Why do I keep getting a slightly different result from the following two ways of determining the center of mass of a rigid, geometrically simple object?

The object is a rectangular 5(x) by 7(y) sheet of uniform rigid material, with a 2(x) by 3(y) rectangle missing from the upper right corner.

Method 1)

The official formula in my textbook:

x center of mass = ( m1x1 + m2x2 ) / ( m1 + m2 )

y center of mass = ( m1y1 + m2y2 ) / ( m1 + m2 )

This gives results of x = 2.18 and y = 3.09.

Method 2)

A torque calculation that would seem more intuitive to me:

The xy coordinate plane is parallel to floor and I watch it from above. Object is in the first quadrant (top right), touching the x and y axes.

I imagine balancing the object on a razor blade parallel to the y-axis. I solve for x by assuming that the torques on the left and right side must cancel out.

I rotate the razor blade so that it's parallel to the x-axis. I balance the object again. I solve for y by assuming again that the torques on each side must cancel out.

This gives results of x = 2.07 and y = 2.9.

The difference between the results given by the two methods is small, but significant. What's going on?

I checked my math several times and even tried a different problem with a simpler geometry. Again the results differed by a small but significant value. I'm fine with having to learn the textbook method, but would like to know why the torque approach gives results that are 3-5 % off.

  • 2
    $\begingroup$ Please post the math you used in "Method 2" $\endgroup$ – Carl Witthoft Jan 13 '14 at 13:45

Why do you assume that in the case of torques, the torques must cancel out, in my opinion the best way to deal with this problem is your method 1, but you can solve by taking torques as follows :

When you are balancing along y-axis (calculating x)

$6A × g × 4 + 29A × g × x = 35A × g × 2.5$
Here A is mass per unit area, 4 is the x coordinate of centre of mass of small cut off block, x is the coordinate of centre of mass of left over block and 2.5 is the coordinate of original block

We have equate the combined torque of cut off and left off pieces with that of original piece, you can do similar operation for y coordinate

$6A × g × 5.5 + 29A × g × y = 35A × g × 3.5$

These operations give the same results as those obtained by method 1, so there is no error.


I would use statics to do this. Very intuitive and easier I think. Simply sum moments about one corner of the composite shape for each sub-area in the x and y directions and solve for the x,y coordinate of the centroid or center of mass if the shape is constant thickness.


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