 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/b/solution

s = sin θ

. The maximum height becomes the concave quadratic function

y_{m a x} = rs + \frac{r ^{2} ω ^{2}}{2 g} (1 - s^{2}), \frac{d y _{m a x}}{d s} = r - \frac{r ^{2} ω ^{2}}{g} s .

Under the stated high-angular frequency assumption,

r ω^{2} > g

, its interior maximum is

sin θ_{*} = \frac{g}{r ω ^{2}}, (x_{*}, y_{*}) = (- r 1 - \frac{g ^{2}}{r ^{2} ω ^{4}}, \frac{g}{ω ^{2}}) .

Thus the optimum release lies on the upper-left quarter of the wheel for the printed clockwise rotation. The largest attained height is

Y (ω) = \frac{r ^{2} ω ^{2}}{2 g} + \frac{g}{2 ω ^{2}} .

If extended to

r ω^{2} \leq g

, the maximum over release positions is the limiting height

r

at the top of the rim. The interior solution is the relevant branch for reaching a ceiling above the wheel.

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