# Find angular momentum using center of mass frame

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

1. How can my intuitive answer be expressed using the above formula
2. How does the above formula explain that for a force parallel to the COM position vector we don't have any angular velocity?

The formula

$$\mathbf L=\mathbf L' +M\mathbf r\times\mathbf v_{\rm com}$$

is used only for finding the value of angular momentum with respect from different locations. It is not a dynamical formula explaining any dynamics of the system. So it can't give you any relation between the torque applied and the subsequent angular momentum. So the answer of

How does the above formula explain that for a force parallel to the COM position vector we don't have any angular velocity?

is that the formula is not at all meant for this task. To answer the above question, you'd need neo more definitions, namely $$\boldsymbol{\tau}=\mathrm d\mathbf L/\mathrm dt$$. Using this you can find the torque and thus the final angular momentum.

How can my intuitive answer be expressed using the above formula?

To do this, you need to use the fact that $$\mathbf L'=I_{\rm com}\boldsymbol{\omega}$$ (where $$I_{\rm com,}$$ is the moment of inertia of the body about the center of mass. Why is this so? Because about the center of mass, the body is only performing rotational motion, so only the term corresponding to rotational motion contributes to the angular momentum. Thus the final expression of angular momentum becomes

$$\mathbf L=I_{\rm com}\boldsymbol{\omega} +M\mathbf r\times\mathbf v_{\rm com}$$

To now find the torque, just differentiate the above equation (I am assuming $$I_{\rm com}$$ and $$M$$ to be constant which is generally true)

$$\dot{\mathbf L}=\boldsymbol{\tau}=I_{\rm com}\dot{\boldsymbol{\omega}} +M\dot{\mathbf r}\times\mathbf v_{\rm com}+M{\mathbf r}\times\dot{\mathbf v}_{\rm com}$$

Generally, for stationary points of refernce, $$\dot{\mathbf r}=\mathbf v_{\rm com}$$. And $$\mathbf F_{\rm net}=M\dot{\mathbf v}_{\rm com}$$. Using these, we get

$$\boldsymbol{\tau}=I_{\rm com}\dot{\boldsymbol{\omega}}+{\mathbf r}\times\mathbf F_{\rm net}$$

• So could you please explain this formula? How can I write $L'$ in a different way (a similar one to $Mr\times v$)? How can I find the angular momentum using the torque (their relation is by integration but it doesn't seem to be the same as $I\omega + r\times p$)? – snatchysquid Jul 20 '20 at 7:58

In the formula $$\vec{L} = \vec{L'}+M\vec{r}_{cm}\times\vec{v}_{cm}$$

Which represents the total angular momentum we have two parts/components:

First let's look at the second term, which is $$M\vec{r}_{cm}\times\vec{v}_{cm}$$:

This represents only the linear motion of the center of mass. Take the vector from the origin to the COM and call it $$r_{cm}$$ and take only the velocity of the COM and call it $$v_{cm}$$. the cross product of these two multiplied by the mass of the whole body is the second term here.

Now let's look at the first term, which is $$L'$$:

This is, as said, the angular momentum in the COM (cetner of mass) frame. This means that now we "move" the origin to the COM point, and calculate the angular momentum for each mass (if it is a rigid body we look at each little piece and take the riemann sum to an integral). since the system is rotating arounf the COM each mass has a speed of $$v=\omega r$$ So we have $$L'=\sum_{i=1}^n(m_ir\times v)=\sum_{i=1}^n(m_i\omega\cdot r^2_i)=\omega I$$.

So In total we have $$L=L'+M\vec{r}_{cm}\times\vec{v}_{cm}$$ meaning $$L=\omega I+M\vec{r}_{cm}\times\vec{v}_{cm}$$ as needed.

Notice also how in the case where the body is not rotating and there's only linear motion, we don't have motion relative to the COM (since all parts are moving together, no rotations) and $$\omega =0$$. To find what omega is, and if it is equal to zero we can many times use torque or conservation of angular momentum.

Both of your formulas say that the total angular momentum is that associated with the rotation of the object around the center of mass plus that associated with the center of mass velocity not passing through some chosen reference point. For the latter contribution, the, r, vector goes from the reference point to the center of mass. If you choose the reference point at the impact point, and the impulse vector passes through the center of mass, then the cross-product of, r, with, p, is zero. Also the torque about the center of mass would be zero.