Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 11C (Dynamics and Relativity)

With total mass $M={\sum }_i{m}_i$, defineThe angular momentum about the centre of mass isExpanding, using ${\sum }_i{m}_i(x_i-R)=0$ and $P=M\dot{R}$, gives

Every particle is initially at distance $R$ from the axis, so

Each freely moving particle has position, relative to its own initial radial and tangential unit vectors,Thus all particles lie on a circle of radiusTheir angular momenta are constant and sum toSince $I(t)=mR(t{)}^2$, the angular velocity isWith $\theta (0)=0$,This is the geometry of a freely expanding rotating particle ring.

For uniform surface density $m_0/(\pi R^2)$,

At time $t$, the remaining water has angular momentumDuring $dt$, the ejected mass is $-dm$, and its signed tangential speed in the inertial frame is $R\omega +u$. Conservation of angular momentum gives, to first order,After simplification, the angular rocket equation isHere $u$ is signed relative to the direction of rotation; reversing the spray changes its sign.

For $u=0$, the equation becomesso $\omega /m$ is constant. With $m(t)=m_0-\mu t$,It reaches zero when the water is exhausted, at