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Center of mass problem

  1. Oct 31, 2004 #1
    A right cyllindrical can with mass M, height H, and uniform density is initially filled with soda of mass m. We punch small holes in the top and bottom to drain the soda; we then consider the height of the center of mass of the can and any soda within it. If x is the height of the remaining soda at any given instant, find x in terms of M, H, and m when the center of mass reaches its lowest point.

    I know that center of mass = (M(H/2) + m(x/2)) / M+m -- except that m will have changed as soda is lost, and I'm not sure how to find it.
     
  2. jcsd
  3. Nov 1, 2004 #2

    Tide

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    At any given instant the amount of soda remaining is the initial mass multiplied by the ratio of x to H.
     
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