Solving a Challenging Problem on Sunday

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In summary, the problem involves two drums, one with a layer of sand on its inner surface, and the other mounted on the same axis with a larger radius. The sand is released from the inner drum and sticks to the outer drum, causing a decrease in its angular momentum. The equations used to solve for the subsequent angular velocities of the two drums involve the conservation of angular momentum and energy, and the solution is different depending on whether the angular momentum of the inner drum is assumed to be constant or not.
  • #1
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Here is a problem I spent my sunday trying to solve. I made progress, but still something is wrong.

Homework Statement



A drum of mass MA and radius a rotates freely with initial angular velocity [tex]\omega_{A}(0).[/tex] 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 [tex]\lambda[/tex] and sticks to the outer drum. Find the subsequent angular velocities of the two drums [tex]\omega_{A}[/tex] and [tex]\omega_{B}.[/tex] Ignore the transit time of the sand.
Ans. clue. If [tex]\lambda t = M_{B}[/tex] and b=2a then [tex]\omega_{B}=\omega_{A}(0)/8[/tex]

Homework Equations



[tex]\frac{dL}{dt}=0[/tex]

[tex]L=mr^{2}\omega[/tex]

The Attempt at a Solution



I begin by looking at the angular momentum of the inner drum: [tex]L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}\omega_{A}(t)[/tex] and at a later time [tex]L_{A}(t+\Delta t)=(M_{A}+M_{S}-\lambda (t+\Delta t))a^{2}\omega_{A}(t+\Delta t)+\lambda \Delta t a^{2}\omega_{A}(t).[/tex] Taking the limit and dividing by dt gives [tex]\frac{dL_{A}}{dt}=(M_{A}+M{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}-\omega_{A}\lambda a^{2}=0.[/tex] This is a differential equation which can be solved by switching a little to give
[tex]\omega_{A}=\omega_{A}(0)\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}[/tex]
I think this might be right.

However, the second part I am not so sure of. First of all some energy is lost when the sand falls in partly perpendicular to the surface of the outer drum. One can show that a fraction a/b of the momentum survives and the rest becomes heat. Therefore the angular momentum of the outer drum at time t will be [tex]L_{B}(t)=(\lambda t + M_{B})b^{2}\omega_{B}(t)+b\frac{a}{b}a\omega_{A} \lambda \Delta t,[/tex] where the second term is the incoming sand losing some mechanical energy. At a later time [tex]L_{B}(t)=(\lambda (t+\Delta t) + M_{B})b^{2}\omega_{B}(t+\Delta t).[/tex] Taking the limit and dividing by dt gives [tex]\frac{dL_{B}}{dt}=(\lambda t+M_{B})b^{2}\frac{d\omega_{B}}{dt}+\lambda(\omega_{B}b^{2}-\omega_{A}(0)a^{2}\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t})=0[/tex] This is a linear differential equation which can be solved to give [tex]\omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}}\ln \lgroup \frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}\rgroup \frac{M_{A}+M_{S}}{M_{B}+\lambda t}[/tex]
This is not in accordance with the clue, nor with my answer sheet that says [tex]\omega_{B}=\omega_{A}(0)\frac{a^{2}(M_{A}+M_{S}-\lambda t)}{b^{2}(M_{B}+\lambda t)}[/tex]
Please tell me if something is unclear.
 
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  • #2


Isn't your first differential equation dWa/Wa=lambda*dt/(Ma+Ms-lambda*t)?
 
  • #3


RTW69 said:
Isn't your first differential equation dWa/Wa=lambda*dt/(Ma+Ms-lambda*t)?

Yes it is [tex]\frac{d\omega_{A}}{\omega_{A}}=\lambda \frac{dt}{M_{A}+M_{S}-\lambda t}[/tex] This leads to [tex]\ln \omega_{A}=-\ln (M_{A}+M_{S}-\lambda t)+C[/tex] which leads to my equation stated in original post.

However I think that the angular momentum of the inner drum might be constant. This is because now looking closer at my equations LA(t) and LA(t+delta*t) I get a sign difference. [tex]L_{A}(t+\Delta t)-L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}(\omega_{A}(t+\Delta t)-\omega_{A}(t))+\lambda a^{2}\Delta t(\omega_{A}(t)-\omega_{A}(t+\Delta t))[/tex] Now the second term in this equation becomes zero when taking the limit, so [tex](M_{A}+M_{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}=0[/tex] with the only solution that the angular momentum is constant.

Now putting this into the differential eguation for the second drum [tex]\frac{d\omega_{B}}{dt}+\frac{\lambda}{\lambda t+M_{B}}\omega_{B}=\omega_{A}(0)\frac{a^{2}}{b^{2}}\frac{\lambda}{\lambda t+M_{B}},[/tex] with the solution [tex]\omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}} \frac{\lambda t}{\lambda t+M_{B}}[/tex] This is correct according to the answer clue but not according to my answer sheet. Do you think this might be right? Will a drum losing mass have a constant angular momentum?
 

1. How can I effectively approach a challenging problem on a Sunday?

The first step in solving a challenging problem on a Sunday is to set aside dedicated time for it. Choose a time when you are well-rested and have minimal distractions. It is also important to have a positive mindset and believe that you are capable of finding a solution.

2. How do I break down a complex problem into manageable parts?

One strategy is to start by identifying the key components of the problem. Then, break each component down into smaller sub-problems. This will help you tackle each part individually and make the overall problem seem less daunting.

3. What should I do if I get stuck while solving a challenging problem?

It is completely normal to encounter obstacles while solving a challenging problem. When you get stuck, take a step back and try approaching the problem from a different angle. You can also take a break and come back to it with a fresh perspective. Don't be afraid to ask for help or seek out resources for additional support.

4. How do I stay motivated while working on a challenging problem on a Sunday?

One way to stay motivated is to set small, achievable goals for yourself. Celebrate each small victory and use it as motivation to keep going. It can also be helpful to remind yourself of the importance or impact of finding a solution to the problem.

5. How can I improve my problem-solving skills for future challenges?

Problem-solving is a skill that can be developed and improved upon with practice. It is important to reflect on your approach and identify areas for improvement. You can also seek out opportunities to solve different types of problems and learn from your experiences. Additionally, continuously learning and expanding your knowledge in various areas can also enhance your problem-solving abilities.

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