Rotating Drums and Sand: Finding Angular Velocities

  • Thread starter geoffrey159
  • Start date
  • Tags
    Rotating
In summary, the conversation revolved around a physics problem involving two drums with different masses and radii, as well as a layer of sand distributed on the inner surface of the smaller drum. The goal was to find the subsequent angular velocities of the two drums after small perforations in the inner drum were opened and the sand started to fly out and stick to the outer drum. The solution involved using the conservation of angular momentum and considering the external forces and torques on the system. The final solution also took into account the hint provided in the problem statement.
  • #1
geoffrey159
535
72

Homework Statement


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

Homework 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 ##

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
 
Physics news on Phys.org
  • #2
geoffrey159 said:
clue. If λt=MB and b=2a then ωA = ωA(0)/8

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
TSny said:
Is there a typo here? Should it read: ωB = ωA(0)/8 ?
Yes it is a typo, sorry :)

TSny said:
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 [...]. What would happen to your rate of rotation if you just released the weights (rather than pull them inward as shown on the right)?

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
with no torque applied, what could cause the inner drum to change angular speed?
What does the sand do with its L ?
 
  • #5
geoffrey159 said:
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 ?

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
TSny said:
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?

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
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?
 

Attachments

  • Ang Momentum.png
    Ang Momentum.png
    4.3 KB · Views: 648
  • #8
If my moment of inertia does not change and angular momentum is conserved, then angular velocity does not change.
 
  • #9
That's right, your angular velocity would not change.
 
  • #10
I don't understand how it relates to the problem, what did you want to tell me with the stool example ?
 
  • #11
Analogy: You're like the drum A and the weights in your hand are like the grains of sand.
 
  • #12
think of it as 3 pieces ... what does the sand do with its L ?
 
  • #13
I've made an attempt at a solution, which is not correct. Can you explain where it starts to be wrong and why ?
 
  • #14
geoffrey159 said:
I've made an attempt at a solution, which is not correct. Can you explain where it starts to be wrong and why ?
You mean, in your original post?
geoffrey159 said:
angular momentum is conserved in both systems ( Drum A + sand and Drum B + sand)
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
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
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
geoffrey159 said:
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 ?
Ok so far, but it's not an argument yet. Where do you go next?
 
  • #18
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
geoffrey159 said:
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.
Ok, and indeed ai suggested in post #14 to look at angular momentum of the whole system.
geoffrey159 said:
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.
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
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?
 
  • #21
geoffrey159 said:
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?
Looks good.
 
  • #22
Thank you very much for your help !
I have this strange feeling that I got mentally blocked, it reminds me a ski tow problem :-)
 
  • Like
Likes poseidon721

1. What is the purpose of using rotating drums in sand testing?

The rotating drum method is commonly used in geotechnical engineering to simulate the effects of soil erosion or abrasion caused by wind or water on sand particles. This allows researchers to study the potential impact of these erosive forces on coastal structures, such as sea walls or breakwaters.

2. How does a rotating drum simulate erosion on sand particles?

The rotating drum is filled with sand and rotated at a controlled speed, causing the sand particles to rub against each other and the drum walls. This mimics the natural process of sand particles being moved and eroded by wind or water. The amount of erosion can be controlled by adjusting the speed and duration of rotation.

3. What types of sand can be used in rotating drum tests?

Generally, any type of sand can be used in rotating drum tests as long as it is well-graded and has a similar particle size distribution to the sand found in the natural environment being studied. The sand should also be clean, with minimal amounts of clay or silt particles, to ensure accurate results.

4. How are the results of rotating drum tests interpreted?

The results of rotating drum tests are typically interpreted by measuring the weight loss or change in particle size of the sand samples before and after testing. This can provide valuable information on the potential erosion rates and mechanisms of different types of sand under varying conditions.

5. What are the limitations of using rotating drums in sand testing?

Rotating drums can provide valuable insights into the erosive forces on sand particles, but they do have some limitations. The results may not fully reflect the complex and dynamic nature of natural erosion processes, and the tests may not accurately simulate the effects of other factors, such as wave action or sediment transport. Additionally, the results may vary depending on the size and shape of the sand particles used in the test.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
735
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
706
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
939
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top