Center of mass in glass of water

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SUMMARY

The discussion focuses on calculating the height of the center of mass (c.o.m) in a glass filled with water, represented by the function h(x) = (m_b*h_b + m_c*h_c + m_w*h_w) / (m_b + m_c + m_w). The parameters include m_b = 10 grams for the bottom, m_c = 50 grams for the cylinder, and m_w = 9πx for the water mass. The critical point occurs at x = 2.550553126 cm, where h(x) reaches its minimum value of 3.550553126 cm. This indicates that the c.o.m is consistently 1 cm above the water level when at its lowest, prompting inquiries about the physical implications of this phenomenon.

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yomakaflo
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Homework Statement


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?

Homework Equations

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?
 
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yomakaflo said:

Homework Statement


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?

Homework Equations

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?
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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