restart:
with(student);
(1)
Pick an (a, b, c) value to place our point mass m in xyz-space.
a:=3;b:=2;c:=6;
(2)
Consider a sphere centered at (1,1,1) with radius 1.
Then (x-1)^2 + (y-1)^2 + (z-1)^2 = 1. Solving for z,
z = 1 +/- sqrt(1 - (x-1)^2 - (y-1)^2) are the upper and lower limits.
Similarly, the projection onto the xy-plane is the circle with radius 1, and center (1,1), i.e., (x-1)^2 +
(y-1)^2 = 1
so y = 1 +/- sqrt(1 - (x-1)^2) are the upper and lower limits for y, with x going from 0 to 2.
If the density is 1, then dM = 1dV (infinitesimal mass at (x,y,z).)
The force of gravity on a point mass m at (a, b, c) from the infinitesmal mass dM at (x, y, z) is dF =
GmdM/r^2*<x-a, y-b, z-c>/r
where r = sqrt( ( x-a)^2 + (y-b)^2 + (z-c)^2 ), so
dF = G*m*dM<x-a, y-b, z-c>/sqrt[ ( x-a)^2 + (y-b)^2 + (z-c)^2 ]^3 dV or
dF = G*m*dM<x-a, y-b, z-c>/[( x-a)^2 + (y-b)^2 + (z-c)^2] ^(3/2) dV
Picking units so that G*m = 1, the force in the x-direction is
dF_x = G*m*dM*(x-a)/[( x-a)^2 + (y-b)^2 + (z-c)^2] ^(3/2) dV
The total force in the x-direction is the triple integral over the solid:
Tripleint((x-a)/((x-a)^2+(y-b)^2+(z-c)^2)^(3/2), z=1-sqrt(1(x-1)^2-(y-1)^2)..1+sqrt(1-(x-1)^2-(y-1)^2), y=1-sqrt(1-(x-1)^2)
..1+sqrt(1-(x-1)^2), x=0..2); aa:=evalf(%,6);
(3)
For a sphere, you can always find the force by putting all the mass at the center of mass.
If p=1, then M=4/3pi(1)^3, and (1,1,1,) is the center (of mass).
So the force in the x direction,
F_x = G*m*M*(x-a)/[( x-a)^2 + (y-b)^2 + (z-c)^2] ^(3/2) becomes
F_x = 4/3*pi*(1-a)/[( 1-a)^2 + (1-b)^2 + (1-c)^2] ^(3/2),
4*Pi/3*(1-a)/((1-a)^2+(1-b)^2+(1-c)^2)^(3/2);evalf(%);
(4)
For the tetrahedron bounded by the coordinate planes and 2x + 3y + z = 6, with p=1, the mass will be
Tripleint(1, z=0..6-2*x-3*y, y=0..-2/3*x+2, x=0..3); m:=value(%)
;
(5)
z_bar = 1.5
Tripleint(z, z=0..6-2*x-3*y, y=0..-2/3*x+2, x=0..3); M_xy:=value
(%);z_bar:=M_xy/m;
(6)
y_bar = 0.5
Tripleint(y, z=0..6-2*x-3*y, y=0..-2/3*x+2, x=0..3); M_xz:=value
(%);y_bar:=M_xz/m;
(7)
x_bar = 0.75
Tripleint(x, z=0..6-2*x-3*y, y=0..-2/3*x+2, x=0..3); M_yz:=value
(%);x_bar:=M_yz/m;
(8)
with(plots);
(9)
(9)
plot with center of mass
p:=plot3d(6-2*x-3*y, y=0..-2/3*x+2,x=0..3,axes=normal,scaling=
constrained,grid=[10,10], style=wireframe,orientation= [136,79])
: sp:=plot3d([1/2+0.03*cos(t)*sin(s),3/4+0.03*sin(t)*sin(s),
3/2+0.03*cos(s)], t=0..2*Pi,s=0..Pi,scaling=constrained,axes=
normal):display(p,sp);
Again,
If the density is 1, then dM = 1dV (infinitesimal mass at (x,y,z).)
The force of gravity on a point mass m at (a, b, c) from the infinitesmal mass dM at (x, y, z) is dF =
GmdM/r^2*<x-a, y-b, z-c>/r
where r = sqrt( ( x-a)^2 + (y-b)^2 + (z-c)^2 ), so
dF = G*m*dM<x-a, y-b, z-c>/sqrt[ ( x-a)^2 + (y-b)^2 + (z-c)^2 ]^3 dV or
dF = G*m*dM<x-a, y-b, z-c>/[( x-a)^2 + (y-b)^2 + (z-c)^2] ^(3/2) dV
Picking units so that G*m = 1, the force in the x-direction is
dF_x = G*m*dM*(x-a)/[( x-a)^2 + (y-b)^2 + (z-c)^2] ^(3/2) dV
The total force in the x-direction is the triple integral over the solid:
Tripleint((x-a)/((x-a)^2+(y-b)^2+(z-c)^2)^(3/2), z=0..6-2*x-3*y,
y=0..-2/3*x+2, x=0..3); F_x:=evalf(%);
(10)
Tripleint((y-b)/((x-a)^2+(y-b)^2+(z-c)^2)^(3/2), z=0..6-2*x-3*y,
y=0..-2/3*x+2, x=0..3); F_y:=evalf(%);
(11)
Tripleint((z-c)/((x-a)^2+(y-b)^2+(z-c)^2)^(3/2), z=0..6-2*x-3*y,
y=0..-2/3*x+2, x=0..3); F_z:=evalf(%);
(12)
Comparing to the values we get by treating all the mass as if it were located at the center of mass (0.75,
0.5, 1.5)
we get for the x-direction (and the true value for comparing);
6*(0.75-a)/((0.75-a)^2+(0.5-b)^2+(1.5-c)^2)^(3/2);F_x;
(13)
Comparing to the values we get by treating all the mass as if it were located at the center of mass (0.75,
0.5, 1.5)
we get for the y-direction (and the true value for comparing);
6*(0.5-b)/((0.75-a)^2+(0.5-b)^2+(1.5-c)^2)^(3/2);F_y;
(14)
Comparing to the values we get by treating all the mass as if it were located at the center of mass (0.75,
0.5, 1.5)
we get for the z-direction (and the true value for comparing);
6*(1.5-c)/((0.75-a)^2+(0.5-b)^2+(1.5-c)^2)^(3/2);F_z;
(15)
(15)
Changing the place the point mass m is {changing (a, b, c) at the beginning}, the sphere will always
behave as if all
the mass is at the center of mass, but not so for other solids like the tetrahedron.