Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Section B / 6

Choose three mutually perpendicular axes through the same point, and label coordinates $x,y,z$. The moments of inertia of any rigid body areSubtracting givesFor a thin planar lamina lying in $z=0$, the right-hand side vanishes, proving the perpendicular axis theorem:The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities arewith equality in the first precisely when all mass lies in the plane perpendicular to its $z$ axis, and cyclically for the others. None of these identities requires principal axes.

Let $I_{\perp }$ denote the moment of inertia normal to the disc and $I_d$ that about any central diameter. Rotational symmetry makes the two perpendicular diameter values equal. By the perpendicular axis theorem, $I_{\perp }=2I_d$. The intended angular impulse has magnitude $J=I_{\perp }{\omega }_0$; delivering that magnitude about a diameter instead gives $J=I_d\omega$. ThusThese are principal axes, so angular momentum is parallel to angular velocity in both cases. Equal delivered angular impulse means equal magnitude of the change in angular momentum, not equal rotational kinetic energy; the actual rotational kinetic energy is twice the intended value.

Take $+x$ along the upper ship's initial velocity, and take $+y$ from the lower ship to the upper ship at closest approach. The upper and lower ships then have initial velocities $v_1\hat{\mathbf{x}}$ and $-v_2\hat{\mathbf{x}}$. Their equal masses put each at distance $l/2$ from the centre of mass, soBy momentum conservation, the centre of mass keeps moving with constant velocityIn that frame the ships have opposite velocities of magnitude $u=(v_1+v_2)/2$. Their initial angular momentum about the centre of mass is nonzero even though their trajectories were straight:The tether exerts central internal forces, so conservation of angular momentum givesEach ship follows uniform circular motion of radius $l/2$ around the translating centre of mass, clockwise viewed from $+z$. More explicitly, put the initial centre at the origin and defineThe kinetic-energy decomposition about the centre of mass verifies conservation of mechanical energy directly:At closest approach the relative radial velocity is already zero, so connecting the tether requires no dissipative radial impulse. Subsequently its tension is perpendicular to each ship's velocity relative to the centre of mass and does no net work on the pair at fixed $l$.

Each ship needs centripetal acceleration ${\omega }^2(l/2)$, supplied entirely by the tension. ThusThe result has units of force, $\mathrm{N}$.

During quasi-static tether contraction, momentum conservation keeps ${\mathbf{V}}_{\mathrm{CM}}$ constant and conservation of angular momentum keeps $L$ constant. To leading order in the slow radial motion, the kinetic energy isIts derivative at fixed $L$, not fixed $\omega$, isTo compute the power supplied by the tension, use the instantaneous unit vector $\mathbf{n}$ from the lower to the upper ship. Their radial velocities relative to the centre of mass are $\pm (\dot{l}/2)\mathbf{n}$, and the tension forces are $\mp T\mathbf{n}$. ConsequentlyThe translational contributions cancel between the two ships and the tangential contributions are zero. The contraction mechanism supplies positive work, increasing the rotational kinetic energy.

The slow-contraction approximation matters: for a prescribed smooth $l(t)$ with a taut tether, the exact kinetic energy and tension include radial terms,These still satisfy $d{E}_{\mathrm{exact}}/dt=-T_{\mathrm{exact}}\dot{l}$. Thus the paper's displayed formula is the leading result for slow, smooth contraction, or is exact when the radial kinetic energy is constant; it is not an identity for every possible shrinking schedule.