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Rope,potential energy

  1. Dec 6, 2011 #1
    Hi everyone

    1. The problem statement, all variables and given/known data

    See picture in attachment.




    2. Relevant equations
    -


    3. The attempt at a solution

    well, the solution should be:

    [tex] V=\rho g[(h-h_{2}) \frac {h+h_{2}} {2}+(h-h_{1}) \frac {h-h_{1}}{2}][/tex]

    where rho is the density per length of the rope. I don't understand where (h+h2)/2 and (h+h1)/2 come from. When I tried to solve it, my guess for the potential energy (just for one 'side' let'say say) was: [tex] \rho g (h-h_{1})*h[/tex] but that's wrong. Can anyone help me out?

    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Dec 6, 2011 #2

    Doc Al

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    Staff: Mentor

    Hint: Where's the center of mass of the left hand side?
     
  4. Dec 6, 2011 #3
    For each side use PE = mgh were h is the position of the centre of mass of m.
     
  5. Dec 6, 2011 #4
    oops!
     
  6. Dec 6, 2011 #5
    There was a mistake in my formula:

    [tex] V=\rho g[(h-h_{2}) \frac {h+h_{2}} {2}+(h-h_{1}) \frac {h+h_{1}}{2}][/tex]

    Thanks for the help. You both say I shall use the center of mass on both sides. I guess that's where the 1/2 comes from. My problem is here, if I want the center of mass of one side, why do I have to add them like (h1+h)/2. I thought If I want to center of mass of the rope on one side I'd had (h-h1)/2. Where's my mistake?
     
  7. Dec 6, 2011 #6

    Doc Al

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    Staff: Mentor

    You want the location of the center of mass measured from the ground.
     
  8. Dec 7, 2011 #7
    Yes thank you.I found me error in reasoning. :)
     
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