\dot {\phi }}^{2}+mgl\sin \theta =0}$ 

and

${\displaystyle {\ddot {\theta }}=\sin \theta \cos \theta {\dot {\phi }}^{2}-{\frac {g}{l}}\sin \theta }$ 

When ${\displaystyle {\dot {\phi }}=0}$  the equation reduces to the differential equation for the motion of a simple gravity pendulum.

Similarly, the Euler–Lagrange equation involving the azimuth ${\displaystyle \phi }$ ,

${\displaystyle {\frac {d}{dt}}{\frac {\partial }{\partial {\dot {\phi }}}}L-{\frac {\partial }{\partial \phi }}L=0}$ 

gives

${\displaystyle {\frac {d}{dt}}\left(ml^{2}\sin ^{2}\theta \cdot {\dot {\phi }}\right)=0}$ .

The last equation shows that angular momentum around the vertical axis, ${\displaystyle |\mathbf {L} _{z}|=l\sin \theta \times ml\sin \theta \,{\dot {\phi }}}$  is conserved. The factor ${\displaystyle ml^{2}\sin ^{2}\theta }$  will play a role in the Hamiltonian formulation below.

The second order differential equation determining the evolution of ${\displaystyle \phi }$  is thus

${\displaystyle {\ddot {\phi }}\,\sin \theta =-2\,{\dot {\theta }}\,{\dot {\phi }}\,\cos \theta }$ .

The azimuth ${\displaystyle \phi }$ , 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 ${\displaystyle {\dot {\theta }}=0}$  and ${\displaystyle {\dot {\phi }}}$  is a constant not depending on time.

The Hamiltonian is

${\displaystyle H=P_{\theta }{\dot {\theta }}+P_{\phi }{\dot {\phi }}-L}$ 

where conjugate momenta are

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

and

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

In terms of coordinates and momenta it reads

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

${\displaystyle {\dot {\theta }}={P_{\theta } \over ml^{2}}}$ 

${\displaystyle {\dot {\phi }}={P_{\phi } \over ml^{2}\sin ^{2}\theta }}$ 

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

${\displaystyle {\dot {P_{\phi }}}=0}$ 

Momentum ${\displaystyle P_{\phi }}$  is a constant of motion. That is a consequence of the rotational symmetry of the system around the vertical axis.^{[dubious – discuss]}

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

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

by noting that the horizontal component of angular momentum ${\displaystyle L_{z}=ml^{2}\sin ^{2}\!\theta \,{\dot {\phi }}}$  is a constant of motion, independent of time.^[1] This is true because neither gravity nor the reaction from the sphere act in directions that would affect this component of angular momentum.

Hence

${\displaystyle E={\frac {1}{2}}ml^{2}{\dot {\theta }}^{2}+{\frac {1}{2}}{\frac {L_{z}^{2}}{ml^{2}\sin ^{2}\theta }}-mgl\cos \theta }$ 

${\displaystyle \left({\frac {d\theta }{dt}}\right)^{2}={\frac {2}{ml^{2}}}\left[E-{\frac {1}{2}}{\frac {L_{z}^{2}}{ml^{2}\sin ^{2}\theta }}+mgl\cos \theta \right]}$ 

which leads to an elliptic integral of the first kind^[1] for ${\displaystyle \theta }$ 

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

and an elliptic integral of the third kind for ${\displaystyle \phi }$ 

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

The angle ${\displaystyle \theta }$  lies between two circles of latitude,^[1] where

${\displaystyle E>{\frac {1}{2}}{\frac {L_{z}^{2}}{ml^{2}\sin ^{2}\theta }}-mgl\cos \theta }$ .

• Foucault pendulum
• Conical pendulum
• Newton's three laws of motion
• Pendulum
• Pendulum (mathematics)
• Routhian mechanics

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