
Angular Momentum of a Rigid Body As shown in [6] in Inertis Tensor, the angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame (frame A) is
Now, let's transforms this angular momentum vector to another reference frame (frame B):
Combination of [1] & [2] leads to
[3] shows that the angular momentum of the object described in frame B is equal to I' times the angular velocity described in frame B. The axis transformation does not alter the nature of any mechanical quantity, but describes it in a different perspective. Since both the angular momentum & the angular velocity vectors are already described in frame B, I' must be the inertia tensor of the object described in frame B:
This type of transformation is called the similarity transformation. Now, let A be a local reference frame (frame L) and B be the global (inertial) frame. Then, the inertia tensor of a body segment described in the global coordinates is
In motion analysis, one can compute the transformation matrix from the global frame to a segmental reference frame based on the marker data, while the inertia tensor is typically first described in the corresponding segmental reference frame. [5] can be used to compute the inertia tensor described in the global frame. From [4]:
where i, j, k, & m = 1 to 3. Or simplifying [6], one can obtain
[7] is identical to [8] in Inertia Tensor. [7] is the requirement of a tensor of the 2nd rank. A scalar is a tensor or zero rank, while a vector is a tensor of the 1st rank. Figure 1 shows two reference frames: the OXYZ system & the O'X'Y'Z' system. The XYZ system is the local reference frame fixed to the body with its origin at the body's center of mass (CM). The X'Y'Z' system, on the other hand, is parallel to the XYZ system with different origin. From [4] of Inertia Tensor, the inertia tensor of the body about the X'Y'Z' system is
Now, let
and therefore,
Substituting [10] into [8] gives
since
where m = the mass of the body. [12] basically means the sum of the first mass moments about the body's CM is always 0. With [11], [8] is reduced to
where
I_{CM} is the inertia tensor of the body about its CM while I_{t} is the additional MOI due to the translation of the reference frame. Matrix 1 shown in [14] is the identity matrix whose diagonal elements are all 1 while the offdiagonal elements are 0. As shown in [13], the moment of inertia of an object about a reference frame parallel to the local reference frame fixed to the body is equal to the sum of the inertia tensor of the body and the additional MOI due to translation of the origin of the frame. Here is an example. Let
where I_{xx}, I_{yy} & I_{zz} = the principal moments of inertia of the body. In other words, the X, Y & Z axes shown in Figure 1 are the principal axes. See Principal Axes for the details of the principal moments of inertia and axes. If the Z' axis shown in Figure 1 is the axis of rotation or:
the angular momentum of the body about the axis of rotation becomes
If we further assume that z_{o} = 0, in other words, both the XYZ and the X'Y'Z' systems share the XY plane,
where I' = the moment of inertia about the axis of rotation (Z' axis), and d = the distance between the axis of rotation & the parallel axis passing through the body's CM (Z axis). I_{zz} in [18] is in fact the moment of inertia of the object about the Z axis. [18] may be generalized to the following form:
where I' = the moment of inertia of the body about the axis of rotation, I_{CG} = the moment of inertia of the body's principal axis which is parallel to the axis of rotation, m = mass of the body, and d = the distance between the two axes. [19] is the socalled the parallelaxis theorem.

© YoungHoo Kwon, 1998 