A direct answer? It doesn't really have a name but has always been written in the literature in 3-vector form as $𝐊$, alongside the angular momentum 3-vector $𝐉$.

Any system, elementary or composite, relativistic or non-relativistic, that possesses a rest frame (i.e. a frame in which the momentum $𝐏$ is $𝟎$), and a center of mass position, $𝐫$, it may be called a "Tardyon" (or "Bradyon"). For such systems, the angular momentum decomposes into $𝐉 = 𝐫×𝐏 + 𝐒$, where $𝐒$ is its internal angular momentum; i.e. its angular momentum taken with respect to its center of mass position. When the system is elementary, and not composite, then this angular momentum component is referred to as its "spin". The other part, $𝐋 = 𝐫×𝐏$ is the "orbital" angular momentum of the system - which is the angular momentum it has by virtue of its motion around the origin $𝐫 = 𝟎$.

The other quantity, for such systems, decomposes as $𝐊 = M𝐫 - 𝐏t + 𝐓$. This has no official name and (as you can see) is explicitly time-dependent, by virtue of the appearance of $t$ for time. For the lack of a better name, you may call it the "moving mass moment", because of the dependence on the time and on the momentum $𝐏$.

The mass here, $M$, is itself a "moving" mass, as well. In non-relativistic theory it is $M = m$, equal to the mass $m$ of the system in its own rest frame - its rest mass. In Relativity, it has a dependence on the momentum given as $$M = m \sqrt{1 + \frac{1}{c^2}\frac{P^2}{m^2}}.$$ Nowadays, the "total energy" $E = Mc^2$ is usually used in the relativistic literature, in place of the moving mass $M$, but this obscures the discussion of what $𝐊$ is; which is just that it's the mass moment that the system would have had at $t = 0$, if its position is projected back to $t = 0$ by treating it as having moved between then and the present at the velocity $𝐯 = 𝐏/M$.

The additional quantity $𝐓$ - an analogue of internal angular momentum - is dependent on $𝐒$. For non-relativistic Tardyons, $𝐓 = 𝟎$, while in Relativity $$𝐓 = \frac{1}{c^2}\frac{𝐏×𝐒}{m + M} = \frac{𝐏×𝐒}{mc^2 + E}.$$ So the fact that it is non-zero and spin-dependent is a purely Relativistic effect.

In both the relativistic and non-relativistic cases, this can be solved $𝐫$ in terms of the canonical quantities $𝐉$, $𝐊$, $𝐏$ and $M$; with the result being the classical version of the Newton-Wigner operator for the center of mass position vector.

For all systems - Tardyons, Luxons, Tachyons - if there is no internal angular momentum - more precisely: if $𝐖 = 𝟎$, where $𝐖 ≡ M𝐉 + 𝐏×𝐊$ is the Pauli-Lubanski 3-vector, then solely by virtue of this fact, there is a decomposition into $𝐉 = 𝐫×𝐏$ and $𝐊 = M𝐫 - 𝐏t$. So, they're called "spin 0", which is an abuse of terminology, if the system is not a Tardyon.

Tachyons belong to a class which possess frames of reference in which $M = 0$, but $𝐏 ≠ 𝟎$ - an infinite speed frame. In that frame, $Π^2 = P^2$ is the square of an impulse, its value being $Π = P \sqrt{1 - M^2c^2/P^2}$ or $Π = P \sqrt{1 - E^2/(Pc)^2}$. The non-relativistic equivalent of this type of system has no name, so I've called it a "Synchron". Correspondingly, the infinite speed frame for a tachyon could be called a "synchron" frame: in it, it does not appear as an entity moving in time, but just as a spatial object existing at an instant. The non-relativistic version, the "synchron" is essentially the instantaneous transfer of non-zero impulse across space: the "-on" for action-at-a-distance dynamics.

The issue is muddy on how $𝐉$ and $𝐊$ decompose, except in the "spin 0" case.

Luxons have neither an infinite speed "Synchron" frame nor a rest frame. This can only happen if $P = Mc = E/c$.

Likewise, here, the issue is muddy on how $𝐉$ and $𝐊$ should decompose, except in the "spin 0" case. But, this time, there is also a marginal exception for the case where $𝐖$ and $𝐏$ align, with a fixed ratio $𝐖 = η𝐏$. This sub-class has no official name, either. So, I've referred to it as the "helical" case - or "Helion". The proportion $η$ is a fixed property of the system and is directly related to the component of the angular momentum $𝐉$ parallel to $𝐏$, which is called "helicity", with its value being $ηc$.

It's possible to write $𝐉 = 𝐫×𝐏 + η𝐏/M$, and $𝐊 = M𝐫 - 𝐏t$ but, this time, $𝐫$ is not quite a bona fide "center of mass position". It's possible to fix it to make it "canonical", but only at the cost of making $𝐫$ a singular function of the other canonical quantities $𝐉$, $𝐊$, $𝐏$ and $M$ (or $E$).

In the spin-0 case, for a non-interacting system, all the canonical quantities are constant, in time - as an expression of the conservation laws for them each. But since $𝐊$ is explicitly time-dependent, its constancy makes $𝐫$ a function of time: $𝐫 = 𝐫_0 + 𝐯t$, expressing a motion in a constant direction at a constant speed, located at $𝐫_0 = 𝐊/M$ at time $t = 0$, moving with constant velocity $𝐯 = 𝐏/M$. So, the conservation law for $𝐊$ is actually just the law of inertia, itself.

You'll have to work out what $𝐫$ looks like, as a function of time $t$ for the case of Tardyons with spin, where $𝐓 ≠ 𝟎$.