 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2021/4/solution

Choose the vertical

z

axis downward, and measure the polar angle

θ

from it. With azimuth

ϕ

and fixed rod length

r

x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ .

The massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yields

v^{2} = \overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2} = r^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) .

V = - m g z = - m g r cos θ

, up to an arbitrary constant. The Lagrangian of this spherical pendulum is therefore

L = \frac{1}{2} m r^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) + m g r cos θ .

m r^{2} \ddot{θ} - m r^{2} sin θ cos θ \dot{ϕ}^{2} + m g r sin θ = 0, \frac{d}{d t} (m r^{2} sin^{2} θ \dot{ϕ}) = 0.

Equivalently, away from the coordinate singularities at the poles,

\ddot{θ} = sin θ cos θ \dot{ϕ}^{2} - \frac{g}{r} sin θ, \ddot{ϕ} + 2 cot θ \dot{θ} \dot{ϕ} = 0.

Since

L

is independent of

ϕ

p_{ϕ} = m r^{2} sin^{2} θ \dot{ϕ}

is the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward

z

axis instead reverses the sign of the cosine term in

L

and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.

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