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Systems of Particles

  1. Oct 18, 2007 #1
    A man of mass m clings to a rope ladder suspended below a balloon of mass M. The balloon is stationary with respect to the ground.

    a) If the man begins to climb the ladder at a speed v (with respect to the ladder), in what direction and with what speed (with respect to the Earth) will the balloon move?

    b) What is the state of motion after the man stops climbing?


    When the man is climbing the ladder, the centre of mass will be moving upwards, so the balloon will be moving upwards aswell with respect to the earth.

    And after the man stops climbing, the balloon will return to a stationary state, with lower altitude since it has greater mass.

    Is my thinking here correct?

    I'm not sure how to begin calculating the v of the balloon, I think I need to calculate the change in centre of mass/the derivate of the centre of mass with respect to time?

    Any help would be much appreciated, thanks :)
     
  2. jcsd
  3. Oct 18, 2007 #2
    So if we are just working in the y direction

    ycm = (m1y1 + m2y2) / (m1+m2)
    ycm = (My1 + my2) / (M+m)

    Vcm = dycm/t = [M(dy1/dt) + m(dy2/dt)] / (M+m)
    Vcm = (Mv1 + mv2) / (M+m)
     
  4. Oct 18, 2007 #3
    Do I have to regard this as a many-particle system? A 2 particle system (man and balloon)? Or 2 many-particle systems?

    If we need to look at it as 2 many particle systems? Will I need to use integral calculus? Because we we're shown many particle systems in the form on integrals but aren't expected to use it to solve problems since it's a first year course.
     
  5. Oct 18, 2007 #4
    Vcm for balloon = (1/M)Σmnvn

    Vcm for man = (1/m)Σmnvn

    or can I do

    Vcm for system = (1/(M+m))Σmnv
     
  6. Oct 18, 2007 #5
    am I doing any of this correctly?
    I'm really confused on the concept of centre of mass, I have read over all my notes and my textbook section on it and I don't understand how to apply it.

    Thanks
     
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