Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 14B / a

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- Solution a

Solution

With the pivot as origin and upward vertical coordinate,

r_{1} = l (sin θ_{1}, - cos θ_{1}), r_{2} = r_{1} + l (sin θ_{2}, - cos θ_{2}) .

Therefore

∣ \dot{r}_{1} ∣^{2} = l^{2} \dot{θ}_{1}^{2},

∣ \dot{r}_{2} ∣^{2} = l^{2} [\dot{θ}_{1}^{2} + \dot{θ}_{2}^{2} + 2 cos (θ_{1} - θ_{2}) \dot{θ}_{1} \dot{θ}_{2}] .

The potential energy is

V = - (M + m) g l cos θ_{1} - m g l cos θ_{2} .

Thus

L = T - V

is

L = \frac{1}{2} M l^{2} \dot{θ}_{1}^{2} + \frac{1}{2} m l^{2} [\dot{θ}_{1}^{2} + \dot{θ}_{2}^{2} + 2 cos (θ_{1} - θ_{2}) \dot{θ}_{1} \dot{θ}_{2}] + M g l cos θ_{1} + m g l (cos θ_{1} + cos θ_{2}) .

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