Atwood's fall machine w. sylinder (again?)

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Atwood's fall machine involves two weights, m1 and m2, connected by a massless rope over a frictionless spinning wheel. The discussion focuses on determining the velocities of the weights as a function of their vertical positions using energy considerations. It is established that the system accelerates in the direction of the heavier weight, m2. The potential energy changes for each mass must be considered, with one mass contributing positively and the other negatively. The key takeaway is that the total potential energy should reflect the changes in height, and the velocities of both masses will be the same due to the constraints of the system.
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Homework Statement



Atwood’s fall machine consists of two weight of mass m1 and m2 attached with a massless
rope running around a spinning wheel of mass M and radius R without slipping. The spinning wheel is attached at its center and rotates around an axis through its center without friction.

Find velocity of each of the weights as a function of their vertical positions.

Homework Equations


E = constant

The Attempt at a Solution


I'm assuming that one m1 < m2 so that the system will accelerate in the direction of where m2 is hanging.

I'm sure I should be using energyconsideration here, but the only thing I'm not sure how to attack is the fact that we have two masses on each side of the wheel. What is then my total potential energy to start with? I tought about focusing my attention on the center of mass, but then I don't see how I get the individual velocity of each of the masses.
 
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Hi center o bass! :wink:
center o bass said:
I'm sure I should be using energyconsideration here, but the only thing I'm not sure how to attack is the fact that we have two masses on each side of the wheel. What is then my total potential energy to start with? …

You're only interested in the change in PE …

just add the changes for each mass (one will be positive, and one negative, of course). :smile:
 
I realize that the velocity of each of the masses must be the same. And that kinetic energy therefore goes into each of the masses individually pluss the rotational kinetic energy of the wheel. But I still don't understand how to think about the potential energy... Just sitting there at the same height, they should have a potential energy of mgh each, but then I don't get the correct andswer.
 
Ah...! You said it ;) Thanks alot!
 
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