 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-b/b8/a/solution

With zero gravitational potential energy at the wheel axis, the release energy and the energy at the apex of the projectile motion are

E_{release} = \frac{1}{2} m r^{2} ω^{2} + m g r sin θ, E_{apex} = \frac{1}{2} m r^{2} ω^{2} sin^{2} θ + m g y_{m a x} .

Only the vertical velocity vanishes at the apex: the horizontal kinetic energy remains. Equating the two energies gives

y_{m a x} (θ) = r sin θ + \frac{r ^{2} ω ^{2}}{2 g} cos^{2} θ .

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