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If a rod is rotating about one end, does it have pure rotation or do you also consider the translation of centre of mass when calculating its angular momentum?

Also, how would one calculate the angular momentum of a body that is both rotating and translating?

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    $\begingroup$ I think this is a valid question, as there are non-trivial things to consider here, like the parallel axis theorem and the moment cross terms (how angular momentum away from the center of mass depends on the linear velocity of the center of mass). $\endgroup$ – ja72 Nov 28 '13 at 23:32
  • $\begingroup$ I wanted to answer yes, but it got closed. If you know the motion of a point A, $\vec{v}_A$, $\vec{\omega}$ and the center of mass was located at $\vec{c}$ relative to A, then the linear and angular momentum at A are $$\vec{L} = m \vec{v}_C = m \vec{v}_A - m \vec{c}\times \vec{\omega} \\ \vec{H}_A = \vec{H}_C + \vec{c}\times\vec{L} = \vec{c}\times m \vec{v}_A + I_C \vec{\omega} - m \vec{c}\times\vec{c}\times\vec{\omega} $$ with $m$ the mass and $I_C$ the mass moment of inertia about the c.m. If $\vec{v}_A=0$ then what you have left for $\vec{H}_A$ is the parallel axis theorem. $\endgroup$ – ja72 Nov 28 '13 at 23:36
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Well, the angular momentum of a rigid body is equal to the sum of the angular momentum of the body around it's center of mass, plus the angular momentum of the center of mass.

Having said that, suppose the rod is rotating about one end (I imagine a pendulum motion; correct me if I'm wrong), you can calculate the angular momentum by $L = I \omega$ if you know the angular velocity $\omega$ and the moment of inertia about the line passing through the axis of rotation.

Suppose you only knew the moment of inertia about the COM. You would then use the parallel axis theorem to calculate the moment of inertia about the new axis. However, most angular momentum tables include moment of inertia about ends of rods also.

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If you know the motion of a point A on a rigid body, with linear velocity $\vec{v}_A$ and angular velocity $\vec{\omega}$ then the formulas below will give you the linear momentum of the rigid body, and the angular momentum about point A. If the body is pivoting about A then $\vec{v}_A=0$, otherwise in the general case $\vec{v}_A \neq 0 $.

Linear momentum is given from the velocity of the center of mass, point C. Consider the c.m. located at $\vec{c}$ from point A.

$$ \vec{L} = m \vec{v}_C = m \left( \vec{v}_A - \vec{c}\times \vec{\omega} \right) $$

Now the angular momentum about the center of mass is $\vec{H}_C = I_C\,\vec{\omega}$ and so by transferring the motion be about A it becomes

$$ \begin{aligned} \vec{H}_A & = \vec{H}_C + \vec{c}\times\vec{L} \\ & = I_C\,\vec{\omega} + m \vec{c}\times \left( \vec{v}_A - \vec{c}\times \vec{\omega} \right) \\ & = \left( I_C\,\vec{\omega}-m (\vec{c}\times \vec{c}\times \vec{\omega}) \right) + \vec{c}\times m \vec{v}_A \end{aligned}$$

In 6×6 matrix form (which I prefer) the above is

$$ \begin{bmatrix} \vec{L} \\ \vec{H}_A \end{bmatrix} = \begin{bmatrix} m \bf{1} & -m [\vec{c}] \\ m [\vec{c}] & I_C-m [\vec{c}][\vec{c}]\end{bmatrix} \begin{bmatrix} \vec{v}_A \\ \vec{\omega} \end{bmatrix} $$

where $[\vec{c}]$ is the cross product matrix operator defined as

$$[\begin{pmatrix}x\\y\\z\end{pmatrix}] = \begin{pmatrix}0&-z&y\\z&0&-z\\-y&x&0\end{pmatrix}$$

This 6×6 matrix is the spatial inertia matrix, seen also in the Spatial equations of motion.

The component $I_A = I_C-m [\vec{c}][\vec{c}]$ represents the parallel axis theorem, if the rotation is not about the center of mass. For the planar case $\vec{c}=(c_x,c_y,0)$ the mass moment about A becomes $I_A = I_C + m \left(c_x^2+c_y^2\right)$. Try it yourself.

Just for kicks I will point out that the instant center of rotation P is located at $$\vec{p} = \frac{\vec{\omega}\times\vec{v}_A}{\vec{\omega}\cdot\vec{\omega}}$$ relative to A

Note that $\vec{v}_P=\vec{v}_A+\vec{\omega}\times\vec{p}$ yields a parallel condition $\vec{v}_P \; \parallel \; \vec{\omega}$

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