# Atwood's Machine using Angular Momentum

1. Jun 19, 2011

### 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?

2. Jun 19, 2011

### Staff: Mentor

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.

3. Jun 19, 2011

### Lil_Aziz1

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.

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?

4. Jun 20, 2011

### Staff: Mentor

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.

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.

5. Jun 20, 2011

### Lil_Aziz1

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: Jun 20, 2011
6. Jun 20, 2011

### Staff: Mentor

Sure. Why not?
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!

Nah. There's nothing wrong with measuring the angular momentum of everything about the axis of the pulley.

As I stated above, you can certainly solve for the acceleration using the same method as in the first problem.

7. Jun 20, 2011

### Lil_Aziz1

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.

8. Jun 20, 2011

### Staff: Mentor

You can. Write a separate equation for each system, then combine them.
Yes, that should work fine.

9. Jun 20, 2011

### Lil_Aziz1

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

10. Jun 20, 2011

### Staff: Mentor

Your first line seems wrong. If you are looking at the pulley, then:
$$T_1R - T_2R = \frac{d}{dt}(I\omega)$$

11. Jun 20, 2011

### Lil_Aziz1

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

Thanks a ton!