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

The central rod occupies the holes in the discs, so the disc masses must be annular rather than counted as solid discs overlapping the rod. Define $C=L_1{\rho }_1=2L_2{\rho }_2$. ThenThe axial moment of inertia isThus $I=\frac{1}{2}M{r}_2^2$, exactly the axial inertia of a uniform disc of outer radius $r_2$ and mass $M$. The cancellation uses the given density–length relation.

The contact radius is $r_1$, so rolling without slipping gives $v=r_1\omega$. With $I=M{r}_2^2/2$, the energy at the minimum of the stated parabolic motion model isConsequentlyThe corresponding spin angular frequency is ${\omega }_{\max }=v_{\max }/r_1=14.8\mathrm{rad}{\mathrm{s}}^{-1}$. It is the rod radius, not the disc radius, that relates translation and rotation; the discs determine the larger rotational inertia.

In the small-amplitude approximation, $\dot{x}$ is the speed along the track, and $y=B{x}^2$ supplies the gravitational potential. The conserved energy isDifferentiating and extending through turning points by continuity givesThis is simple harmonic motion with ${\omega }_0^2=4gB/\{(r_2/r_1{)}^2+2\}$. HenceThe frequency is lower than for a sliding point mass because energy is also stored in rotation.

For weak linear damping, write the amplitude envelope as $A(t)=A_0{e}^{-{\omega }_{0}t/(2Q)}$. The amplitude e-folding time is $2Q/{\omega }_0$, and one period is approximately $2\pi /{\omega }_0$. ThusEnergy, being proportional to amplitude squared, decays twice as rapidly in its exponent. Using its e-folding time instead would introduce a factor-of-two error.

At the initial turning point the oscillation energy is $E=Mg{h}_0=0.04905\mathrm{J}$. In the high-quality factor regime, $Q=2\pi E/\Delta E$, so the required input per period isThis replaces the small dissipated energy each cycle, rather than supplying the entire stored energy again.