Bubba has a barrel in the shape of a cylinder of mass 39.4 kg. The barrel has a diameter of 62.4 cm and is 1.32 m tall. He fills the barrel to a depth of 49.5 cm with loose packed grain that has an effective density of 0.676 g/cm^3. Where is the center of mass of the barrel partially filled with grain?
Density = Mass/Volume
Volume = h x pi x r^2
Volume = 49.5cm x pi x 31.2cm^2 = 151,378.52cm^3
Density x Volume = Mass
Mass = 0.676 g/cm^3 x 151,378.52cm^3 = 102,231.88g
Total Mass = Mass of Barrel + Mass of Grain
Total Mass = 39.4kg + 102.33kg = 141.73 kg
I think my method to finding total mass was correct.
Now, in order to find the center mass in terms of x,y, and z. I will need to integrate each coordinate from 1 to 3.
So, $X_{com}$ = ${1}\over M$ $\sum_{i=1}^3 (m_1x_1) $ = ${m1x1 + m2x2 + m3x3} \over M$
But, I'm not sure what to use for these variables.
This looks a little more complicated than it needs to be. Find the center of mass of the grain and the center of mass of the barrel, then take a weighted average of the two using the mass of each as the scaling factor.
I'm going to assume the barrel is of constant width (if the material is thicker in some areas that makes the calculation harder) and that it has a lid that has the same mass as the base.
For each point x, you can find y such that dist(x, center line) = dist(y,center line) and x and y lie on opposite sides of the center. Therefore, the center of mass for the grain, the barrel and the grain barrel system will lie on the line between the center of the lid and the center of the base.
We also have a height symmetry in that for either the barrel or the grain (but not both together) any disk at a height h < half the total height can be paired with a disk of the same mass at height (half total height) + h, where total height refers to the height of the material in question.
Therefore, by these two symmetric arguments, I claim the center of mass of the barrel is at height 66 cm along the center line and the center of mass of the grain is at height 24.75 cm along the same center line.
The center of mass of the system is therefore at (66 *39.4 + 24.75 * 102.23188)/141.73 = 36 cm