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There seem to be two definitions for the angular momentum of a particle with respect to a moving point.

See the figure below. Let $O$ be the origin of a coordinate system in an inertial frame. Let $Q$ be a point moving with respect to $O$. Let $K$ be the position of the particle. $\vec r_Q $ is the vector from $O$ to $Q$, $\vec r_K$ is the vector from $O$ to $K$, and $\vec r_{KQ} = \vec r_K - \vec r_Q$ is the vector from $Q$ to $K$. The velocity of $K$ in the inertial frame is $\vec v_K$, the velocity of Q in the inertial frame is $\vec v_Q$, and the velocity of $K$ with respect to $Q$ is $\vec v_ {KQ} = \vec v_ K - \vec v_Q$. Some physics texts seem to define the angular momentum of $K$ with respect to $Q$, $\vec L_{KQ}$, as $$(1) \vec L_{KQ} = m[(\vec r_K - \vec r_Q) \times (\vec v_K - \vec v_Q)]$$ For example, see Symon, Mechanics

Engineering texts seem to use the definition $$(2) \vec L_{KQ} = m[(\vec r_K - \vec r_Q) \times \vec v_K] $$ For example, see the following text by Kochmann online: https://ethz.ch/content/dam/ethz/special-interest/mavt/mechanical-systems/mm-dam/documents/Notes/Dynamics_LectureNotes.pdf.

Updated Information

I recently found an MIT physics class lecture online that uses relationship (2). See https://www.youtube.com/watch?v=NHedXxUO-Bg. The professor specifically emphasizes that for $\vec L_{KQ} = (\vec r_K - \vec r_Q) \times m\vec p$ where $\vec p$ is the linear momentum of the particle, the momentum must be evaluated using the velocity of $K$ with respect to $O$, that is $\vec p = m\vec v_K$.

Is relationship (1) ever used in practice?

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