# Expression for angular momentum

I wanted to derive an expression for angular momentum of a rigid body in the frame of a particle on the rigid body
I started out this way.
$$L=r x p =r x mv= m (r x v) = m (r x (w x r))=m( r^2(w)-(r.w)(r).$$ summation of this over all particles should give me the answer, but
now the issue is I don't know under what cases it reduces to the form Iw
also the r which I have used is from the point not from the axis.
My teacher told me it does so only for fixed axis rotation. Also I don't really know how to define an axis. If it is the locus of all centres of circles then why is it a straight line? Please explain the cases when angular momentum reduces to $$Iw$$.

• if you turn something, first just one point, than you have always an axes of rotation , if a second point is on the same rigid body it has the same axes. Try to turn any object and you see it. Commented Aug 11, 2023 at 14:13
• FYI - Don't use x for the cross-product. It is confusing. Use \times which renders as $\times$. Similarly, don't use . for the dot-product, but rather \cdot which renders as $\cdot$ Commented Aug 11, 2023 at 15:22

Perhaps using vectors carefully might make it easier to answer your question. $$$$\vec{L} = \vec{r} \times \vec{p} = m\vec{r} \times \vec{v}.$$$$ Since $$\vec{v} = \vec{\omega} \times \vec{r}$$, we have $$$$\vec{L} = m(r^2\vec{\omega} - \vec{r}(\vec{r}\cdot\vec{\omega})),$$$$ If the axis of rotation is such that $$\vec{\omega}\cdot\vec{r} = 0$$, that is if the plane of rotation is always perpendicular to the axis of rotation you get the familiar expression $$$$\vec{L} = mr^2\vec{\omega} = I\vec{\omega}.$$$$ You may want to consult a book like Halliday and Resnick for an excellent explanation of this topic.

The vector $$\vec{r}$$ is the position of the particle with respect to (your choice of) origin and $$\vec{\omega}$$ is the angular velocity vector.

## System of particles

Consider a rigid body as a system of particles, each of which is fixed on some rotating reference frame. Consider that the frame is rotating with rotational velocity $$\vec{\omega}$$ about the origin.

Each particle of mass $$m_i$$ has translational velocity $$\vec{v}_i= \vec{\omega} \times \vec{r}_i \tag{1}$$ where $$\vec{r}_i$$ is location of the particle at some time-slice.

Note that since $$\vec{\omega}$$ is shared among all the particles, the total translational momentum is zero if the center of mass is also at the origin

$$\vec{p} = \sum_i m_i \vec{v}_i = \sum_i m_i ( \vec{\omega}\times \vec{r}_i) = \vec{\omega} \times \underbrace{ \sum_i m_i \vec{r}_i}_{\text{defn. as }0} = \vec{0} \tag{2}$$

Now we look at total rotational momentum

\begin{aligned} \vec{L} & = \sum_i \vec{r}_i \times (m_i \vec{v}_i) = \sum_i m_i ( \vec{r}_i \times( \vec{\omega} \times \vec{r}_i))= \\ & = \sum_i m_i \left( (\vec{r}_i \cdot \vec{r}_i) \vec{\omega} - \vec{r}_i ( \vec{r}_i \cdot \vec{\omega}) \right) = \\ &= \left[ \sum_i m_i \left( ( \vec{r}_i \cdot \vec{r}_i ) {\bf 1}_{3×3} - \vec{r}_i \odot \vec{r}_i \right) \right] \vec{\omega} = \\ & = {\rm I}\; \vec{\omega} \end{aligned} \tag{3}

where $$\cdot$$ is the vector inner product (dot-product), $$\odot$$ is the vector outer product and $${\bf 1}_{3×3}$$ is the identity matrix.

The above leads to the explicit definition of

$${\rm I} = \sum_i m_i \begin{vmatrix} y_i^2+z_i^2 & -x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2+z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2 \end{vmatrix} \tag{4}$$

You need the following linear algebra manipulations

• $$\boldsymbol{a} \cdot \boldsymbol{b} = \boldsymbol{a}^\intercal \boldsymbol{b}$$
• $$\boldsymbol{a} ( \boldsymbol{b} \cdot \boldsymbol{c}) = \boldsymbol{a} ( \boldsymbol{b}^\intercal \boldsymbol{c}) = (\boldsymbol{a} \boldsymbol{b}^\intercal) \boldsymbol{c} = (\boldsymbol{a} \odot \boldsymbol{b}) \boldsymbol{c}$$