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.

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:

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:

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 - x^{4} /4] ^{a }_{0}

= [ a^{ 4} /2 - a^{4} /4]-
0

= a^{4} /4

__x__= a^{4} /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.

_{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)

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