 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/h/1/b/solution

- a cos θ /2

. With gravitational potential energy zero at the pivot level,

V = - \frac{M g a}{2} cos θ .

Hence the Lagrangian is

L = T - V = \frac{M a ^{2}}{6} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) + \frac{M g a}{2} cos θ .

Adding a constant to the potential would change the numerical energy reference but would not change the motion.

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