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
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
~ 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
Post a Comment