Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Conservation laws typically hold only for closed systems. In this case the inner drum is losing mass so it can't be considered a closed system; instead, you must also include the angular momentum of the outer drum. On the other hand, you're right that there is no torque on the inner drum, and that therefore its angular momentum - that of the drum itself, ignoring the sand - is also constant. Together, these two constraints are enough to determine the two final velocities.

share|improve this answer
    
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
    
That is correct. The outer drum and the sand must therefore share the initial angular momentum of the sand. –  Emilio Pisanty Jan 9 at 23:42
    
I understand, and got the right answer, many thanks for clearing that up. –  Astrum Jan 9 at 23:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.