Two identical spinning pucks on an air table are drawn together by a spring and collide completely inelastically, with friction preventing any slip at the contact. Starting from no orbital motion, the lecture shows that the central spring force leaves both spins untouched, then splits the impact into a normal impulse that stops the approach and a tangential friction impulse that launches an orbit and slows both spins equally. The no-slip condition fixes that impulse: the pair orbits at one sixth of the summed initial spins, each puck gives up a third of that sum, and the result is checked against angular momentum about the centre of mass. It closes on the gear-like meaning of the answer and three limiting cases.
Two identical pucks sit on an air table with their centres at rest. Each one spins about its own centre, the first at omega one nought and the second at omega two nought. A spring pulls them together until they collide. The impact is completely inelastic, and friction stops the surfaces slipping while it lasts. Our question is what they do afterwards: two final spins, and the rate at which the pair orbits. Each puck is a uniform disc of mass m and radius R, so its moment of inertia about its centre is one half m R squared. The spring force lies along the line of centres. It passes straight through each centre, so it exerts no torque on either puck, and the spins arrive at contact unchanged. The pull is internal, so the momenta stay equal and opposite and the centre of mass never moves. Every velocity lies along the line of centres, so there is no orbital angular momentum at all. So just before impact, the total angular momentum about the centre of mass is spin alone: I times the sum of the two initial spins. The collision may share that amount out differently, but it cannot change it.
Freeze the instant of contact. The centre of mass sits exactly at the touching point Q, with the line of centres horizontal. Two impulses act there, equal and opposite on the two pucks. The normal impulse P is the inelastic part. It wipes out the approach speed entirely, so P equals m v, and the pucks stay pressed together afterwards. Friction supplies a tangential impulse J. It kicks puck two upward and puck one downward, each with speed J over m, so the total momentum is still zero. Two centres, each a distance R from Q, moving in opposite directions across the line. The pair has started to orbit, at Omega equal to J over m R. The same friction impulse also has a lever arm R about each centre. Work out the two cross products and both pucks receive the same angular impulse, minus R J, a clockwise kick. Divide by I, which is one half m R squared, and each spin drops by two J over m R. That is exactly twice Omega, the same drop for both pucks. One more condition fixes J. Friction forbids slipping, so the two surfaces at Q must move together. Puck one's surface there moves at minus R Omega plus R omega one, and puck two's at R Omega minus R omega two. Set them equal and divide by R. The rule is clean: the two final spins add up to twice the orbital rate. Now insert the spin drop for each puck. The initial spins, minus four Omega, must equal two Omega. So Omega is one sixth of the sum of the initial spins. Check it against angular momentum. The orbit carries two m R squared Omega, the spins carry I times two Omega, three m R squared Omega in all. Since I is half m R squared, the initial I times the sum of the spins matches that exactly when Omega is one sixth of the sum.
Put that back into the spin drops, and the whole final state is written in terms of the initial spins. The pair orbits at one sixth of their sum. Puck one ends at two omega one nought minus omega two nought, all over three, and puck two is the mirror image. Subtract the orbital rate and something simple appears. Measured against the turning line of centres, the two pucks spin at plus and minus half their initial difference. They roll on each other like meshed gears, and friction has nothing more to do. Take the plainest start: puck one spinning at six units and puck two not spinning at all. Then the pair orbits at one unit, puck one keeps four, and puck two, which began at rest, is driven backwards at two. Watch one full revolution of the pair. In that time puck one turns four times forward, while puck two turns twice the other way. The spring now only supplies the inward pull that keeps them together. Three starting cases are worth keeping. One spinning puck, as we just watched: a sixth of its spin goes into the orbit, and the idle puck is driven backwards. Equal spins: every rate becomes one third of the original, so the two pucks lock and turn together as a single rigid body. Opposite spins: the sum is zero, there is no orbit, and nothing changes at all, because the pucks were already rolling on each other like gears. So the sum of the spins is shared out by angular momentum, the difference survives untouched, and friction only enforces the rolling. The kinetic energy that disappears was taken by the impact itself.
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