# Angular Momentum Conservation

There's a question I've come across that I've got some confusion on.

A drum of mass $M_A$ and radius $a$ rotates freely with initial angular speed $\omega _0$. A second drum of radius $b>a$ and mass $M_B$ is mounted on the same axis, although it is free to rotate. A thin layer of sand $M_s$ is evenly distributed on the inner surface of the smaller drum (drum $A$). At $t=0$ small perforations in the inner drum are opened, and sand starts to fly out at a rate $\dfrac{dM}{dt}=\lambda$ and sticks to the outer drum. Find subsequent angular velocities of the two drums. Ignore the transit time of the sand

In this question, we start off with an initial angular momentum of $L_0 = I_A \omega = (M_A +M_s ) a^2 \omega _0$ (where the moment of inertia for a drum is $I=mr^2$). If my understanding of conservation of angular momentum holds, this should mean that at all times, this system must maintain this initial value of $L$.

I understand that as long as there is no external torque, $\frac{dL}{dt}=0$, and hence be constant.

Here's where my confusion comes in. Say we only include drum $A$ to be in our system. It's starts off with $L_0$, and as it loses mass, it must continue to rotate faster in order to keep $L_0$ constant. But if we now consider both drum $A$ and $B$ to be in our system, this can't possibly be the case. If drum $A$ continues to "compensate" for this loss in mass by having an angular acceleration, drum $B$ would have to remain stationary so that our value of $L_0$ remains stationary. In both cases, we have zero external forces, and so that means $\tau = 0$, but in the second system, we have an internal torque being applied to drum $B$ by the sand.

Where's my error?

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Thanks for the quick response. That makes sense to me, but I'm a little bit confused about how the angular velocity of drum $A$, minus the sand must remain constant. This means that the total angular momentum of our system (both drums), must equal our initial angular momentum (angular momentum of drum $A$ + angular momentum of the sand). –  Astrum Jan 9 at 23:39