Center of Mass and Linear Momentum?

Question

A metal soda can of uniform composition has a mass of 0.140 kg and is 12.0 cm tall. The can is filled with 1.31 kg of soda. Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height h of the center of mass of the can and contents (a) intially 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 its lowest point.

Answer 1

Since the can is of uniform construction, its COM is at the point in the center of the can (i.e., half its height and at the center of its area, assuming a cylindrical can). The COM of the liquid is going to be 1/2 the remaining height of liquid in the can, again at the center of the area.

a, b) The height of the COM in both cases is 6cm. The height of the COM of the can itself is 6cm, no matter what the liquid level. When all the liquid is present, the height of its COM is 6cm. When there is no liquid, you only have the can.

c) As the liquid drains, the height of the COM decreases. The soda is about 0.11 kg/cm of height. If M is the mass of remaining soda and h is its height at any given time, the height of the COM (x) will be:
x = [6*.140 + (h/2)*(0.11*h)] / (.140 + .11*h)
= (.84+.055*h^2) / (.140+.11h)

d) To find the minimum value of x, you could either determine the derivative of the the function above or plot a graph. As my calc is very rusty, I did a quick graph in Excel. The minimum occurs at about 2.84cm.

Answer 2

In this kind of problems, air resistance is ignored. From your data, both fragments have the same vertical velocity (0), so they will reach ground at the same time (as well as the center of mass). But the center of mass moves as if no explosion took place (conservation of momentum) and reaches ground (assumed flat) at distance d. First fragment drops directly, from top of trajectory, so it travels d/2. Since the center of mass travels distance d, it follows that the second fragment travels distance 3d/2.