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Center of mass of a Canoe

  1. May 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A 55 kg man is standing on a 65 kg and 4.0 m length canoe that floats without friction on water.
    The man walks from a 0.75m from one end of the canoe to another point 0.75 from the other end of the canoe.
    What distance does the canoe move?

    b87gug.png
    2. Relevant equations
    Center of mass equation.
    3. The attempt at a solution
    A tried looking for the ΔXCM, the change on the center of mass, but I do not know how to compute the canoe's CM.

    Mod note: Fixed misspelling of "canoe."
     
    Last edited by a moderator: May 25, 2015
  2. jcsd
  3. May 25, 2015 #2

    Orodruin

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    The center of mass for the canoe does not matter, only its displacement does.
     
    Last edited by a moderator: May 25, 2015
  4. May 25, 2015 #3
    So; I should compute CMo = CMf?
    CMo=(1/M)⋅∑xn ⋅ mn
    CMf=(1/M)⋅∑xn ⋅ mn

    And calculate ΔCM=Cf - Co

    With this I get ΔCM= -0.33 but I do not know how to proceed forward.
     
  5. May 25, 2015 #4

    Orodruin

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    Well, first of all, you should define all of your quantities, I cannot read your mind to know what Cf and Co are and what sets them apart from CMo and CMf.

    Second, the difference in the center of mass is a length and must be given in terms of a length unit.
     
  6. May 25, 2015 #5
    Sorry :frown: Here goes the complete reasoning;

    CMo=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅0.75 kg⋅m)+(65⋅2 kg⋅m)] = 1.43 m
    CMf=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅ 3.25 kg⋅m)+(65⋅2 kg⋅m)] = 2.57 m

    And calculate ΔCM=CMf-CMo= (2.57-1.43)m = 1.14 m

    Would be this the displacement of the boat?
     
  7. May 25, 2015 #6

    Orodruin

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    What you have computed is the change in the center of mass in the boat frame. How does this relate to how far the boat moves? What assumptions go into that relation?
     
  8. May 25, 2015 #7
    So, If I applied the reference frame, I guess I should undo it. Then by substracting it to each center of mass I should be given the result; so:
    dcanoo= (xocanoo-ΔCM)= (2 - 1.14)m = 0.86 m
    Would it be correct?
     
  9. May 25, 2015 #8

    Orodruin

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    No. What is true about the center of mass of both man and boat (assuming that there is no transfer of momentum to the water)?
     
  10. May 25, 2015 #9
    I think the momentum is conserved, no externa forces are acting on the system.
     
  11. May 25, 2015 #10

    Orodruin

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    So, assuming that there is no momentum to begin with, what happens with the center of mass?
     
  12. May 25, 2015 #11
    The conoe does not move? :oldeek:
     
  13. May 25, 2015 #12

    Orodruin

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    No, the center of mass does not move. You have computed that the center of mass moves by 1.14 m (1.15 m with correct rounding) it a frame that is attached to the canoe. How far must the canoe move for this to happen at the same time as the CoM does not move in the lake frame?
     
  14. May 25, 2015 #13
    What if I tried a different approach, as CMo=CMf=1.43 m And calling the displacement "x";
    1.43 ⋅ 120 m⋅kg = 55 ⋅ 3.25 m⋅kg + 65 ⋅ x kg ; solving for x i get x ≈-0.11 m
    And this would be the displacement of the canoe, I think.
     
  15. May 25, 2015 #14

    Orodruin

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    No, this is not correct. Why switch methods when you are basically done?
     
  16. May 25, 2015 #15
    I do not understand what is actually happening within the lake's reference frame. I do not know where is the origin for that frame :(
     
  17. May 25, 2015 #16

    Orodruin

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    You do not need an origin (you can pick one wherever you like).

    It may be beneficial to start looking in the lake frame, where the center of mass does not move. What does this tell you about the displacements of the man and boat, respectively?
     
  18. May 25, 2015 #17
    So, if in the lake's frame the CM of the man and the boat does not move even if they do. This may suggest the displacement of the man is somehow proportional to the boat's?
    I mean, if the man walks to the right, then the boar moves to the left?
     
  19. May 25, 2015 #18

    Orodruin

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    Correct, but how much?
     
  20. May 25, 2015 #19
    The distance the man walks minus the change in the center of mass (in the boat's frame)?
     
  21. May 25, 2015 #20

    Orodruin

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    I suggest you only look at the lake frame. What is the relation between how far the canoe moves and how far the man moves in that frame?
     
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