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Center of Mass and Translational Motion

  1. Mar 14, 2013 #1
    Suppose two friends, with masses m2 = 1.3 m1, are on a perfectly smooth, frictionless, frozen lake.
    They are both holding the end of a rope of length Lo .
    a. Find the position of the center of mass, in terms
    of Lo , from the smaller person.
    b. If the two pull half of the rope in such that the final length of the rope is L = Lo/2, find the new position of the center of mass from the smaller person.
    c. Find the distance each person moves from their original positions.




    2. Relevant equations



    3. Ive got no clue
     
  2. jcsd
  3. Mar 14, 2013 #2

    haruspex

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    What is the definition of centre of mass?
     
  4. Mar 14, 2013 #3
    I am not sure of what exactly you are asking but center of mass is equal to

    Xcm= (m1X1 + m2X2) / (m1 + m2)
     
  5. Mar 14, 2013 #4

    ehild

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    You have to find the distance of the centre of mass from the smaller person. So put the origin at the smaller person. What are X1, X2 then?

    ehild
     
  6. Mar 14, 2013 #5
    I believe X1 would be 0 and that would make X2 the whole length making it equal to Lo?
     
  7. Mar 14, 2013 #6

    ehild

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    Yes. So where is the centre of mass?

    ehild
     
  8. Mar 14, 2013 #7
    This is that part of the equation that has been tricking me up.

    I want to say that
    Xcm=(m2X2) / (m1+m2)

    So I believe I can take out m2 from each side of the equation giving me
    Xcm=(X2)/(m1) (?)
     
  9. Mar 14, 2013 #8

    HallsofIvy

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    That's just bad arithmetic: 5/(6+ 5) is NOT equal to 1/6.
     
  10. Mar 14, 2013 #9
    So I cannot cancel out the m2.

    I am not positive where to go on from Xcm=(m2X2) / (m1+m2)
    Would this be my final equation or is there more I can do?
     
  11. Mar 14, 2013 #10

    mfb

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    That is your final equation for the center of mass, you cannot simplify it any more.

    Edit: Oh right. You can use your given ratio for the masses, of course.
     
    Last edited: Mar 14, 2013
  12. Mar 14, 2013 #11

    ehild

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    X2=Lo and m2=1.3 m1. So what is the position of the CM in terms of Lo?
     
  13. Mar 14, 2013 #12
    I think I have it now.
    Xcm= (1.3m1*Lo) / 2.3m1

    Xcm=.57Lo (which makes this the distance from the smaller person) ?

    And for the second part, would this mean i set it up like

    Xcm=(1.3m1*.5L) / 2.3m1
     
    Last edited: Mar 14, 2013
  14. Mar 14, 2013 #13

    haruspex

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    yes.
     
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