a) The center of mass of the can is halfway up; the center of mass of the soda in the can is also halfway up when the can is full of soda, so the center of mass is
(0.135*(12.4/2) + 1.25*(12.4/2))/(0.135+1.25)
b) After the can loses all the soda, the center of mass is the center of mass of the can, which is halfway up or 12.4/2
c)
x/12.4 is the percentage of soda that has been drained out.
The center of mass of the remaining soda is x/2. The mass of the remaining soda is 1.25*(x/12.4)
Therefore, the COM of the system of the can + soda is:
h = (0.135*(12.4/2) + 1.25*(x/12.4)*(x/2))/(0.135+1.25*(x/12.4)
Graph this function, and see what happens to it as a function of h
d) Evaluate dh/dx = 0
A uniform soda can of mass 0.135 kg is 12.4 cm tall and filled with 1.25 kg of soda (figure below). Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. (Initially the soda can is full.)(a) What is the height h of the com (center of mass) of the can and contents initially. cm (b) What is the height h of the com of the can after the can loses all the soda? cm (c) What happens to h as the soda drains out? ---Select--- rises to the top decreases then rises again stays the same decreases to the bottom (d) If x is the height of the remaining soda at any give instant, find x when the com reaches its lowest point. cm
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