# Rotating drums and sand

1. Feb 10, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
A drum of mass MA and radius a rotates freely with initial angular velocity ωA(0). A second drum with mass MB and radius b > a is mounted on the same axis and is at rest, although it is free to rotate. A thin layer of sand with mass Ms is distributed on the inner surface of the smaller drum. At t=0, small perforations in the inner drum are opened. The sand starts to fly out at a constant rate λ and sticks to the outer drum. Find the subsequent angular velocities of the two drums ωA and ωB. Ignore the transit time of the sand.

clue. If λt=MB and b=2a then ωA = ωA(0)/8

2. Relevant equations

Mass of sand into drum A : $m_a(t) = M_s -\lambda t$
Mass of sand into drum B : $m_b(t) = \lambda t$

Angular momentum in the system Drum A + sand : $L_a = (M_a + m_a) a^2 \omega_a +m_b b^2 \omega_b$

Angular momentum in the system Drum B + sand : $L_b = m_a a^2 \omega_a +(M_b+m_b) b^2 \omega_b$

3. The attempt at a solution

Hello, I have just finished reading the chapter dealing with rotational motion of rigid bodies. I am very confused for the moment. I have spent a lot of time thinking about this problem but did not manage to get to the right answer. Can you help me please ?

------ attempt :
I need two coupled equations in order to find $\omega_b$ from $\omega_a$.
Since the motion is rotational around a fixed axis, it is natural to use angular momentum.
Furthermore, angular momentum is conserved in both systems ( Drum A + sand and Drum B + sand) because in both cases, external torque is due to :
(i) Sand weight in both drums. Rings of sand concentrate their weight on the axis of rotation so the torque due to sand weight is 0.
(ii) Radial push from the other drum (that is not in the system). A radial force has 0 torque.

So,
$L_a(t) = L_a(0) = (M_a+M_s) a^2 \omega_a(0)$
$L_b(t) = L_b(0) = M_s a^2 \omega_a(0)$

Subtracting second equation to first equation, I get:

$\omega_b = \frac{M_a}{M_b}\frac{a^2}{b^2} (\omega_a - \omega_a(0))$

Replacing $\omega_b$ by its expression in $L_a$, I get:

$\begin{array}{lcr} \omega_a = \frac{M_a+M_s+\lambda t \frac{M_a}{M_b}}{M_a+M_s+\lambda t (\frac{M_a}{M_b} - 1 ) } \omega_a(0) & & \omega_b = \frac{M_a}{M_b}\frac{a^2}{b^2} \frac{\lambda t}{M_a+M_s+\lambda t (\frac{M_a}{M_b} - 1 ) } \omega_a(0) \end{array}$

but it does not match with the hint given in the problem statement

2. Feb 10, 2015

### TSny

Is there a typo here? Should it read: ωB = ωA(0)/8 ?

It might help to consider a simpler case first. Suppose you are sitting on a rotating stool holding a weight in each hand with arms outstretched, as shown on the left here: http://hyperphysics.phy-astr.gsu.edu/hbase/rstoo.html . What would happen to your rate of rotation if you just released the weights (rather than pull them inward as shown on the right)?

3. Feb 10, 2015

### geoffrey159

Yes it is a typo, sorry :)

In this situation, I think that angular momentum is conserved because there are no external forces in the plane of motion.
If I drop the masses, my moment of inertia will decrease, so the angular velocity must increase in order to satisfy conservation of angular momentum, right ?

4. Feb 10, 2015

### lightgrav

with no torque applied, what could cause the inner drum to change angular speed?
What does the sand do with its L ?

5. Feb 10, 2015

### TSny

That's not correct. It's kind of tricky. When considering conservation of angular momentum, you have to be clear on the "system". Suppose you take just yourself as the system. So, the weights in your hand are objects external to "the system". Let's forget about gravity as it is irrelevant to the problem. While you are rotating and holding the weights in your hands, do the weights exert any force on you? If so, in what direction? Do these forces from the weights exert a torque on you? While you release the weights do the weights exert a torque on you? When you release the weights, does the moment of inertia of the system (remember, that's you alone) change?

