What happens to angular speed when an object lands on a spinning disc?

The disc slows when the arriving object adds moment of inertia without bringing angular momentum about the axle. Moment of inertia measures how strongly an object resists a change in its rotation. During the brief landing, conserve angular momentum about the axle; do not assume rotational kinetic energy is conserved.

For a uniform disc of mass $M$ and radius $R$, initially rotating at $\omega_i$,

$$I_d=\frac12MR^2.$$

A compact object of mass $m$ that sticks a distance $r$ from the axis adds approximately $mr^2$ to the moment of inertia. Suppose it falls vertically, with no initial sideways motion, onto a horizontal disc. Its initial angular momentum about the vertical axle is then zero. If the external torque about that axle is negligible, through conservation of angular momentum

$$I_d\omega_i=(I_d+mr^2)\omega_f,$$

so

$$\omega_f=\frac{I_d}{I_d+mr^2}\omega_i.$$

Friction produces opposite torques, or turning effects, on the object and disc. It transfers angular momentum from the disc to the object without changing the system's total. A fixed axle can exert a force without exerting a torque about its own axis. The presence of this friction causes losses in the rotational kinetic energy.

Consider a $2.0\,\mathrm{kg}$ disc of radius $0.30\,\mathrm{m}$ rotating at $12\,\mathrm{rad\,s^{-1}}$. A $0.50\,\mathrm{kg}$ object lands at $r=0.20\,\mathrm{m}$ and sticks. Then

$$I_d=0.090\,\mathrm{kg\,m^2},\quad I_f=0.110\,\mathrm{kg\,m^2},$$ $$\omega_f=\frac{0.090}{0.110}(12)=9.82\,\mathrm{rad\,s^{-1}}.$$

Check energy separately:

$$E_{Ki}=\tfrac12(0.090)(12)^2=6.48\,\mathrm{J},$$ $$E_{Kf}=\tfrac12(0.110)(9.82)^2\approx5.30\,\mathrm{J}.$$

About $1.18\,\mathrm{J}$ of the initial rotational kinetic energy becomes other forms during slipping and sticking, such as internal energy. This calculation only compares rotational energy: the energy from the object's vertical fall must be accounted for separately.

There is a useful algebraic check: at fixed angular momentum, $E_K=L^2/(2I)$, so $E_{Kf}/E_{Ki}=I_d/I_f=\omega_f/\omega_i$ here. Increasing $I$ reduces both speed and rotational energy.

Landing nearer the rim causes a larger slowdown because the added inertia depends on $r^2$. Landing exactly at the centre adds no inertia in the point-mass approximation, which treats all the object's mass as concentrated at one point. If the arriving object already has tangential velocity (motion along the circular path), include its signed initial angular momentum $mrv_{\mathrm{tan}}$ before solving; the disc need not slow down.

Have a go at the disc-and-dropped-object problem in the IB Physics space, time and motion questions (Question 162). Can you use the object's landing radius to work out its added moment of inertia and choose the final angular speed?