# 6.1: Linking Linear and Angular Momentum

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## Rotational Impulse-Momentum Theorem

By now we have a very good sense of how to develop the formalism for rotational motion in parallel with what we already know about linear motion. We turn now to momentum. Replacing the mass with rotational inertia and the linear velocity with angular velocity, we get:

The vector \(L\) is called *angular momentum*, and it has units of:

\( \left[ L \right] = \dfrac{kg \cdot m^2}{s} = J \cdot s \)

Continuing the parallel with the linear case, the momentum is relates to the force through the impulse-momentum theorem, which is:

While there is no need to append "*cm*" to the angular momentum as we do with the linear momentum, we do have to keep in mind that all of the quantities in the rotational case must be referenced to the same point. That is, the net torque requires a reference point, and the angular momentum contains a rotational inertia, which also requires a reference point.

Recall that the impulse-momentum theorem is just a repackaging of Newton's second law, and so it is with the rotational case, though there is a twist, as we will see shortly:

\[ \overrightarrow F_{net} = \dfrac{d\overrightarrow p_{cm} }{dt} = \dfrac{d\left(m\overrightarrow v_{cm}\right) }{dt} = m\overrightarrow a_{cm} \;\;\; \iff \;\;\; \overrightarrow \tau_{net} = \dfrac{d \overrightarrow L}{dt} = \dfrac{d\left(I\overrightarrow \omega\right) }{dt} = I\overrightarrow \alpha_{cm} \]

## Link Between Angular and Linear Momentum

When there are several particles in a system, we find the momentum of the system by adding the momenta of the particles:

\[ \overrightarrow p_{cm} = \overrightarrow {p_1 }+ \overrightarrow {p_2} + \dots \]

We have a definition for the angular momentum of a rigid object, but can we define the angular momentum of a single particle, and then add up all of the angular momenta of the particles to get the angular momentum of the system, in the same way that we do it for linear momentum? The answer is yes, but we have to be careful about our reference point. That is, to add the angular momentum of every particle together to get a total angular momentum, the individual angular momenta must be measured around the same reference.

So how do we define the angular momentum of an individual particle around a certain reference point? Let's look at a picture of the situation. The particle has a mass \(m\), a velocity \(\overrightarrow v\), and is located at a position \(\overrightarrow r\) with the tail of that position vector at the reference point.