Atwood's Machine using Angular Momentum

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Discussion Overview

The discussion revolves around the application of angular momentum in solving an Atwood's machine problem, specifically addressing the forces acting on the pulley and the treatment of tension in the system. Participants explore the differences between using angular momentum and Newton's second law in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions why the textbook uses the force of gravity acting on the pulley instead of tension, noting a previous problem solved using Newton's second law.
  • Another participant explains that tension is internal to the system and that the torques it exerts on the pulley and hanging masses cancel each other out.
  • There is a discussion about why the torques are attributed to gravity rather than tension, with some participants expressing confusion over this distinction.
  • Participants debate whether the first problem can be solved using angular momentum while treating the masses as separate systems, with differing opinions on the feasibility of this approach.
  • One participant suggests that treating the entire Atwood's machine as one system is necessary for calculating angular momentum, while others argue that separate systems can also be used.
  • Concerns are raised about the implications of treating tension as an external force in different problem contexts, with some participants unsure about the validity of their reasoning.
  • There is a correction regarding algebraic errors in the calculations of linear acceleration, with participants encouraged to double-check their work.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of tension and the applicability of angular momentum in solving the Atwood's machine problem. No consensus is reached on the best approach, and confusion remains regarding the cancellation of forces and the classification of systems.

Contextual Notes

Participants highlight limitations in their understanding of the system's dynamics, particularly regarding the treatment of internal versus external forces and the implications for angular momentum calculations. There are unresolved questions about the algebraic steps leading to different results for linear acceleration.

Who May Find This Useful

This discussion may be useful for students and educators interested in the application of angular momentum in mechanical systems, particularly in the context of Atwood's machines and the interplay between forces and torques.

Lil_Aziz1
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Hey everyone. I'm kind of stumped on this example from my textbook. It uses angular momentum to solve an Atwood's machine problem. Here is how the problem and the solution goes:

angmomentum.png


My question is, why does the book use force of gravity as the force acting on the pulley instead of the tension? the book did this problem previously using only Newton's second law, i.e.,
atwoodmachine.png


I did the angular momentum problem but replaced m_{1}g and m_{2}g with T_1=m_1(g-a) and T_2=m_2(g+a), respectively. Consequently, for linear acceleration I got a=\frac{m_1-m_2}{m_1+m_2+0.5M}\frac{g}{2} (notice the g/2)

Can anyone explain to me why it's doing that?
Thanks in advance.
 
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Lil_Aziz1 said:
My question is, why does the book use force of gravity as the force acting on the pulley instead of the tension?
The force acting on the pulley is the tension, but you're not just finding the angular momentum of the pulley. You need to consider the torques acting on the entire Atwood's machine, including the hanging masses. Note that the tension is internal to the system and that the torque it exerts on the pulley is equal and opposite to the torque it exerts on the hanging mass--thus it cancels out.
 
Doc Al said:
The force acting on the pulley is the tension, but you're not just finding the angular momentum of the pulley.
That makes sense but this doesn't explain (at least not to me) why the torques are the forces of gravity and not the tension. I have no problem with the angular momentum part of the question, it's just the torques.

Doc Al said:
You need to consider the torques acting on the entire Atwood's machine, including the hanging masses. Note that the tension is internal to the system and that the torque it exerts on the pulley is equal and opposite to the torque it exerts on the hanging mass--thus it cancels out.
This was my first surmise but why didn't the tensions cancel in the second problem? It didn't because tension was considered an external force (right?). Why can't we make tension an external force in the first problem?
 
Lil_Aziz1 said:
That makes sense but this doesn't explain (at least not to me) why the torques are the forces of gravity and not the tension. I have no problem with the angular momentum part of the question, it's just the torques.
You need the total torque due to all the forces acting on the system. It's just that the torques due to the tension forces add up to zero, since they are internal to the system as a whole.

This was my first surmise but why didn't the tensions cancel in the second problem? It didn't because tension was considered an external force (right?). Why can't we make tension an external force in the first problem?
They solve the second problem by treating the pulley and mass as separate systems. With respect to those systems, the tension is an external force. In the first problem, they choose the entire Atwood's machine as the system, which makes the tension forces internal.

