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I am reading this book Vector Mechanics for Engineers. On page 1332, I read about the Conservation of the angular momentum about the axis of precession and Conservation of the angular momentum about the axis of spin.

How am I supposed to apply these conservation theorems? I can't find the definition of angular momentum about an axis in the book. Throughout the book, only definition of angular momentum about a point O or center of mass G is stated. What were the authors thinking to put the readers in so much confusion? I have read this question but it does not help much.

Again: In a Newtonian reference frame, we have an axis (fixed or moving) and a moving rigid body. How do we defined the angular momentum of the rigid body about this axis at an instant when looking from this Newtonian reference frame?

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Angular momentum is a vector, with a magnitude and a direction.

For a general 3D shape rotating in space with rotational velocity vector $\boldsymbol{\omega}$, it is calculated by

$$ \boldsymbol{L} = \mathbf{I}\, \boldsymbol{\omega} $$

$$ \pmatrix{L_x \\ L_y \\ L_z} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{bmatrix} \pmatrix{ \omega_x \\ \omega_y \\ \omega_z } \tag{1}$$

where the 3×3 symmetric matrix $\mathbf{I}$ is the mass moment of inertia tensor expressed in the same orientation as $\boldsymbol{\omega}$.

Now to say that the (scalar) angular momentum is to be conserved along a particular direction, say $\boldsymbol{n}$ means that

$$ L_n = \boldsymbol{n} \cdot \boldsymbol{L} = n_x L_x + n_y L_y + n_z L_z = \text{(const)} \tag{2}$$

That is all there is to it. I assume the vector $\boldsymbol{n}$ is a unit vector above, in order to correctly project the momentum vector along the target direction and get its magnitude.

You can interpret the above as $L_n = \| \boldsymbol{L} \| \cos \theta$ where the angle $\theta$ is the angle between the $\boldsymbol{L}$ vector and the $\boldsymbol{n}$ vector.

The above value for angular momentum is only about the center of mass. To measure angular momentum about any other point you need a transformation law, just as you need a transformation law for velocities

$$ \boldsymbol{L}_A = \boldsymbol{L}_O + \boldsymbol{r}_{O/A} \times \boldsymbol{p} $$ where $\boldsymbol{p} = m \,\boldsymbol{v}_O$ is the linear momentum vector and $\boldsymbol{r}_{O/A}$ is the position of O relative to the reference point A. You interpret this as when you measure from any point away from the center of mass, angular momentum increases as the mass is further away.

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  • $\begingroup$ Does the axis that we calculate the angular momentum about must go through a point A? Do we calulate the angular momentum about point A first then project this vector to the axis to achive angular momentum about this axis? $\endgroup$
    – Dat
    Commented Apr 26, 2020 at 5:06
  • $\begingroup$ The vector $\boldsymbol{n}$ is just a direction, so if you apply the projection to $\boldsymbol{L}_O$ then it is the components along that direction at O and if you apply the projection to $\boldsymbol{L}_A$ then the components along that direction at A are found. So you do the transformation first to the point you want, and then you do the projection, just like you mention in the 2nd part of the comment. $\endgroup$ Commented Apr 26, 2020 at 19:43
  • $\begingroup$ ok, then if we calculate the angular momentum about an axis, this value does not only depend on the direction of the axis, it also depend on the point where the axis goes through? $\endgroup$
    – Dat
    Commented Apr 27, 2020 at 2:16
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    $\begingroup$ This seems similarly to caculate moment of a force about an axis. Have you read the book I put in the link. I wonder why the authors did not show us how to caculate angular momentum about an axis. They just told us to use conservation angular momentum theorem about an axis. $\endgroup$
    – Dat
    Commented Apr 27, 2020 at 4:31
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    $\begingroup$ ok, so I conclude: angular momentum about an axis is the scalar quantity, it is the projection of angular momentum about a point which the axis goes through. Now, what are the general conditions to apply conservation of angular momentum about an axis? $\endgroup$
    – Dat
    Commented Apr 30, 2020 at 11:18
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The angular momentum of a point $M$ about a point $A$ is defined by: $$\vec L_{A}(M)=\vec{AM}\times m\vec v=\vec{AM}\times\vec p$$ If $\Delta$ is an axis through $A$, with a unit direction vector $\vec u$, then: $$\vec L_\Delta(M)=\vec L_{A}(M)\cdot\vec u$$ Also, note that for $A'\in\Delta$: $$\begin{align*} \vec L_{A}(M)\cdot\vec u &=\left[\left(\vec{AA'}+\vec{A'M}\right)\times\vec p\right]\cdot\vec u\\ &=\left[\vec{AA'}\times\vec p+\vec{A'M}\times\vec p\right]\cdot\vec u\\ &=\vec u\cdot\left(\vec{AA'}\times\vec p\right)+\left(\vec{A'M}\times\vec p\right)\cdot\vec u\\ &=\vec p\cdot\left(\vec u\times\vec{AA'}\right)+\left(\vec{A'M}\times\vec p\right)\cdot\vec u\\ &=\left(\vec{A'M}\times\vec p\right)\cdot\vec u\\ &=\vec L_{A'}(M)\cdot\vec u \end{align*}$$ Taking the derivative of the angular momentum about $A$: $$\begin{align*} \frac{d}{dt}\left(\vec{AM}\times\vec p\right) &=\left(\frac{d}{dt}\vec{AO}+\frac{d}{dt}\vec{OM}\right)\times\vec p+\vec{AM}\times\frac{d}{dt}\vec p\\ &=-\vec v_A\times\vec p+\vec{AM}\times\vec F\\ &=\vec p\times\vec v_A+\vec{AM}\times\vec F \end{align*}$$ If $A$ is fixed, then: $$\frac{d}{dt}\vec L_{A}(M)=\vec{AM}\times\vec F$$

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  • $\begingroup$ Could we apply conservation of angular momentum about a moving axis? If yes, what are the conditions of the moving axis? Does the moving axis have to be through a fixed point O and center mass of the rigid body? $\endgroup$
    – Dat
    Commented Apr 30, 2020 at 15:40
  • $\begingroup$ Look at $\frac{d}{dt}\left(\vec{AM}\times\vec p\right)$, I have directly substituted but it is $\frac{d}{dt}\vec L_{A}(M)$. Since the derivative is not zero, the angular momentum is not conserved, per se. For it to be conserved you need either $\vec p\times\vec v_A=-\vec{AM}\times\vec F$ or that both are $\vec0$, or, if A is fixed, $\vec{AM}\times\vec F=\vec 0$. In this, I assume that the direction vector of your axis, $\vec u$ in the above, is constant. Try to differentiate $\vec L_\Delta(M)$ and see what happens if the axis is itself rotating. $\endgroup$
    – GDGDJKJ
    Commented Apr 30, 2020 at 16:38
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The angular momentum of a rigid body rotating about a fixed axis is L = Iω , where I is the rotational inertia of the body (relative to the axis), and ω is the angular velocity of rotation. The L and ω are generally defined as vectors directed along the axis of rotation, using a right hand rule. (Wrap your right hand fingers around the axis in the direction of rotation, and our thumb gives the direction of the vectors.)

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  • $\begingroup$ Your definition is only for a rigid body rotating about a fixed axis. How about a general moving body in three dimension space. If you read page 1332 in the book, the axis (axis of spin) is not fixed and the rigid body is not rotating about any axis! $\endgroup$
    – Dat
    Commented Apr 25, 2020 at 18:37

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