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The angular momentum of an object about an arbitrary point $O$ is given by, $$ \mathbf{H_O}=[I_O]\mathbf{\omega}$$ where $[I_O]$ is the moment of inertia tensor about $O$. And for the center of mass $G$, $$ \mathbf{H_G}=[I_G]\mathbf{\omega} $$

My book says that a new array of moments and products of inertia would be obtained if a different system of axes were used. The transformation characterized by this new array, however, would still be the same. The angular momentum $\mathbf{H_G}$ about the mass center corresponding to a given angular velocity $\mathbf{\omega}$ is independent of the choice of the coordinate axes.

So if I setup the axes with origin at $O$, according to the first equation $\omega $ would transform into $\mathbf{H_O}$ and according to the book that would equal $\mathbf{H_G}$. But if that is the case then won't the angular momentum about any arbitary point be equal to the angular momentum about the center of mass?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jul 12 at 23:10
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The first equation is wrong. Angular momentum of a rigid body is first defined at the center of mass G as

$$ \boldsymbol{H}_G = \mathbf{I}_G \boldsymbol{\omega} \tag{1}$$

and then transferred to another point O with the following law

$$ \boldsymbol{H}_O = \boldsymbol{H}_G + \boldsymbol{r}_{G/O} \times \boldsymbol{p} \tag{2} $$where $\boldsymbol{p}$ is linear momentum, and $\boldsymbol{r}_{G/O}$ is the location of the center of mass G with respect to the reference point O.

Proof

Consider a rigid body as a system of particles moving together. The total mass of the body is $m = \sum_i^n m_i$ and the position of each particle as measured from the reference point O is decomposed into two parts, one the location of the center of mass G and then the relative location of the particle to the center of mass.

$$ \boldsymbol{r}_i = \boldsymbol{r}_{G} + \boldsymbol{d}_i \tag{3} $$

For brevity I drop the O subscripts, and thus $\boldsymbol{r}_G = \boldsymbol{r}_{G/O}$ and later $\boldsymbol{v}_G = \tfrac{\rm d}{{\rm d}t} \boldsymbol{r}_{G/O}$.

The center of mass is defined as the location which makes $\sum_i^n m_i \boldsymbol{d}_i=0$ and thus

$$ \sum_i^n m_i \boldsymbol{r}_i = \boldsymbol{r}_G \sum_i^n m_i + \sum_i^n m_i \boldsymbol{d}_i = m\, \boldsymbol{r}_G \tag{4}$$

The time derivative of each position is velocity and therefore the total momentum of the rigid body is derived from the time derivative of (4)

$$ \boldsymbol{p} = \sum_i^n m_i \boldsymbol{v}_i = m\,\boldsymbol{v}_G \tag{5}$$

There is a little trickery in the above expression because the time derivative of (3) is

$$ \boldsymbol{v}_i = \boldsymbol{v}_G + \boldsymbol{\omega} \times \boldsymbol{d}_i \tag{6} $$

which means that $\require{cancel} \boldsymbol{p} = \sum_i^n m_i \boldsymbol{v}_i = (\sum_i^n m_i) \boldsymbol{v}_G + \boldsymbol{\omega} \times (\cancel{\sum_i^n m_i \boldsymbol{d}_i}) = m \boldsymbol{v}_G $

The same trickery is to be applied for angular velocity about the reference point

$$ \require{cancel} \begin{aligned}\boldsymbol{H}_O & =\sum_{i}^{n}\boldsymbol{r}_{i}\times(m_{i}\boldsymbol{v}_{i})=\sum_{i}^{n}(\boldsymbol{r}_{G}+\boldsymbol{d}_{i})\times m_{i}(\boldsymbol{v}_{G}+\boldsymbol{\omega}\times\boldsymbol{d}_{i})\\ & =\sum_{i}^{n}\left(\boldsymbol{r}_{G}\times m_{i}(\boldsymbol{v}_{G}+\boldsymbol{\omega}\times\boldsymbol{d}_{i})+\boldsymbol{d}_{i}\times m_{i}(\boldsymbol{v}_{G}+\boldsymbol{\omega}\times\boldsymbol{d}_{i})\right)\\ & =\boldsymbol{r}_{G}\times(\sum_{i}^{n}m_{i})\boldsymbol{v}_{G}+\boldsymbol{r}_{G}\times\left(\boldsymbol{\omega}\times\cancel{\sum_{i}^{n}m_{i}\boldsymbol{d}_{i}}\right)\\ & +\left(\cancel{\sum_{i}^{n}m_{i}\boldsymbol{d}_{i}}\right)\times\boldsymbol{v}_{G}+\sum_{i}^{n}\boldsymbol{d}_{i}\times m_{i}(\boldsymbol{\omega}\times\boldsymbol{d}_{i})\\ & =\boldsymbol{r}_{G}\times\boldsymbol{p}+\mathbf{I}_{G}\,\boldsymbol{\omega} = \boldsymbol{r}_{G}\times\boldsymbol{p} + \boldsymbol{H}_G\;\;\; \checkmark \end{aligned} \tag{7}$$

Given the definition $$ \boldsymbol{H}_G = \mathbf{I}_{G}\,\boldsymbol{\omega}=\sum_{i}^{n}m_{i}\left(-\boldsymbol{d}_{i}\times(\boldsymbol{d}_{i}\times\boldsymbol{\omega})\right) \tag{8} $$

or expanded out

$$ \mathbf{I}_G = \sum_i^n m_i \begin{bmatrix} y_i^2+z_i^2 & -x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2+z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2 \end{bmatrix} \tag{9} $$ where $\boldsymbol{d}_i = \pmatrix{x_i \\ y_i \\z_i}$.

