 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015-resit/10/b/solution

ϕ

\partial L / \partial ϕ = 0

. Its Euler-Lagrange equation gives conservation of the vertical component of angular momentum:

h = p_{ϕ} = \frac{\partial L}{\partial ϕ ˙} = m l^{2} sin^{2} θ \dot{ϕ} = constant .

This

h

is the actual angular momentum, not specific angular momentum. The Lagrangian also has no explicit time dependence. The conserved energy is

E = \dot{θ} \frac{\partial L}{\partial θ ˙} + \dot{ϕ} \frac{\partial L}{\partial ϕ ˙} - L = \frac{1}{2} m l^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) - m g l cos θ .

Along solutions,

\dot{E} = - \partial L / \partial t = 0

. These are the conserved quantities associated with rotational symmetry about the vertical and time-translation symmetry, respectively.

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