A perfectly cylindrical container standing upright with a top and bottom (an oil drum
or coke can for example) has an empty mass of M and a height of H. It is filled with
a liquid of uniform density of variable mass m up to height h. When the container is
full the centre of gravity is in the centre (H/2). As the container is emptied and m
reduces the centre of gravity moves down. Once the container is empty the centre
of gravity is again at the centre (height H/2). Using calculus calculate the value of h
in terms of H, m and M when the centre of gravity is at its lowest position.
See:
http://www.jiskha.com/display.cgi?id=1305630971
To calculate the value of h when the center of gravity is at its lowest position, we need to find the location of the center of gravity as a function of h.
Let's assume that the diameter of the cylindrical container is D.
First, let's define the variables:
- M: mass of the empty container (constant)
- H: total height of the container (constant)
- m: mass of the liquid in the container (variable)
- h: height of the liquid inside the container (variable)
To determine the location of the center of gravity, we'll consider two main components: the empty container and the liquid inside it.
1. Center of gravity of the empty container:
The center of gravity of the empty container can be assumed to be at its geometrical center, which is at H/2.
2. Center of gravity of the liquid:
The center of gravity of the liquid can be determined by considering it as a collection of infinitely small horizontal layers. Each layer has a height dh and is at a distance y from the top.
To find the position of the center of gravity of each layer, we'll assume that the density of the liquid is constant and equal to ρ (rho). The mass of each layer can be expressed as dm = ρ * A * dh, where A is the cross-sectional area of the layer.
Now, let's derive an expression for the position of the center of gravity of the liquid in terms of h.
The distance y from the top to the layer can be given as y = h - dh/2.
The distance of the center of gravity of each layer from the top can be obtained by integrating the product of the position of each layer (y) and its mass (dm) over the entire height h:
G = ∫(y * dm) = ∫(y * ρ * A * dh)
We need to express y in terms of h. Using the previously defined equation, y = h - dh/2, we can substitute it in the integral:
G = ∫((h - dh/2) * ρ * A * dh)
Now, let's solve this integral:
G = ρ * A * ∫(h * dh) - ρ * A * ∫(dh/2 * dh)
G = ρ * A * (h * h/2) - ρ * A * (dh/2 * h)
G = ρ * A * (h^2/2) - ρ * A * (h * dh/2)
The first term, ρ * A * (h^2/2), represents the moment due to the height of the liquid. The second term, ρ * A * (h * dh/2), represents the moment due to the slope of the liquid surface.
Now, we need to find the minimum position of G by setting its derivative with respect to h equal to zero and solving for h:
dG/dh = ρ * A * h - ρ * A * dh/2 = 0
Simplifying the equation:
ρ * A * h - ρ * A * dh/2 = 0
ρ * A * h = ρ * A * dh/2
h = dh/2
Therefore, h = 2*dh
Since we want to find the lowest position of the center of gravity, we need to find the value of h that minimizes G. From the equation above, we can see that h is directly proportional to dh.
Therefore, the lowest position of the center of gravity occurs when the liquid is completely empty, and h = 0.
Thus, when the center of gravity is at its lowest position, h = 0.
Note: This solution assumes a uniform density for the liquid and neglects the thickness of the container walls.