Center of mass involving changing positions

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Homework Statement



Richard, mass 80 kg, and Judy, who is less massive, are enjoying Lake George at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they change seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. Richard notices that the canoe moved 35 cm relative to a submerged log and calculates Judy's mass. What is it?


Homework Equations



Center of mass equation: ([tex]\Sigma[/tex]m*x)/M

M = total mass


The Attempt at a Solution



Ok so since they shifter positions, their center of mass did not change at all. So I set the coordinate system at the center of the canoe.

So after the canoe shifted and after they change their positions, i set both the center of masses together.

m(richard)*-1.5 + m(judy)*1.5 + m(canoe)*0 = m(richard)*1 + m(judy)*-2 + m(canoe)*.5

solving for m(judy) i got 60 kg.

Can anyone tell me what I did wrong?
 

Answers and Replies

  • #2
Doc Al
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The center of mass doesn't change position with respect to the log. It does change with respect to the center of the canoe.
 
  • #3
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The center of mass doesn't change position with respect to the log. It does change with respect to the center of the canoe.
I'm not quite understanding what you are saying. Doesn't the center of mass shift to the opposite side of where its original position?
 
  • #4
Doc Al
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Doesn't the center of mass shift to the opposite side of where its original position?
Yes--with respect to the center of the canoe. Relate the change in its position to the change in position of the canoe.
 
  • #5
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Yes--with respect to the center of the canoe. Relate the change in its position to the change in position of the canoe.
I still don't really understand what you are saying.......Sorry but bear with me, center of mass is not my forte
 
  • #6
Doc Al
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I still don't really understand what you are saying.......Sorry but bear with me, center of mass is not my forte
You were partly on track. The left hand side of your equation made sense, but not the right hand side.

Just as you say, the center of mass shifts from one side of the canoe to the other. Draw yourself a diagram of how the canoe must move to keep the center of mass fixed. Set that change in canoe position equal to the amount given in the problem and then you can solve for the unknown mass.

(Don't be too hard on yourself--this kind of problem is a bit tricky.)
 
  • #7
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So when it says that it moves 50 cm relative to a log, is that the distance that the log moved to keep its center of mass at a fixed point?
 
  • #8
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Ahhhh you're offline............crap, can anyone else help me?
 
  • #9
Doc Al
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So when it says that it moves 50 cm relative to a log, is that the distance that the log moved to keep its center of mass at a fixed point?
The log doesn't move--think of it as just sitting quietly in a calm lake. The canoe moves in order to keep the center of mass of "canoe + Richard + Judy" at a fixed point. (I thought the canoe moved 35 cm, not 50.)

Hint: Where is the center of mass with reference to the center of the canoe? Say it starts out at a distance X to the left of the center. (Write an expression for that distance in terms of the masses.) When they swap positions where does the center of mass end up? Of course we know that it's the canoe that moved, not the center of mass--so how much did the canoe's center move? Use that to set up your equation.
 

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