 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2021/s03/3/a/solution

v = r \dot{ϕ}

I = m r^{2}

L = I \dot{ϕ} = m r v

. Hence its kinetic energy is

K = \frac{1}{2} m v^{2} = \frac{1}{2} I \dot{ϕ}^{2} = \frac{L ^{2}}{2 I} = \frac{L ^{2}}{2 m r ^{2}} .

There is no potential energy variation along the circle in the stated model.

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