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Center of mass and momentum in boat question

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A person with mass m is sitting on one end of a lifeboat of uniform density. Mass of lifeboat is 3m. The person then walks a distance of L to the other side of the lifeboat. Which of the following statements best characterizes the motion of the lifeboat? "

    1. The lifeboat shifts by the same distance traversed, but in the opposite direction of his motion.

    2. The lifeboat shifts by L/4 in the opposite direction of the persons motion.

    3. The lifeboat shifts in the same direction as the persons motion with constant velocity.

    4. The lifeboat does not shift forward or backward as a result of the persons motion.
    Specifically what I dont understand is what is actually happening in the problem and also why. Also what is the underlying assumption that is made for a boat in water? Ths problem confuses me.

    The thing i want to know before applying any formulas is what is happening in the problem conceptually. Would it be the same idea if you walked on a boat that was in space, so its about no friction?

    2. Relevant equations
    center of mass formula: x = m1x1 + m2x2 / m1m2
    conservation of momentum: m1v1 = m2v2

    3. The attempt at a solution
    I need some explanation of what Im being asked in the problem before I apply the formulas. Thanks
     
  2. jcsd
  3. Feb 16, 2016 #2

    JBA

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    Consider what happens at each end of the trip and how they relate
     
  4. Feb 16, 2016 #3

    TSny

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    Yes, you are to assume that there is no significant friction between the boat and the water.
     
  5. Feb 16, 2016 #4
    oK thanks. So one of the reasons they are asking about person in boat is that it does away with friction right?
    I have considered each end of the trip.
    Before the guy goes anywhere the center of mass is a combination of the boats mass and his mass, then apply the forumla, but ihave question about the formula first
    You get the distance from each mass to some arbitrary point. But why is it an arbitrary point? Is this because it is the ratio that matters?
     
  6. Feb 16, 2016 #5

    TSny

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    The arbitrary point just corresponds to an arbitrary choice of origin for your coordinate system. Once the coordinate system is fixed, each mass in the system will have coordinates that you can plug into the formula to get the coordinates of the center of mass. If you choose a different origin, then the coordinate values of the masses and the center of mass would change. But the location of the center of mass relative to the masses would not change. So, you can choose the origin at any convenient location.
     
  7. Feb 16, 2016 #6
    OK thanks i undrstand so far
    so the convenient location for origin could be the starting point (the point before the guy walks across boat) right? If yes than center of mass before he goes anywhere is:
    center of mass = m1x1 + m2x2 / (m1 + m2)
    = m*0 + 3m*(L/2) / (4m)
    = 3/2L

    is this correct so far? thanks
     
  8. Feb 16, 2016 #7

    TSny

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    OK except for the very last step. Note that an answer of (3/2) L would put the CM of the system beyond the far end of the boat.
     
  9. Feb 16, 2016 #8
    I mean (2/3)L
     
  10. Feb 16, 2016 #9

    TSny

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    In simplifying 3m*(L/2) / (4m), you should get an answer less than L/2. The center of mass of the system should lie between the person and the center of the boat.

    Note that

    ##3m(L/2) / (4m) = 3m\frac{L}{2} \cdot \frac{1}{4m}##
     
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