You can certainly solve the first problem by treating the pulley and the two masses as separate systems, in which case the tensions are external forces. That's the way the problem is usually solved.
 
Ah that kind of makes sense. I'll sleep on it tonight. :)

One more question: can we solve the first problem using angular momentum like done above but treat the masses as different systems?
You can't solve the second problem using tension and choosing the entire Atwood's machine as the system because that wouldn't help at all. I can't really describe why it won't work but my gut says it won't. Ahh now I have the feeling you can because I can't describe why one cannot solve it using only one system.

EDIT: Okay I think I know the answer to the question on why we can't use separate systems for the angular momentum problem.
One cannot use separate systems because he would not be able to get the angular momentum with respect to the z axis drawn on the diagram. He would be unable to compute the angular momenta of the two masses in respect to O without treating the whole Atwood's machine as one system. We would have to get the angular momentum of mass m_1 with respect to itself, which it zero, so that doesn't help.

Now I would be obliged if someone could explain to me why we can't solve the second problem (using t = I\alpha and stuff) treating the whole Atwood's machine as one system.
 
Last edited:
Lil_Aziz1 said:
One more question: can we solve the first problem using angular momentum like done above but treat the masses as different systems?
Sure. Why not?
You can't solve the second problem using tension and choosing the entire Atwood's machine as the system because that wouldn't help at all. I can't really describe why it won't work but my gut says it won't. Ahh now I have the feeling you can because I can't describe why one cannot solve it using only one system.
Since you are asked to find the tension, and the tension is internal to the entire system (when you treat disk + hanging mass as the system), you won't be able to solve for the tension directly using that method. But you can certainly solve for the acceleration that way. Try it!

EDIT: Okay I think I know the answer to the question on why we can't use separate systems for the angular momentum problem.
One cannot use separate systems because he would not be able to get the angular momentum with respect to the z axis drawn on the diagram. He would be unable to compute the angular momenta of the two masses in respect to O without treating the whole Atwood's machine as one system. We would have to get the angular momentum of mass m_1 with respect to itself, which it zero, so that doesn't help.
Nah. There's nothing wrong with measuring the angular momentum of everything about the axis of the pulley.

Now I would be obliged if someone could explain to me why we can't solve the second problem (using t = I\alpha and stuff) treating the whole Atwood's machine as one system.
As I stated above, you can certainly solve for the acceleration using the same method as in the first problem.
 
Doc Al said:
Sure. Why not?
Nah. There's nothing wrong with measuring the angular momentum of everything about the axis of the pulley.
Oh. Then why can't we solve the first problem using angular momentum when treating the Atwood's machine as three different systems? Actually, isn't that what I did when I replaced m_1g and m_2g with T_1=m_1(g−a) and T_2=m_2(g+a), respectively. Consequently, for linear acceleration I got
a=\frac{m1−m2}{m1+m2+0.5M}\frac{g}{2}
(notice the g/2)
I got a different acceleration.
 
Lil_Aziz1 said:
Oh. Then why can't we solve the first problem using angular momentum when treating the Atwood's machine as three different systems?
You can. Write a separate equation for each system, then combine them.
Actually, isn't that what I did when I replaced m_1g and m_2g with T_1=m_1(g−a) and T_2=m_2(g+a), respectively.
Yes, that should work fine.

Consequently, for linear acceleration I got
a=\frac{m1−m2}{m1+m2+0.5M}\frac{g}{2}
(notice the g/2)
I got a different acceleration.
Double check your algebra.
 
Sorry about that. I did mess up on the algebra, but I still got a different acceleration:
angmom.png
 
  • #10
Lil_Aziz1 said:
Sorry about that. I did mess up on the algebra, but I still got a different acceleration:
angmom.png
Your first line seems wrong. If you are looking at the pulley, then:
T_1R - T_2R = \frac{d}{dt}(I\omega)
 
  • #11
:O It all makes sense. You are a genius.

Thanks a ton!
 

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