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.
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:
[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?