Centre Of Mass Physics?

Question

Ricardo, of mass 81 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 29 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 2.6 m apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moves 46.5 cm horizontally relative to a pier post during the exchange and calculates Carmelita's mass. What is it?

Answer 1

To solve this, we have to assume that there are no external forces acting in the horizontal direction on the canoe. We will also neglect any viscous drag force of the water on the canoe, which would be present in any real situation. In this case, the center of mass of the canoe cannot move horizontally relative to the pier. What is happening, though, is that the center of mass of the canoe is changing position relative to a reference frame anchored to the canoe (i.e., relative to a fixed point in the canoe itself).

For simplicity (though this is not a necessary assumption, it just makes the math a bit easier -- the selection of the coodinate origin is arbitrary), let's say the center of the canoe is initially alongside the pier post, and that the mass distribution of the canoe is symmetric about the middle of the canoe (so that the center of mass of the empty canoe coincides with the middle of the canoe). Call the location of the pier post x = 0 Let the seats be located at coordinates x1 and x2 relative to the pier. Let m1 and m2 be the masses of Carmelita and Ricardo, respectively, and let m3 be the mass of the empty canoe.

Then, relative to the pier post, the center of mass of the loaded canoe is located at:

C = (x1*m1 + x2*m2 + 0*m3)/(m1 + m2 + m3)

After the passengers switch positions, and the center of the boat has moved a distance d relative to the post, the center of mass of the loaded canoe is given by:

C = ((x1+d)*m2 + (x2+d)*m1 + d*m3)/(m1 + m2 + m3)

But, relative to the post, the position of the center of mass of the loaded canoe cannot have changed, so we can equate the right hand sides of the above equations:

(x1*m1 + x2*m2 + 0*m3) = ((x1+d)*m2 + (x2+d)*m1 + d*m3)

m1 = (m2*(x1 - x2 + d) + m3*d)/(x1 - x2 - d)

In this case, x1 = -x2 = -1.3 m
m2 = 81 kg and m3 = 29 kg

We are not told which direction the boat moves when the passengers switch places (i.e., does it move to the right or left? Is d = 46.5 cm or -46.5 cm). We are, however, told that Carmelita has less mass than Ricardo, and this is how we can distinguish between the two possibilities for the value of d:

Plugging in all the values above, we find that

if d = 46.5 cm, m1 = 52.02 kg
if d = -46.5 cm, m1 = 122.6 kg

Carmelita's mass is 52.02 kg.


See sources for other examples of this type of problem.
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The previous answerer's response is incorrect, and the analysis he cites is flawed when applied to this case.


It is true that the total horizontal momentum of the canoe/boat must always be zero, because no external horizontal forces are acting on the system (we are neglecting drag forces). By equating the position of the center of mass of the canoe before and after the passengers chance positions, I have essentially used the same principle. However, the momentum of the system is given by the mass of the loaded boat multiplied by the velocity of the center of mass of the boat, not the velocity of a fixed point on the boat. The previous answer used the latter and came to the wrong conclusion.

Answer 2

Hmmm, I'm not an expert but perhaps the location of the center of mass doesn't change with respect to the pier but does change with respect to the canoe's center. The boat moves 46.5 cm in order to keep the location of the center of mass unchanged with respect to the pier. If this is so, calculating Carmelita's mass should be straightforward.

Answer 3

I believe that "hfshow" gave the correct answer from textbook point of view but he did not understand the answer from Alexander. That is the point, that if mu != 0 then the center of mass DOES move and the boat does not. Center of mass moves because there are external forces on the boat.
Alexander, you mentioned yourselves that there is a discontinuity at mu->0. I believe that school problem assumes that mu = 0, so from this point of view the latter answer is correct. And second, I did not read your problem solution but does the result hold if the drag is not linear?

Answer 4

Dear Kyle,
your teacher does not know what he's teaching.

Please, print the following solution by Prof. Shimrod of your problem, and submit it to your incompetent teacher:

http://answers.yahoo.com/question/index?qid=20070425122736AAnWr64