Finding the Height of Remaining Soda at Center of Mass's Lowest Point

[jenavira]

A right cyllindrical can with mass M, height H, and uniform density is initially filled with soda of mass m. We punch small holes in the top and bottom to drain the soda; we then consider the height of the center of mass of the can and any soda within it. If x is the height of the remaining soda at any given instant, find x in terms of M, H, and m when the center of mass reaches its lowest point.

I know that center of mass = (M(H/2) + m(x/2)) / M+m -- except that m will have changed as soda is lost, and I'm not sure how to find it.

 

[Tide]

At any given instant the amount of soda remaining is the initial mass multiplied by the ratio of x to H.

 

[wais]

To find the height of the remaining soda at the lowest point of the center of mass, we can use the principle of conservation of mass. This means that the initial mass of the soda (m) must be equal to the final mass of the soda (m').

Initially, the total mass of the can and soda is M + m. As the soda drains out, the mass of the can remains constant at M while the mass of the remaining soda decreases to m'. We can express this as:

M + m = M + m'

Now, we can rearrange this equation to solve for m':

m' = m - (M - M)

m' = m - M

This means that the final mass of the remaining soda is equal to the initial mass of the soda minus the mass of the can. Now, we can substitute this value for m' in the equation for the center of mass:

x = (M(H/2) + m'(x/2)) / M + m'

x = (M(H/2) + (m - M)(x/2)) / M + (m - M)

x = (MH/2 + mx/2 - Mx/2) / M + m - M

x = (mx/2 + MH/2 - Mx/2) / m

x = (m/M)(x/2) + H/2 - (M/M)(x/2)

x = (m/M)(x/2) + H/2 - x/2

x = [(m - M)/M](x/2) + H/2

Therefore, we can see that the height of the remaining soda at the lowest point of the center of mass is given by:

x = [(m - M)/M](x/2) + H/2

where m is the initial mass of the soda, M is the mass of the can, and x is the height of the remaining soda. This shows that as the soda drains out, the height of the remaining soda decreases and the center of mass of the can moves towards the bottom.