# How are these two definitions of angular momentum related? [duplicate]

I've seen angular momentum defined as:

$$\ L=I \omega\$$

In dynamics, the notation is different and states:

$$\ L_o = r × (mv) \$$

How are these definitions related, if they are describing the same thing?

$\mathbf{L} = \mathbf{r} \times (m\mathbf{v})$ is the general one.

For the special case of a rigid body (i.e. not point mass) rotating freely about an axis`with angular velocity $\omega$, then $\mathbf{L} = I\mathbf{\omega}$ where $\mathbf{\omega}$ points in the axis of rotation. The moment of intertia $I$ is a scalar.

This follows from $v = \omega /r$ and from $\mathbf{r} \times \mathbf{v}$ pointing in the direction of $\mathbf{\omega}$. $I = mr^2$ for a point mass, and it can be discretised to then be integrated over for a rigid body.

In the case of a forced rotation (under an external force), then $I$ becomes a tensor and $\mathbf{L}$ and $\omega$ are not parallel anymore.

diregarding the vectors we have: $$L=I ω$$ $$ω = \frac{(r v)}{|r^2|}$$ $$I=mr^2$$ So. $$Iω= rmv= H_o$$

• I'm not sure I understand why you have defined angular velocity (ω), the way you did. I haven't seen this before. Could you explain? – Siddart Fredrick Oct 28 '17 at 18:34
• it follows by definition of cross product. Check the link:en.wikipedia.org/wiki/Angular_velocity – Vaggelis Kyrilas Oct 28 '17 at 18:41

A rigid body can only describe circular motion respect to an axis, because of its definition (rigid ↔ the distances between molecules are always the same; and the distance is invariant under rotations). It can also move linearly, but that's just the CoM.

Consequently, $v$ is that of the circular motion $v=\omega r$, and so

$L=mrv=mr^2\omega$

Then, you call $I\equiv mr^2$, so you have $L=I\omega$

You can also do it with vectors.

The angular momentum, with respect to some point $O$, of a particle is defined as $$\vec L=\vec r\times\vec p,$$ where $\vec r$ is the particle position with respect to $O$.

For any system of particles, the total angular momentum is the sum of the individual ones. For a discrete system $$\vec L=\sum_i\vec r_i\times\vec p_i,$$ whereas for a continuum distribution $$\vec L=\int\vec r\times\vec vdm.\tag1$$

For a rigid body rotating with angular velocity $\omega$ about an axis, the velocity of any mass element can be written as $\vec v=\vec\omega\times\vec r$. Plugging this into Eq. (1) we obtain the relation $$\vec L=\mathbb I\vec\omega,\tag2$$ where $\mathbb I$ is the inertia tensor, a $3\times 3$ matrix with components $$\mathbb I_{ij}=\int(r^2\delta_{ij}-x_ix_j)dm.$$ As you can see from its construction, Eq. (2) is quite general. For some special cases however, see this answer, the last equation simplifies to $$\vec L=I\vec\omega,$$ where $I$ is a scalar. In that case, angular momentum and angular velocity point in the same direction.