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A pulley of mass m using conservation of energy

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider a massive pulley of mass m and radius r shown in the figure below, with two objects hanging off it having masses m and 2m, respectively. The pulley can be considered as a uniform disk, and the string is massless, does not stretch and does not slip.
    By considering conservation of energy, determine the acceleration of the masses.
    Screen Shot 2017-03-06 at 9.42.22 PM.png
    2. Relevant equations
    $$I=\frac{1}{2}mr^2$$
    $$K=\frac{1}{2}mv^2$$
    $$K=\frac{1}{2}I\omega^2$$

    3. The attempt at a solution
    I think the total kinetic energy of the system is $$K=\frac{1}{2}mv^2+\frac{1}{2}(2m)v^2+\frac{1}{2}I\omega^2$$ but I'm not sure how to calculate the total potential energy of the system.
     
    Last edited: Mar 6, 2017
  2. jcsd
  3. Mar 6, 2017 #2

    BvU

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    Don't see any equation involving energy in your relevant equations ?
    Also don't see anything with acceleration ?
    Do see unexplained variable ##r##, though. Unknown ?
     
    Last edited: Mar 6, 2017
  4. Mar 6, 2017 #3
    I don't know how to get the acceleration without using Newton's Laws, which is what the question is asking me to do.
     
  5. Mar 6, 2017 #4

    BvU

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    DId you render the problem statement completely, litterally ? I read it asks you to do it with energy conservation.
     
  6. Mar 6, 2017 #5
    Sorry that is what I meant to say. I am supposed to do it with conservation of energy rather than Newton's laws but I am unsure how to do so.
     
  7. Mar 6, 2017 #6

    BvU

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    I repeat posts 2 and 4.
     
  8. Mar 6, 2017 #7
    I'm unsure of what you're asking.
     
  9. Mar 6, 2017 #8

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    From post #2:
    What is ##r## ?
    Is it known or unknown ?
    You now edited in ##K##. What about potential energy ?
    What about acceleration ?

    From post #4:
    What is the complete problem statement ?
     
  10. Mar 6, 2017 #9
    r is unknown. I'm not sure about potential energy because I don't know where the ground is defined. I'm also not sure about acceleration.

    I copied the problem statement exactly as I saw it in my homework problems.
     
  11. Mar 6, 2017 #10

    BvU

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    Ok, means you are supposed to formulate the answer in terms of ##m## and ##r##. Funny they don't even mention ##r## -- perhaps it divides out ultimately; we'll see.

    You still need more equations (there are various relationships you haven't listed).
    Ground can be anywhere. So pick something (you can even pick the ceiling if you want). More importantly: what coordinate system do you plan to define and use ?
     
  12. Mar 7, 2017 #11
    I think I'll use Cartesian coordinate system with h as the initial height from the ground in both masses. Then the finish height would be 0 for mass 2m? Although I'm not sure about the final height for the second mass.

    I suppose another relevant equation would be $$v^2-v_0^2=2a\Delta x$$
     
  13. Mar 7, 2017 #12

    BvU

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    That is indeed the energy conservation equation. You can take ##v_0 = 0##.

    m and 2m are connected by a rope (you implicitly use that already with the single variable ##v##)
    the rope doesn't slip on the pulley



    To be honest, I still don't see how this can be answered using energy considerations only. Maybe you can get something containing acceleration by differentiating a form that contains ##v^2## :rolleyes: .
    If I were you I would embark on a solution attempt with a more conventional approach and then see if it can be bent to accommodate the instruction in the last line of the problem statement.
     
  14. Mar 7, 2017 #13
    I think I have a solution. First define the two masses as being h metres off the ground initially. Then by the conservation of energy,
    $$E_{initial}=E_{final}$$
    $$\implies mgh+(2m)gh=\frac{1}{2}mv^2+\frac{1}{2}(2m)v^2+\frac{1}{2}I\omega^2+mg(2h)$$
    $$\implies mgh=\frac{3}{2}mv^2+\frac{1}{4}mr^2\frac{v^2}{r^2}$$
    $$\implies gh=\frac{7}{4}v^2$$
    Now we use the kinematic equation: $$v^2-v_0^2=2a\Delta x$$ which gives
    $$\frac{4gh}{7}-0^2=2ah$$
    and finally, $$a=\frac{2g}{7}$$
     
  15. Mar 7, 2017 #14

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    Consistent with ##mg## having to cause 3.5 ##\times {1\over 2}mv^2##. Well done !

    I like this exercise very much: both h and r divide out -- which I hoped for, but it wasn't obvious at the start -- and something simple remains.
     
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