Does Changing Material Density Affect the Center of Mass in a Composite Object?

[1MileCrash]

My textbook is giving awfully complicated formulas for centers of mass for actual objects (if they are a system of massive points, that's simpler for me.)

Intuitively though, it just seems like it would be a weighted average??

So,

If I have a solid metal cube, divided into two halfs, one iron, and one with a density of less than that of iron, does it have the same center of mass as if the cube was pure iron, but the half that was "formerly" a less dense metal was "smushed" along the axis perpendicular to the line the cube was divided among so that it has the same mass as it did when the lighter metal was there?

 

[daqddyo1]

The centre of mass of an object is the point at which it can be balanced (in a gravitational field). It is usually difficult though to balance something exactly at its C of M as it is usually inside the object.
If you could balance the object on a point, then the C of M would be directly above that point.
If you suspend the object using a (massless) string, then the C of M would be directly below and in line with the string.
In your examples, think about trying to balance the cube at different points on its surface. What would you discover?

Does the C of M stay fixed or does its position change?

A neat C of M experiment is to take two forks and stick them together at their tines so that the reulting shape is a rough parabola. Where is the centre of mass of the pair of forks?
To find it, determine where it can be balanced - by sticking a toothpick into the forks so that you can balance them by resting the other end of the toothpick on your finger.

 

[sophiecentaur]

My textbook is giving awfully complicated formulas for centers of mass for actual objects (if they are a system of massive points, that's simpler for me.)

Intuitively though, it just seems like it would be a weighted average??

So,

If I have a solid metal cube, divided into two halfs, one iron, and one with a density of less than that of iron, does it have the same center of mass as if the cube was pure iron, but the half that was "formerly" a less dense metal was "smushed" along the axis perpendicular to the line the cube was divided among so that it has the same mass as it did when the lighter metal was there?

The CM of the two different blocks can be calculated by first working out where the CMs of the two individual blocks is. Then using those two CMs and the two masses, you can work out the resulting CM position. That's not too complicated.
Many of these problems can be made simpler by doing them in stages.

Your "weighted average" description is just about exactly what I described.

 

[256bits]

1Milecrash,
Just take note that the CM is computed for the x, y and z axis seperately, and where they meet is the real CM. Take a donut - CM right in the middle of the donut since it is symetrical on all 3 axis. Half a donut - well it is symmetrical on 2 axis but not the third, so you have to do your calculation for that 3rd axis. Most statistics books at the back have tables listing different shapes and you can use these to obtain your weighted averages. If it is a really oddball shape than you just have to go through with all the calculations