ADONGO'S CENTRAL POINT THEORY

CENTER OF MASS

The Central(or Gravitational) Point Theory has application to many practical situations in Physics. Some of its practical areas are Center of Mass and Moment of Initial. In practical situation it is convenient to regard thin sheets of material, such as copper stripping, as two-dimensional. The area of plane that represents a two-dimensional distribution of matter is called lamina.

In general, however, substances are not homogeneous and so the mass density is variable, suppose a lamina is represented by a region R of the xy-plane and its mass density ϱ= ϱ(x, y), varies continuously over the region R.

The center of mass (or center of gravity) of a lamina, is defined as;

 mx_g = M_y …(1)

my_g = M_x …(2)

 The  later is a purely geometric property of the lamina and coincides with the center of mass in the case of a variable density will almost always  have its center of mass off-center. The variable mass density of a lamina which is directly proportional to the distance from the vertex opposite any side due to symmetric character of the region R is given by

 ϱ(x, y) =kx^m + ky^m ……..(3)

 Also, the mass density of lamina which is directly proportional to the distance from the vertex opposite any side to non-symmetric character is given by

ϱ(x, y) = k_xx^m + k_yy^m ……(4)

Further analysis of the formulae above, we have

ϱ(x, y)*m = k(M_x^m + M_y^m) …..(5)

ϱ(x, y)*m=k_xM_x^m + k_yM_y^m……..(6)

If the degree is equal to 1, then for symmetric character, we have;

M_x = [ϱ(x, y)*m]/2k ……..(7)

M_y = [ϱ(x, y)*m]/2k………(8)

Also, if the degree is equal to 1, then for non-symmetric character, we have

M_x = [ϱ(x, y)*m]/2k_x …….. .(9)

M_y = [ ϱ(x, y)*m]/2k_y………(10)

If the mass density ϱ is directly proportional to the  square (second degree) of the distance from the vertex opposite any side, then for symmetric character, we have

M_x = [√ϱ(x, y)*√m]/√2k ……..(11)

M_y = [√ϱ(x, y)*√m]/√2k………(12)

For non-symmetric character, we have

M_x = [√ϱ(x, y)*√m]/√2k_x ……..(13)

M_y = [√ϱ(x, y)*√m]/√2k_x………(14)

EXAMPLE

Find the moment of the mass with respect to x-axis and y-axis if the mass density and entire mass a lamina is 3 and 8 in the shape of an isosceles right triangle  to the square of the distance from the vertex opposite the hypotenuse.

SOLUTION

The mass density of the lamina is given as;

ϱ(x, y) = kx^m + ky^m

but ϱ(x, y)=3

Due to symmetric character of the region R and the mass density we apply equation (7) and (8).

M_x=√(3*8)/√2k = 3.464/k

M_y=√(3*8)/√2k = 3.464/k

The center of mass (x_g, y_g) must lie on the line y=x. Consequently, in equation (1) and (2), we have

x_g = M_x/m=0.43/k

y_g = M_y/m=0.43/k


NOTE: The symbols k_x, k_{y and k is denoted as non-symmetry constant in x-axis, non-symmetry constant in y-axis and symmetry constant respectively.}



 REFERENCE
*Adongo Ayine William(Me). Transcript(2008), posted(EMS-Bolgatanga Branch) to Mathematical Association of Ghana( Tittle: New Mathematical Concept..........,) in the year 2008. 
*Rene Descartes.(1637). "The Geometry".