# Does $L = I \omega$ apply for an axis not passing through the rigid body? [duplicate]

I have seen in Kleppner & Kolenkow that for a rigid body moving with a fixed point on it, the angular momentum about an axis passing through that fixed point is given by the product of inertia tensor and angular velocity. Does that apply for an axis not passing through the rigid body too?

• Yes the point doesn't have to be on it - it could be in a hole for example - but it has to be the point you are using as the axis – Martin Beckett Dec 6 '17 at 5:34
• You have to specify the location of $L$ and $I$ in your equation for it to apply in general. – John Alexiou Dec 6 '17 at 5:52

While Kleppner & Kolenkow derive this equation for an axis passing through the center of mass, they haven't explicitly used the fact that it passes through the center of mass, so it can be applied to any axis, as long as you use the corresponding angular velocity and moment of Inertia.

• Yes. But the book also derived it for rotation about a fixed point on the body right. So I thought whether the equation applies to a point which is not on the body. – Akhil Dec 6 '17 at 5:44

In general the equations of motion are tied together with the center of mass.

• Linear Momentum is defined as mass times velocity of the center of mass. $$\mathbf{p} = m\, \mathbf{v}_C$$

• Angular Momentum about the center of mass is defined as the tensor product of mass moment of inertia (about the COM) and angular velocity. $$\mathbf{L}_C = \mathrm{I}_C {\boldsymbol \omega}$$

• Net force is the time derivative of momentum $$\mathbf{F} = \frac{{\rm d}}{{\rm d}t} \mathbf{p} = m\, \mathbf{a}_C$$

• Net torque about the center of mass is the time derivative of angular momentum $${\boldsymbol \tau}_C = \frac{{\rm d}}{{\rm d}t} \mathbf{L}_C = \mathrm{I}_C {\boldsymbol \alpha} + {\boldsymbol \omega} \times \mathrm{I}_C {\boldsymbol \omega}$$

• To get the angular momentum at a different point A you need to transform the angular momentum vector with $$\mathbf{L}_A = \mathbf{L}_C + (\mathbf{r}_C - \mathbf{r}_A) \times \mathbf{p}$$

• This is similar to the transformation of velocities $$\mathbf{v}_A = \mathbf{v}_C + (\mathbf{r}_C-\mathbf{r}_A) \times {\boldsymbol \omega}$$

• Finally to calculate momentum at a location not at the center of mass you need the following relationships

\begin{aligned} \mathbf{p} & =\underbrace{ m \mathbf{v}_A}_{\mbox{momentum}} -\underbrace{ m\, \mathbf{r} \times {\boldsymbol \omega}}_{\mbox{cross term}} \\ \mathbf{L}_A & =\underbrace{ \mathrm{I}_C {\boldsymbol \omega} - m \, \mathbf{r} \times \mathbf{r} \times {\boldsymbol \omega}}_{\mbox{parallel axis thrm}} + \underbrace{ \mathbf{r} \times m \mathbf{v}_A }_{\mbox{cross term}} \end{aligned}

where $\mathbf{r} = \mathbf{r}_C - \mathbf{r}_A$.

So when calculating momentum not at the center of mass you need not only the parallel axis theorem, but also you need to include the appropriate cross terms.