7—Operators and Matrices 151Put this in words and it says that the tensor of inertia about any point is equal to the tensor of inertia
about the center of mass plus the tensor of inertia of a point massMplacedatthe center of mass.
As an example, place a disk of massMand radiusRand uniform mass density so that its center
is at(x,y,z) = (R, 0 ,0)and it is lying in thex-yplane. Compute the components of the inertia
tensor. First get the components about the center of mass, using Eq. (7.17).
x
z
y
The integrals such as
−∫
dmxy, −
∫
dmyz
are zero. For fixedyeach positive value ofxhas a corresponding negative value to make the integral
add to zero. It is odd inx(ory); remember that this is about thecenterof the disk. Next do theI 33
integral. ∫
dm(x^2 +y^2 ) =
∫
dmr^2 =
∫
M
πR^2
dAr^2
For the element of area, usedA= 2πrdrand you have
I 33 =
M
πR^2
∫R
0dr 2 πr^3 =
M
πR^2
2 π
R^4
4
=
1
2
MR^2
For the next two diagonal elements,
I 11 =
∫
dm(y^2 +z^2 ) =
∫
dmy^2 and I 22 =
∫
dm(x^2 +z^2 ) =
∫
dmx^2
Because of the symmetry of the disk, these two are equal, also you see that the sum is
I 11 +I 22 =
∫
dmy^2 +
∫
dmx^2 =I 33 =
1
2
MR^2 (7.23)
This saves integration.I 11 =I 22 =MR^2 / 4.
For the other term in the sum (7.22), you have a point mass at the distanceRalong thex-axis,
(x,y,z) = (R, 0 ,0). Substitute this point mass into Eq. (7.17) and you have
M
0 0 0
0 R^20
0 0 R^2
The total about the origin is the sum of these two calculations.