Center of Mass of draining Soda can

AI Thread Summary
The discussion focuses on calculating the center of mass of a soda can filled with liquid as it drains. Initially, the center of mass is at 6 cm, and it remains at this height after the soda is completely drained. As the soda drains, the center of mass decreases until it reaches a minimum point before rising again. The participants discuss the need to express the mass of the liquid as a function of its height to find the minimum center of mass. The conversation emphasizes the relationship between the height of the liquid and the center of mass, aiming to derive the equation for the center of mass during the draining process.
jpalmer91
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Homework Statement


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.


Homework Equations


rcom=1/M\Sigmamiri

M=total mass of the system

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= \Sigmacanmiri+\Sigmasodamiri

\Sigmacanmiri = constant = (.140kg)*(.120m)= .0168kg*m

I think I need to rewrite a \Sigmasodamiri 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.
 
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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

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

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

AM
 
still missing this somehow. isn't ml varying as well as y?
 
jpalmer91 said:
still missing this somehow. isn't ml varying as well as y?
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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