 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-b/8/b/solution

r_{1}

v = r_{1} ω

. With

I = M r_{2}^{2} /2

, the energy at the minimum of the stated parabolic motion model is

M g h_{0} = \frac{1}{2} M v_{m a x}^{2} + \frac{1}{2} \frac{M r _{2}^{2}}{2} \frac{v _{m a x}^{2}}{r _{1}^{2}} .

Consequently

v_{m a x} = \frac{2 g h _{0}}{( r _{2} / r _{1} ) ^{2} + 2} = 0.148 m s^{- 1} .

The corresponding spin angular frequency is $ω_{m a x} = v_{m a x} / r_{1} = 14.8 rad s^{- 1}$ . It is the rod radius, not the disc radius, that relates translation and rotation; the discs determine the larger rotational inertia.

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