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

Conservation of energy states that the total energy of an isolated system is constant. In a mechanical problem with conservative forces and no dissipation, it gives $K+U=\text{constant}$. It often determines a speed or turning height directly, without solving the full time-dependent equation of motion. With external work or dissipation, use the corresponding energy balance instead.

After a drop leaves the wheel, neglecting air resistance, the conservative force of the Newtonian gravitational field acts. Put the origin at the wheel axis and choose positive horizontal displacement to the right. In the printed diagram $\theta$ is measured upward from the leftward horizontal radius, and rotation is clockwise. The release position and velocity are thereforeFor upward release, $\cos \theta >0$. These conventions fix the signs used below.

With zero gravitational potential energy at the wheel axis, the release energy and the energy at the apex of the projectile motion areOnly the vertical velocity vanishes at the apex: the horizontal kinetic energy remains. Equating the two energies gives

Set $s=\sin \theta$. The maximum height becomes the concave quadratic functionUnder the stated high-angular frequency assumption, $r{\omega }^2>g$, its interior maximum isThus the optimum release lies on the upper-left quarter of the wheel for the printed clockwise rotation. The largest attained height isIf extended to $r{\omega }^2\le g$, 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.

To avoid contact with the ceiling, require $Y(\omega )<h$. Writing $X=r^2{\omega }^2$ giveswhose roots are $g(h\pm \sqrt{h^2-r^2})$. Because $h>r$, the lower root is less than $gr$, while the assumed branch has $X>gr$. The lower inequality is therefore automatic on this branch, leavingEquality is the first grazing contact. This branch check is essential: retaining both algebraic roots as independent lower and upper speed limits would introduce a spurious lower limit on avoiding the ceiling.

At first contact the globally highest trajectory just touches the ceiling at its apex. Its flight time to that point is $t_{\ast }=r\omega \cos {\theta }_{\ast }/g$. The horizontal displacement from the wheel axis isHence the first point struck is directly above the rotation axis, independently of $h$. The release starts to the left of the axis, but the drop's horizontal travel exactly cancels that offset at the optimum apex. The first-contact angular velocity satisfies $r^2{\omega }_{\ast }^2=g(h+\sqrt{h^2-r^2})$.