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Help settle an argument! Mass system on frictionless pulleys

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    I got into a disagreement with a fellow student today. He insisted that it's possible to simplify the system on the right as the system on the left. I agree with him that the difference in potential energy (and, therefore, the kinetic energy) is the same but I insist that the final velocities of the two systems will differ. We ended up surveying 6 different physics / engineering students and all agreed with him at a glance. One did text me tonight that he's on my side after having done the problem on paper.

    All of the engineering and physics profs were at lunch and couldn't be reached. I intend to ask at least one of them in Physics 2 in the morning.

    Given:
    Qv0RuzG.png

    Release both systems from rest (initial velocity = 0). Assume no moment of inertia for pulleys, massless, non-stretchy cable. You can simplify the box with mass=0 as an empty rope end if you wish.

    2. Relevant equations

    I am basing my predicted results on energy theory, since there are no outside forces acting on the system, Ei = Ef.

    3. The attempt at a solution

    I worked this out on paper and believe it confirms my intuition that you cannot equate the two systems to simplify a problem.
    S0vZRwN.png

    Thanks for your input!
     
  2. jcsd
  3. Apr 18, 2013 #2

    TSny

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    Your analysis is correct except that your final equation relating ##v_1## and ##v_2## is missing a square root in part of the equation.

    Definitely, ##m_1## will aquire more speed as it falls. It is essentially in free fall with acceleration ##g##. Masses ##m_3## and ##m_4## will have much less acceleration (about .005##g##). The net force accelerating the two systems is the same, but the system on the right has much more inertia.
     
  4. Apr 18, 2013 #3

    ehild

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    You are right, the problem can not be "simplified". Both the acceleration and the speed of the blocks depend both on the sum and the difference of the masses.

    ehild
     
  5. Apr 18, 2013 #4

    Simon Bridge

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    Depends on the problem - depends what you mean by equate".

    By context - the problem is to find the speed of the left-hand mass after falling, from rest, through a height h.

    You can see that the second case is cannot be equivalent because, if m4 were removed, then the final speed of m3 would be the same as the final speed of m1 (acceleration of gravity does not depend on mass remember?) Adding any mass to the right hand side can only reduce the final speed.

    No need for calculations - nothing but physics :approve:

    You nay be able to convince the other, without appeal to authority, by starting out with the first case ... they should agree.
    Staying with the first case - ask what will happen if any mass is added to the RHS.
    Then go to a bigger mass on the left and zero on the right ... ask what the final speed will be compared with the first case... (correct if needed) ... then ask what would happen if any mass is added to the RHS.
     
  6. Apr 18, 2013 #5
    Thank you! That's what I kept saying! Yes that's a typo on the bottom, I was considering substituting one expression into the other so was changing around the square roots and stopped halfway.
     
  7. Apr 18, 2013 #6
    The original problem is to find the velocity right before impact. Originally there was a moment of inertia for the pulley, on which the rope could not slip, so angular velocity had to be considered.

    The student wanted to simplify by subtracting the two masses, (he came up with the 0, 1, 100, 101 in the diagram) and I said he couldn't because the total mass affects the system's inertia and therefore affects the final velocity.

    Everyone else agreed with him. Their rationale was that it was gravity doing the acceleration, and gravity is a fixed acceleration (not dependent on mass), which is true enough on earth's surface. But in this problem Tension is involved.

    We discussed it to death and nobody would agree with me. The root of the problem was that the acceleration was gravity and everyone seemed to think that gravity didn't care how much total mass there was given the same Δm.
     
  8. Apr 18, 2013 #7

    Simon Bridge

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    I think that these sorts of misconceptions need to be destroyed quickly.
    What I would do is set up an experiment - a teacher should have access to stopwatches, pulleys, string, and weights, for class use? I usually just improvise... they can time the fall.

    i.e. I'd start them with a 150g and 100g mass, compare with a 550g and 500g mass - see they are different ... then get them to stay with 500g on one side and plot fall-time vs counter-mass for increments of 50g.

    That should drive it home.
    Then use the guiding questions from the last paragraph post #4, on the board, to sum up.

    The last nail should be an observation about the power of empiricism.
    It is good that they are not just taking your word for it though.

    (Note: there is tension involved in a pendulum as well but the period does not depend on the mass there either - it is the weight acting in different directions that makes the difference in this case.)
     
    Last edited: Apr 18, 2013
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