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Center of Mass of draining Soda can

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform soda can of mass 0.140kg is 12.0cm tall and filled with 1.31kg of soda. Then small holes are drilled in the top and the bottom (with negligible loss of metal) to drain the soda. The can is placed upright such that the soda only drains from the bottom hole. What is the height h of the center of mass of the can and contents (a) initially and (b) after the can loses all the soda? (c) What happens to h as the soda drains out? (d) If x is the height of the remaining soda at any given instant, find x when the center of mass reaches it lowest point.


    2. Relevant equations
    rcom=1/M[tex]\Sigma[/tex]miri

    M=total mass of the system

    3. The attempt at a solution
    a) hi = 6cm = .060m
    b) hf = 6cm = .060m
    c) h will decrease as the soda begins the drain from the can. However, as more and more soda drains over time, there will eventually be a point when h reaches a minimum and then begins to increase. Once all the soda is drained h will return to 6cm.
    d) Move 1/M to the other side of the equation, such that:
    MRcom= [tex]\Sigma[/tex]canmiri+[tex]\Sigma[/tex]sodamiri

    [tex]\Sigma[/tex]canmiri = constant = (.140kg)*(.120m)= .0168kg*m

    I think I need to rewrite a [tex]\Sigma[/tex]sodamiri as a function such that the height of the draining soda varies with the height?

    Then I could find xmin and then get the center of mass of the system.

    Can anyone help me through the rest of this?

    Thanks.
     
  2. jcsd
  3. Nov 1, 2009 #2

    Andrew Mason

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    Homework Helper

    Write out the equation of the centre of mass. You can use the symmetry of the container and column of liquid to determine it.

    Let h be the height of the liquid. Let y = 0 be the bottom and y=12 be the top. The cm of the can is at y=6. The cm of the liquid is h/2. The mass of the liquid is mL = m0(h/12). So the centre of mass of the can + liquid is a point y such that

    [tex]m_L(h/2 - y) + m_c(6-y) = 0[/tex]

    See if you can find the value h for minimum y from that.

    AM
     
  4. Nov 6, 2009 #3
    still missing this somehow. isnt ml varying aswell as y?
     
  5. Nov 6, 2009 #4

    Andrew Mason

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    Yes, of course. I even gave you the expression for m_L. All you need to do is solve for h at minimum y. hint: what is the rate of change of y with h? What is dy/dh at minimum y?

    AM
     
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