# Can we write $\vec L=\vec r\times \vec p$ for an extended object?

Say we have a rod with a fixed axis of rotation. If we give it a push at some point other than the axis, it will start to rotate. Say our force is $F$, then we might write: $\Delta P=F\Delta t$. I’m wondering how to incorporate angular momentum here. We know that for a single particle it holds that $L=r\times p$. And for an extended object with a fixed axis of rotation, we generally might write $L=I\omega$. However, can I write $L=r\times p$ too for the extended object? I would think that I would need to integrate. So we have $p_\alpha=m_\alpha v_\alpha=m_\alpha r_\alpha\omega$, where we consider an infinitesimal mass $\alpha$. So should I write $L=\int r^2\omega\,dm$? I’m confused whether to work with this integral, or just use $L=r\Delta P$ at once.

The reason I'm asking, is because I'm trying to calculate the sweet spot of a baseball bat. So this is my attempt at solving the exercise.

The way you normally do this is that you specify that your body is rigid and undergoing rotations about a given point, which on geometrical grounds means that the velocity of each particle is given by $$\mathbf v = \boldsymbol \omega \times \mathbf r,$$ where $\boldsymbol \omega =\omega \hat{\mathbf n}$ is the angular velocity of the body. Once you've done that, the angular momentum of a bit of body of mass $\mathrm dm$ at position $\mathbf r$ is given by $$\mathrm d\mathbf L = \mathbf r \times \mathrm d\mathbf p = \mathbf r \times (\mathrm dm \ \boldsymbol \omega \times \mathbf r),$$ and the total angular momentum is given by $$\mathbf L = \int\mathbf r \times ( \boldsymbol \omega \times \mathbf r)\mathrm dm,$$ where $\mathrm dm$ is the mass element, normally given by $\rho(\mathbf r) \mathrm dV$.
The problem there, of course, is that you've written down $\mathrm d\mathbf L = \mathbf r \times \mathrm d\mathbf p$ for each bit of the body, but the momentum changes across the body (it's rotating, so some parts will move one way and other parts the opposite way) and the angular momentum only makes sense when you integrate over the whole body.
Now, that integral looks a bit tricky, but you can try to simplify it somewhat - at least by factoring out the dependence on $\boldsymbol \omega$. To do that, you first apply the vector triple product identity, to get $$\mathbf L = \int \left[ (\mathbf r\cdot\mathbf r) \boldsymbol \omega - (\mathbf r\cdot\boldsymbol \omega)\mathbf r\right]\mathrm dm,$$ which is a slight improvement but doesn't get you that much closer. What really matters is realizing that the mapping $$\boldsymbol \omega \mapsto (\mathbf r\cdot\mathbf r) \boldsymbol \omega - (\mathbf r\cdot\boldsymbol \omega)\mathbf r$$ is a linear mapping, so it must be representable by a matrix: in this case, given by $$(\mathbf r\cdot\mathbf r) \boldsymbol \omega - (\mathbf r\cdot\boldsymbol \omega)\mathbf r = \left[ (\mathbf r\cdot\mathbf r) \mathbb I - \mathbf r \otimes \mathbf r \right]\boldsymbol\omega,$$ where $\mathbb I$ is the identity matrix and $\mathbf r \otimes \mathbf r$ is the product of $\mathbf r$, seen as a column vector, multiplied on the right by $\mathbf r$ seen as a row vector, to give a $3\times 3$ matrix.
Once you've done that, then you get \begin{align} \mathbf L & = \int \left[ (\mathbf r\cdot\mathbf r) \boldsymbol \omega - (\mathbf r\cdot\boldsymbol \omega)\mathbf r\right]\mathrm dm \\ & = \int \left[ (\mathbf r\cdot\mathbf r) \mathbb I - \mathbf r \otimes \mathbf r \right]\boldsymbol\omega \:\mathrm dm \\ & = \left(\int \left[ (\mathbf r\cdot\mathbf r) \mathbb I - \mathbf r \otimes \mathbf r \right] \:\mathrm dm \right) \boldsymbol\omega \\ & = I \boldsymbol\omega \end{align} by factorizing $\boldsymbol \omega$ out of the (now matrix-valued) integral, and calling the resultant matrix $I$, which is the moment of inertia.
So, that's how the standard $L=I\omega$ comes out as just the body-integrated version of $\mathbf L = \mathbf r \times \mathbf p$.