6. Feb 10, 2015

### geoffrey159

When I am rotating with the weights in my hand they exert a downward force on me. That force is parallel to the axis of rotation so there will be no torque about this axis.
When I release the weight, the moment of inertia does not change if I am the system (lol, it sounds crazy)

7. Feb 10, 2015

### TSny

I think we should ignore gravity. It is not relevant. Nevertheless, the weights do exert outward "centrifugal" forces on your hands. But you can see that these outward forces on your hands do not produce any torque on your body. And when you release the weights, they still don't exert any torque on you. So, there is never a torque on you. As you say, your moment of inertia does not change. So, does your angular velocity change?

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8. Feb 10, 2015

### geoffrey159

If my moment of inertia does not change and angular momentum is conserved, then angular velocity does not change.

9. Feb 10, 2015

### TSny

That's right, your angular velocity would not change.

10. Feb 10, 2015

### geoffrey159

I don't understand how it relates to the problem, what did you want to tell me with the stool example ?

11. Feb 10, 2015

### TSny

Analogy: You're like the drum A and the weights in your hand are like the grains of sand.

12. Feb 10, 2015

### lightgrav

think of it as 3 pieces ... what does the sand do with its L ?

13. Feb 10, 2015

### geoffrey159

I've made an attempt at a solution, which is not correct. Can you explain where it starts to be wrong and why ?

14. Feb 10, 2015

### haruspex

You mean, in your original post?
The A system is losing momentum because it is losing mass, and there is no means by which that loss of mass will lead to an increase in rotation rate. The departing sand exerts no torque on the drum.
Likewise, the B system gains momentum because the sand hitting it has a moment about the axis.

So... what about the angular momentum of the whole system?

15. Feb 10, 2015

### lightgrav

Your "Drum A + sand" sub-system LA includes sand that has moved to drum B.
in the next line, your "Drum B + sand" LB includes sand that is still in drum A.

16. Feb 11, 2015

### geoffrey159

Thanks ! But why is it wrong?
"Drum A + sand" sub-system includes drum A, the ring of sand that is into drum A, and the ring of sand that is into drum B.
So the angular momentum of this sub-system about the axis of rotation is :
$L_a = L_{\text{drum A}} + L_{\text{Sand in A}} + L_{\text{Sand in B}} = M_a a^2 \omega_a + m_a a^2 \omega_a + m_b b^2 \omega_b$
What is incorrect with this argument ?

17. Feb 11, 2015

### haruspex

Ok so far, but it's not an argument yet. Where do you go next?

18. Feb 11, 2015

### geoffrey159

I did symetrically with drum B, so that I get and expression of La and Lb.
Then, the argument is that there is conservation of angular momentum for both La and Lb.
From the point of view of "drum A + sand" subsystem, the only exterior forces acting on it are the weight of the sand rings into both drums, and the radial push of drum B onto the ring of sand that is inside it. Both have 0 torque.

19. Feb 11, 2015

### haruspex

Ok, and indeed ai suggested in post #14 to look at angular momentum of the whole system.
No, there is a tangential, retarding, torque from the outer drum onto the sand that reaches it from A. The sand lands on drum B with a greater angular velocity than drum B and the sand already there.
Given that the total angular momentum is constant, that drum A has constant angular speed, and that drum A sand is losing mass at a known rate, you can compute the mass and angular momentum of drum B (plus its sand) at time t.

20. Feb 11, 2015

### geoffrey159

Ok, with your explanations and those from TSny (stool example), maybe I see clearer.

a)The sand into drum A exerts a force on it, the centrifugal force TSny mentioned, radially. Drum A would rotate freely without this applied force, so the external torque of Drum A is 0. Then, the angular momentum of Drum A is conserved, so its angular velocity $\omega_a$ is constant.

b) Following your suggestion, I now consider the the whole system 'drum A + drum B + total mass of sand'.
I don't have to care about contact forces anymore, only the weight of both rings of sand contribute to the external torque (because the drums would rotate freely without the presence of sand). Then the angular momentum is conserved:

\begin{align} L(t) = L(0) \Rightarrow &\ (M_a + m_a) a^2 \omega_a + (M_b+m_b) b^2 \omega_b = (M_a+M_s) a^2 \omega_a(0) \\ \Rightarrow &\ \omega_b = \frac{a^2}{b^2} \frac{\lambda t}{M_b+\lambda t} \omega_a(0) \end{align}

which is in agreement with the hint given in the problem statement. Is it correct?