 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-a/3/solution

The string's no-slip condition gives

v = R ω

. Conservation of mechanical energy from release to the stated height drop yields

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v ^{2}}{R ^{2}} .

Solving for the spool's moment of inertia,

I = \frac{2 m g h R ^{2}}{v ^{2}} - m R^{2}, A = 2 m g h R^{2}, B = m R^{2} .

The axle does no work because its point of application is fixed, and the ideal string and axle introduce no losses. The spool mass

M

is not needed to infer

I

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