Using Integration to Find the Center of Mass of a Solid Object

 Obtaining the center of mass for a system or object is an important one in general physics.  Indeed, we've already seen the technique for obtaining the center of mass of a binary star system with the line of centers treated as a 1-dimensional problem, e.g.

Revisiting Binary Stars 

From calculus-based general physics (or classical mechanics) , the general equations for the center of mass of a 3D solid as a whole are given by:

x  =     ò x dm /   ò dm,    y  =     ò y dm / ò dm,   

z  =     ò z dm /   ò dm

The derivation, which will not be done here, is generally by adding the moments (for basic principles of moments see:

• Mechanics Problems Solutions (Simple Machines: Pt. 9)

Of all the concentrated masses in all the volume elements  D V  and taking the limit as the  D V 's approach 0.

An interesting introductory problem for a planar 3D object, say a wafer-thin, homogeneous triangular plate with height h and base dimension b, is laid out as shown in the sketch of mind below:

The basic procedure entails dividing the triangle into strips of width dy parallel to the x-axis. The mass of the strip then is approximately:

dm =  d dA

Where dA =    ℓ  dy

With  ℓ   the width of the triangle at the distance y above its base. Using similar triangles we get:

ℓ  / b  =  (h - y) / h

Then:

ℓ  =  b/h  (h - y) 

So the element of mass will be expressed:

dm =   (d )   b/h  (h - y) dy

For the entire plate we can then write for the y-coordinate of the center of mass:

y  =     ò y dm /   ò dm

=  ò ^b ₀   y db/h  (h - y) dy/  ò ^b ₀   d b/h  (h - y) dy

= h  / 3

Thus, the center of mass of the uniform triangular plate lies at a distance one third of the way toward the opposite vertex.

An interesting introductory problem  for a 3D object is to find the position of the center of mass for a uniform solid hemisphere – as shown in the graphic below.

  This is take to have a radius a, and a uniform density,r,  and we note the center of the plane face of the hemisphere is the origin.

The x-axis is perpendicular to this plane strip or face.  By symmetry the center of mass will lie on the x-axis.  We begin by considering an elemental strip of thickness  d x with its plane face perpendicular to the x-axis and distance x from the origin.  If its radius is given by y the volume of the disk is:

d V  =  p y ²  d x

With the element of mass given by:

d m  =  p y ²  d x r

And so we can write for the x -coordinate of the center of mass for the elemental disk:

x   ~  å ^a  _x=o   x p y ² r d x/  å d m

Or:

~   å ^a  _x=o   x p y ²  r d x/  Total mass

 

In the limit as  d x  -> 0

x =  ò ^a ₀   x p y ² r dx 

And:   2 p a ³ r/ 3 =

p r   ò ^a ₀   x y ² dx 

 

x 2 a ³ / 3 =  

ò ^a ₀   x (a ² -   x ²) dx 

=   ò ^a ₀   (xa ² -   x ³)  dx 

= [ a ² x ² /2  -   x⁴  /4] ^a ₀

= [ a ⁴ /2  -   a⁴  /4]-  0  

=  a⁴  /4 

  x=   a⁴  /4    ( 3/2 a ³ )  =   3 a /8

Suggested Problems:

1) A straight, thin rod PQ of length 2a of constant cross sectional area A is of variable density, r.  

Given r  = r_o (1 + 2x ² / a ²)  where x is the distance along the rod from the end of P and r_o  is a constant.  Find the position of the center of mass. 

2) Find the center of mass of a solid hemisphere of radius r if the density at any point P is proportional to the distance of P from the base of the hemisphere.  (Hint:  Add another lower latitutde layer to the upper latitude circle below to form a thin slice of thickness dy.  Then imagine the whole hemisphere cut into slices of thickness dy, i.e. by planes perpendicular to the y-axis)