Now that angular momentum at the center of mass $\boldsymbol{H}_G$, mass moment of inertia at the center of mass $\mathbf{I}_G$ and angular momentum at the reference point $\boldsymbol{H}_O$ is defined, we can derive the rule of transferring the mass moment of inertia tensor to the reference point, such that $$ \boldsymbol{H}_O = \underbrace{ \left( \mathbf{I}_G - m [\boldsymbol{r}_G \times] [\boldsymbol{r}_G\times] \right) }_{\mathbf{I}_O} \boldsymbol{\omega} = \mathbf{I}_O\, \boldsymbol{\omega} \tag{10}$$

Proof

Consider the body rotating about the reference point, setting the kinematics of the center of mass as $\boldsymbol{v}_G = \boldsymbol{\omega} \times \boldsymbol{r}_G$ and thus momentum as $$\boldsymbol{p} = m ( \boldsymbol{\omega} \times \boldsymbol{r}_G) = m (-[\boldsymbol{r}_G \times] \boldsymbol{\omega}) \tag{11}$$ where $[\boldsymbol{r}\times] = \begin{bmatrix}0 & -z & y \\ z & 0 & -x \\ -y & x &0 \end{bmatrix}$ is the 3×3 cross product matrix operator.

Now use equations (1) and (2) to define angular momentum about the reference point, and factor out the rotational velocity vector to create a vector = matrix * vector equation.

$$ \begin{aligned} \boldsymbol{H}_O & = [\mathbf{I}_G] \boldsymbol{\omega} + [\boldsymbol{r}_G \times] \boldsymbol{p} \\ & =[\mathbf{I}_G] \boldsymbol{\omega} + [\boldsymbol{r}_G\times] (-m [\boldsymbol{r}_G \times] \boldsymbol{\omega} ) \\ & = \left( [\mathbf{I}_G] - m [\boldsymbol{r}_G \times] [\boldsymbol{r}_G\times] \right) \boldsymbol{\omega} \;\;\checkmark \end{aligned} \tag{12} $$

Parallel Axis Theorem

So the transfer of mass moment of inertia to a different location is described by the so-called parallel axis theorem, which in vector form is

$$ [\mathbf{I}_O] = [\mathbf{I}_G] - m [\boldsymbol{r}_G \times][\boldsymbol{r}_G \times] \tag{13} $$

which in turn allows us to find angular momentum at the reference point as just simply

$$ \boldsymbol{H}_O = [\mathbf{I}_O]\, \boldsymbol{\omega} \tag{14} $$

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  • $\begingroup$ I thought that the equation that you wrote could be used but is not how $\mathbf{H_O}$ is defined, as here on page 3 they have written the first equation. $\endgroup$ – Akshat Sharma Jul 13 at 1:00
  • $\begingroup$ I had made some notation errors in my answer. I edit some fixes. The above is the parallel axis theorem $\boldsymbol{H}_O = \mathbf{I}_O \boldsymbol{\omega}$ if the center of mass is rotating about the reference axis, and thus $\boldsymbol{p} = m \boldsymbol{v}_G = m ( \boldsymbol{\omega} \times \boldsymbol{r}_G)$ or $$\boldsymbol{H}_O = \boldsymbol{H}_G + m (- \boldsymbol{r}_G \times( \boldsymbol{r}_G \times \boldsymbol{\omega})) $$ and $$ \mathbf{I}_O = \mathbf{I}_G - m [\boldsymbol{r}_G \times] [\boldsymbol{r}_G\times] $$ $\endgroup$ – John Alexiou Jul 13 at 1:53
  • $\begingroup$ So $\mathbf{H_O}=\mathbf{I_O}\mathbf{\omega}$ is not always true? $\endgroup$ – Akshat Sharma Jul 15 at 6:41
  • $\begingroup$ It is true, as the definition of $\mathbf{I}_O$ using the parallel axis theorem. Consider $$ \boldsymbol{H}_O = \mathbf{I}_G \boldsymbol{\omega} + m (- \boldsymbol{r}_G \times( \boldsymbol{r}_G \times \boldsymbol{\omega})) = \mathbf{I}_O \boldsymbol{\omega} $$ where $$ \boldsymbol{H}_O = \underbrace{ \left( \mathbf{I}_G - m [\boldsymbol{r}_G \times] [\boldsymbol{r}_G\times] \right) }_{\mathbf{I}_O} \boldsymbol{\omega}$$ $\endgroup$ – John Alexiou Jul 15 at 11:50
  • $\begingroup$ @AkshatSharma, I updated the answer with more details on how the transfer law for rot. momentum leads to the transfer law for mass moment of inertia. $\endgroup$ – John Alexiou Jul 15 at 12:14

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