This is more a of a theory question of the atwood and energy

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SUMMARY

The discussion centers on the mechanics of an Atwood machine involving two masses, m_1 and m_2, where m_1 is greater than m_2. The key conclusion is that both masses will meet at the midpoint (h/2) due to their connected nature, ensuring they accelerate at the same rate. The analysis confirms that if one mass descends by a certain distance, the other ascends by the same distance, reinforcing the principle of equal and opposite forces in this system. The conversation clarifies that the assumption of a non-stretchable rope is crucial for this behavior.

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  • Understanding of Newton's laws of motion
  • Familiarity with basic kinematics
  • Knowledge of Atwood machine mechanics
  • Concept of tension in inextensible ropes
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flyingpig
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Homework Statement



I picked this up somewhere and I couldn't get a good answer

Let's say I have m_1 and m_2 hanging by a pulley on an atwood. Let m_1 > m_2 and m_2 on ground and m_1 hanging in the air at a height h

So if I ask you, what is the speed when they meet, what is the set up?

The Attempt at a Solution



Now for some reason, they will always meet at h/2, I don't know why. Why would it be even h/2?? I am not awake right now so I am probably overlooking something
 
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The blocks are connected, and hence they accelerate at the same rate. If they met anywhere other than the midpoint, it would imply that one of the objects is accelerating faster than the other.
 
Hi flyingpig! :smile:

m_1 and m_2 are connected by a rope.
When m_1 descends by h/2, m_2 ascends by h/2.
 
So something like meeting at h/3 never happens?
 
No.

If m_1 ascends by h/3,
then m_2 descends by h/3 ending up at (2/3)h.
 
Look at it this way: if two objects are subject to the same force in opposing direction, and are separated by a distance h, they will always meet in the middle. It is when the forces are are not coordinated and the acceleration of one is not equal and opposite to the other that the objects will meet anywhere else than 0.5h. Hope that helps.
 
I like Serena said:
No.

If m_1 ascends by h/3,
then m_2 descends by h/3 ending up at (2/3)h.

Yet the masses are different
 
flyingpig said:
Yet the masses are different

If the rope were elastic that would matter, but we're assuming a rope that does not stretch.
 
I like Serena said:
No.

If m_1 ascends by h/3,
then m_2 descends by h/3 ending up at (2/3)h.

I had to read this again and I think I understand now from your mistake (yeah who would've thought?) from my mistake

I stated that m_2 is on ground level, so m_2 can't descend. But the better way to understand is that if m_1 goes down h/3, m_2 climbs up h/3

sandy.bridge said:
Look at it this way: if two objects are subject to the same force in opposing direction, and are separated by a distance h, they will always meet in the middle. It is when the forces are are not coordinated and the acceleration of one is not equal and opposite to the other that the objects will meet anywhere else than 0.5h. Hope that helps.


Thisis just kinematics then. v0 = 0, but the acceleration has to be the same since they are tied to a rope like you said

Thank you all for answering
 

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