Spherical pendulum explained

In physics, a spherical pendulum is a higher dimensional analogue of the pendulum. It consists of a mass moving without friction on the surface of a sphere. The only forces acting on the mass are the reaction from the sphere and gravity.

Owing to the spherical geometry of the problem, spherical coordinates are used to describe the position of the mass in terms of, where is fixed such that

r=l

.

Lagrangian mechanics

See main article: Lagrangian mechanics.

Routinely, in order to write down the kinetic

T=\tfrac{1}{2}mv2

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
x=l\cos\theta\cos\phi
\theta-l\sin\theta\sin\phi\phi
y=l\cos\theta\sin\phi
\theta+l\sin\theta\cos\phi\phi
z=l\sin\theta\theta
.Thus,
2=x
v
2+y
2+z

2 =l

2\left(\theta

2+\sin

2\theta\phi

2\right)

and

T=\tfrac{1}{2}mv2 =\tfrac{1}{2}ml

2\left(\theta

2+\sin

2\theta\phi

2\right)

V=mgz=mgl(1-\cos\theta)

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

L=1
2

ml2\left(

\theta

2+\sin

2\theta\phi

2 \right) +mgl\cos\theta.

The Euler–Lagrange equation involving the polar angle

\theta

d
dt
\partialL-
\partial\theta
\partial
\partial\theta

L=0

gives
d
dt
2\theta
\left(ml
2\sin\theta\cos\theta\phi
\right) -ml

2+ mgl\sin\theta=0

and
\ddot\theta=\sin\theta\cos\theta\phi
2-g
l

\sin\theta

When
\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

,
d
dt
\partialL-
\partial\phi
\partial
\partial\phi

L=0

gives
d
dt

\left(ml2\sin2\theta

\phi

\right) =0

.The last equation shows that angular momentum around the vertical axis,

|Lz|=l\sin\theta x ml\sin\theta

\phi
is conserved. The factor

ml2\sin2\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

\theta\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

\theta=0
and
\phi
is a constant not depending on time.

Hamiltonian mechanics

See main article: Hamiltonian mechanics.

The Hamiltonian is

H=P
\theta\theta

+

P
\phi\phi-L

where conjugate momenta are

P
\theta=\partialL
\partial
\theta

=ml2 ⋅

\theta

and

P
\phi=\partialL
\partial
\phi

=ml2\sin2\theta

\phi
.

In terms of coordinates and momenta it reads

H = \underbrace_ + \underbrace_=+-mgl\cos\theta

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

\theta

={P\theta\overml2}

\phi

={P\phi\overml2\sin2\theta}

P\theta
2\over
={P
\phi

ml2\sin3\theta}\cos\theta-mgl\sin\theta

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[1
2
2\theta
ml

2+

1
2

ml2\sin2\theta

\phi
2\right]}
T

+\underbrace{[-mgl\cos\theta]}V

by noting that the horizontal component of angular momentum

Lz=ml2\sin2\theta

\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=1
2
2\theta
ml

2+

1
2
2
L
z
ml2\sin2\theta

-mgl\cos\theta

\left(d\theta
dt
2=2
ml2
\right)\left[E-
1
2
2
L
z
ml2\sin2\theta

+mgl\cos\theta\right]

which leads to an elliptic integral of the first kind for

\theta

2}\int\left[E-1
2
t(\theta)=\sqrt{\tfrac{1}{2}ml
2
L
z
ml2\sin2\theta
-1
2
+mgl\cos\theta\right]

d\theta

and an elliptic integral of the third kind for

\phi

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

The angle

\theta

lies between two circles of latitude, where
E>1
2
2
L
z
ml2\sin2\theta

-mgl\cos\theta

.

See also

References

  1. Book: Landau, Lev Davidovich. Course of Theoretical Physics: Volume 1 Mechanics. Evgenii Mikhailovich Lifshitz. Butterworth-Heinenann. 1976. 0750628960. 33–34.

Further reading