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Center of mass in glass of water

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data
    So we have a glass consisting of a 10 cm long cylinder on top of a 1 cm long bottom standing on a table. The radius is 3 cm. For every x in [0, 10] we let h(x) be the height of the center of mass when we fill the glass with x cm water. That is, h(x) is the distance from the table to the center of mass.

    We have h(x) = (m_b*h_b + m_c*h_c + m_w*h_w) / (m_b + m_c + m_w), with
    m_b = mass of bottom = 10 gram
    h_b = height of center of mass in bottom = 0.5 cm
    m_c = mass of cylinder = 50 gram
    h_c = height of center of mass in cylinder = 6 cm
    m_w = mass of water = 9*pi*x
    h_w = height of center of mass in the water = 1 + x/2 cm

    Now the task is to find where h(x) decreases and increases for x in [0, 10].
    Which height of water gives the lowest center of mass in the glass, and what is h(x) for that x-value?


    2. Relevant equations


    3. The attempt at a solution
    By differentiating h(x) I found that x = 2.550553126 makes h(x) be the lowest. h(2.550553126) = 3.550553126. This puzzled me. Why is the center of mass exactly 1 cm over the water level when h(x) is at its lowest? Does someone have a mathematical or physical explanation to this?
     
  2. jcsd
  3. Oct 28, 2015 #2

    SteamKing

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    When the glass is empty, it has a certain c.o.m. located above the table. Putting a little water in the bottom of the glass will lower the overall c.o.m. of the glass + water a little, but the c.o.m. may still be located above the surface of the water.

    You can't really do an analysis of this system unless you know more about the glass and its construction.
     
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