# Atwood's Machine using Angular Momentum

• Lil_Aziz1
In summary, the conversation discusses using angular momentum to solve an Atwood's machine problem. The book uses the force of gravity as the force acting on the pulley instead of the tension because the tension is an internal force and cancels out. Treating the pulley and masses as separate systems can also be done but may make the problem more complex. However, tension can still be considered an external force in the second problem. It is also possible to solve the first problem using angular momentum by treating the masses as separate systems. In the second problem, treating the entire Atwood's machine as one system would not work because the tension is internal and cannot be directly solved for. The conversation also discusses the difficulty of solving the problem using separate systems
Lil_Aziz1
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:

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.,

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?

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.

Sorry about that. I did mess up on the algebra, but I still got a different acceleration:

Lil_Aziz1 said:
Sorry about that. I did mess up on the algebra, but I still got a different acceleration:
Your first line seems wrong. If you are looking at the pulley, then:
$$T_1R - T_2R = \frac{d}{dt}(I\omega)$$

:O It all makes sense. You are a genius.

Thanks a ton!

## 1. What is an Atwood's Machine?

An Atwood's Machine is a simple mechanical system consisting of two masses connected by a string or rope over a pulley. It is used to demonstrate the principles of classical mechanics, specifically the laws of motion and conservation of energy.

## 2. How is angular momentum used in Atwood's Machine?

Angular momentum is used in Atwood's Machine to explain the motion of the masses as they move towards equilibrium. As the system moves, the angular momentum of the masses and the pulley remains constant, according to the law of conservation of angular momentum.

## 3. What factors affect the angular momentum in Atwood's Machine?

The angular momentum in Atwood's Machine is affected by the masses of the objects, their distance from the center of rotation, and their angular velocity. The length and tension of the string also play a role in determining the angular momentum.

## 4. How does Atwood's Machine demonstrate the conservation of energy?

Atwood's Machine demonstrates the conservation of energy by showing how the potential energy of the system is converted into kinetic energy as the masses move. As the masses accelerate towards equilibrium, the potential energy decreases while the kinetic energy increases, but the total energy remains constant.

## 5. What real-life applications use similar principles to Atwood's Machine?

Atwood's Machine principles are used in various real-life applications, such as elevators, cranes, and even amusement park rides. These systems use pulleys and ropes to lift and move heavy objects, and the principles of conservation of energy and angular momentum are essential for their safe and efficient operation.

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