 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1/1/c/solution

Let

ψ

be the small hoop's absolute spin angle with the same angular sign convention. Equating the two contact-point velocities cancels the large hoop centre's translation. The remaining tangential equality is

a \dot{ϕ} + a \dot{ψ} = 2 a \dot{θ}, \dot{ψ} = 2 \dot{θ} - \dot{ϕ} .

M a^{2} (2 \dot{θ} - \dot{ϕ})^{2} /2

. At quadratic order its cross term cancels the translational cross term in part a. The potential is unchanged, so

L_{2} = \frac{1}{2} M a^{2} (16 \dot{θ}^{2} + 2 \dot{ϕ}^{2}) - \frac{1}{2} M g a (4 θ^{2} + ϕ^{2}) .

Static friction enforces rolling; ideal no-slip contact dissipates no energy.

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