= Solution
Let $t=0$ be the start of deposition and $I_0=MR^2/2$ be the original <moment of inertia of a uniform disc>. Assume that all deposited sand sticks and remains at radius $r$, with negligible thickness and no losses. The sand's incoming <angular momentum> about the vertical axis is zero. Neither the bearing nor <gravity> supplies an axial <torque>, so <conservation of angular momentum> gives
$$
I(t)=I_0+\alpha r^2t,\qquad
I(t)\omega(t)=I_0\omega_0.
$$
Consequently the <angular velocity> and <angular acceleration> are
$$
\boxed{\omega(t)=\frac{I_0\omega_0}{I_0+\alpha r^2t}
=\frac{MR^2\omega_0}{MR^2+2\alpha r^2t},}
\qquad
\boxed{\dot\omega(t)=-\frac{\alpha r^2I_0\omega_0}{(I_0+\alpha r^2t)^2}.}
$$
The <rotational kinetic energy> is
$$
\boxed{K(t)=\frac12I(t)\omega(t)^2
=\frac{I_0^2\omega_0^2}{2(I_0+\alpha r^2t)}
=K(0)\frac{I_0}{I_0+\alpha r^2t}.}
$$
Thus, for $\alpha r^2>0$,
$$
\boxed{t_{1/2}=\frac{I_0}{\alpha r^2}=\frac{MR^2}{2\alpha r^2}.}
$$
This is <mass-loaded rotation>: the sticking process dissipates <mechanical energy> even though <angular momentum> is conserved. Indeed $\dot K=-\alpha r^2\omega^2/2$. If $r=0$ or $\alpha=0$, no additional axial <moment of inertia> is deposited and the nonzero initial <rotational kinetic energy> never halves.
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