Suppose I have a square-shaped plate getting hit by a ball as shown in the picture below (notice how the force vector applied by the ball is not parallel to the $r$ vector). Let's set the origin to be in the point of the collision.
I know that the plate will now have both angular and linear speed so we can write the angular momentum as a compsition of those two speeds (and receieve something along the lines of $L=L_{1}+L_{2}=r\times p + I\omega$).
Now this (although correct) is an intuitive explanation. My question is how to recieve this in a more "formal" way, perhaps using this formula:
$$ \vec{L} = \vec{L'}+M\vec{r}_{cm}\times\vec{v}_{cm}$$
Where $L'$ is the angular momentum in the COM frame, and $\vec{r}_{cm},\vec{v}_{cm}$ are the position and velocity vectors of the COM.
I know also that in the case where the force vector (from the ball) is parallel to the position vector of the COM we don't have any rotation ($\omega=0$). How can this be explained using the above formula?
In summary
- How can my intuitive answer be expressed using the above formula
- How does the above formula explain that for a force parallel to the COM position vector we don't have any angular velocity?