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Center of mass after particles move

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    a 90 kg block 1 and 360 kg block 2 are on opposite ends of a 300 kg slab of length 12.0 m which is on a frozen lake. assume no friction between slab and ice. block 2 moves 4.0 m towards block 1 while block 1 moves 2.50 m towards block 2. how far does the slab move?

    2. Relevant equations

    c.m.= (Ʃmr)/Ʃm


    3. The attempt at a solution

    i am kind of lost on how to approach the problem. i know the center of mass does not change because there are only internal forces. i dont know how the motion of the blocks relates to the motion of the slab. any help would be appreciated.
     
  2. jcsd
  3. Dec 11, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    A method that I find works for me with this sort of problem is to first consider things in the slab's frame of reference -- ignore the ice for now. Draw a picture showing the blocks in their initial position on the slab. If you set an origin at the center of the slab it will coincide with the slab's center of mass, which is handy since the slab will then disappear from the numerator of the center of mass calculations (because the distance of the slab's center of mass from the origin will always be zero). Now locate the horizontal center of mass for the two remaining objects in this coordinate system.

    Then redraw the picture with the blocks in their new locations and repeat the operation, locating the new center of mass in the slab frame of reference.

    Since, as you've stated, the overall center of mass of the system can't move with respect to the ice frame of reference, then the slab must have shifted by the distance between the centers of mass you calculated in the slab's frame.
     
    Last edited: Dec 11, 2012
  4. Dec 11, 2012 #3
    thank you very much for your help. i thought of setting the reference point in the center of the slab. However, I never thought of isolating the slab from the ice. I like this method far more than how I have been taught
     
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