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Momentum problem#2

  1. Apr 11, 2007 #1
    1. The problem statement, all variables and given/known data

    A(50 kg) and B(60 kg) are sitting at the two extremes of a 4m long boat,standing still in water.They came at the middle.Neglecting water friction,find the displacement of the boat

    2. Relevant equations



    3. The attempt at a solution

    I did it in the following way.Check if I am right.

    I first found the initial position of CM=(32/15)m [A at 0,B at 4m,Boat CM at 2m]
    I then found the final position of CM=2m
    So,displacement of the CM=[(32/15)-2](-i) m assuming x axis pointing right.

    This gives displacement~13.3 m which is close to given answer 13cm.
    But I am not convinced.Since,there is no external force,the CM should be static...
    Please help.
     
  2. jcsd
  3. Apr 11, 2007 #2

    malawi_glenn

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    Have you checked the significant values gicen in the problem and checked that your answer is ok? If only two digits of significance is given in the problem, the answer then should contain two.. and not three. for example
     
  4. Apr 11, 2007 #3
    that plus theres a decimal point issue, either Its .13m or 13cm, not 13.3 m
     
  5. Apr 11, 2007 #4
    OK,what I got previously was 13.3 cm,not 13.3 m.
    I do not really suppose I am right.Why the CM will be displaced at all?There is no external force acting on the system and the system was initially at rest.
    The question asks to determine r_bw=displacement of the boat relative to water.
    I think I may do it if you can tell me what the word "middle" of the boat mean in this problem.Does it mean the geometrical middle?I think it is not.After all,when two people of different masses come closure,it is more likely that one would move more than another.Then I may have some prescription for it.
     
  6. Apr 11, 2007 #5

    malawi_glenn

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    If you change the mass distribution the CM must change.
     
  7. Apr 11, 2007 #6
    are you sure?Change in CM involves the motion of AM,and hence an acceleration.But,here we have external force F=0.
    I have posted another thread on momentum problemRefer to it also.
     
  8. Apr 11, 2007 #7

    malawi_glenn

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    You can look up the definition of the Center of mass if you like and see for yourself.
     
  9. Apr 11, 2007 #8
    Ok,whenever distributionm of mass changes,the CM will in general change.
    But that's a geometrical problem.Here you are dealing with DYNAMICS.
    And since CM was stationary w.r.t. water initially,and F_ext acting on the total system=0,the position of CM w.r.t. water will be the same all the way.
    Because,to change the position of CM relative to water,you are to make it move.So,ther must be an initial acceleration,so that F_ext is not zero.If you are still not convinced,cross this point.

    I think what I did was correct...but,the logic should be different.

    Let me choose the origin of co-ordinates as the geometrical midpoint of the boat.The boat moves relative to water and the origin of co-ordinates will change relative to a frame attached to water.Whereas the position of CM will be stationary relative to water.That is,the boat will move,but, CM will not move relative to water.It is also clear from the statement of the problem.

    So they key to this problem is to find the position of the CM relative to the center of the boat, both before and after A and B move. The difference in the positions of CM relative to the center of the boat must be the distance the boat moved in the water to keep the CM at the same location in the water.
    So... let's do "after" first. When they are both at the center of the boat, the CM is in the middle, relative to the middle, its position is zero!

    Ok, now lets find the cm when they are at opposite ends of the boat. We need to consider the masses and positions of all three objects: A, B and M (the mass of the boat) relative to the center of the boat.

    calc the pos of cm with:

    x_cm = ( 1 / total mass) ( m1x1 + m2x2 + m3x3 )

    person A m = 50, x = -2 meters
    person B m = 60, x = 2 meterrs
    boat 40 kg,x=0

    xcm = ( 1 / 50 + 60 + 40) ( 50*-2 + 60*2 + 40*0 ) =

    = ( 1 / 150 ) ( 20 ) = 0.133 meters

    So,here it is the way to do it.The approach really does not make any problem because,we are interested in the difference.We have used a frame relative to which the CM changes,but,relative to water CM does not change at all.
     
  10. Apr 11, 2007 #9
    thats also how I tried to do it but didn't know the mass of boat until now.
     
  11. Apr 11, 2007 #10
    denverdoc,I am sorry,as I forgot to provide it.
    another method:X_Ab=X_Aw-X_bw
    X_Bb=X_Bw-X_bw
    Now use the definithion of CM,and use that delta R_CM=o
    In RHS,replace X_Aw and X_Bw by (X_Ab+X_bw) and (X_Bb+X_bw)
    It is probably the best method to do it.
     
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