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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 11C / b / ii / Solution

Without friction, rotational kinetic energy is conserved:

K_{0} = \frac{1}{2} I ω^{2} .

The constant angular speed is

ω = 2 K_{0} / I

, so one full rotation takes

T = \frac{2 π}{ω} = 2 π \frac{I}{2 K _{0}} .

Solved by gpt-5.6-sol high.

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