Consider an irregular rigid object of mass ##M## simultaneously rotating and translating in space in the absence of external forces and torques. This means that both its linear and angular momentum are conserved. We will calculate its angular momentum about an arbitrary point A (see diagram below) given that the velocity of the center of mass is ##\vec V_{cm}## and its rotational angular velocity is ##\vec \omega.##
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First we point out that the axis of rotation must pass through the CM. That's because linear momentum is conserved which means that ##\vec V_{cm}= \rm{const.}## If the axis of rotation did not pass through the CM, the CM would rotate about that hypothetical axis and its velocity would change direction which means that it cannot be constant. Thus, the linear velocity relative to the CM of an arbitrary point at position ##\vec {r}'## from the CM is ##\vec {v}'=\vec{\omega}\times \vec {r}'.##
Now consider mass element ##dm## at an arbitrary point P the position vector of which relative to A is ##\vec r##. Let ##\vec {r}' ## be the position vector of ##dm## relative to the CM and ##\vec R## be the position of the CM relative to A. Finally, let ##\vec v_P## be the instantaneous velocity of point P relative to point A. Addition of velocities requires that the velocity of point P relative to A is ##\vec v_P=\vec V_{cm}+\vec{\omega}\times \vec {r}'.##
The angular momentum contribution of ##dm## about point A is $$d\vec L=(dm)\vec r\times \vec v_P=(dm)(\vec R+\vec {r}')\times(\vec V_{cm}+\vec{\omega}\times \vec {r}')$$ We multiply out the cross product on the RHS and consider each of the four resulting terms separately.
1. ##~(dm) \vec R \times \vec V_{cm}##
The contribution from the first term to the total angular momentum is $$\vec L_1=\int (dm) \vec R \times \vec V_{cm}=\left( \int dm \right) \vec R \times \vec V_{cm}=M\vec R \times \vec V_{cm}.$$ This is often called the angular momentum
of the center of mass.
2. ##~(dm)\vec R\times (\vec{\omega}\times \vec {r}')##
Using the
triple-product rule,
$$\vec R\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec R\cdot \vec {r}') -\vec {r}'(\vec R \cdot \vec{\omega}).$$The second term on the RHS is zero because ##\vec R## is perpendicular to ##\vec {\omega}##. To find the contribution of the first term, we integrate $$\vec L_2=\int (dm)\vec{\omega}(\vec R\cdot \vec {r}')=\vec{\omega}\left(\vec R\cdot \int (dm)\vec {r}'\right)=0$$ Note that by definition of the CM coordinates, ##\int (dm)\vec {r}'=0.##
3. ##~(dm)\vec {r}'\times \vec V_{cm}##
This term also vanishes upon integration for the same reason when we take constant ##\vec R## out of the integral. $$\vec L_3=\int (dm)\vec {r}'\times \vec V_{cm}=\left(\int (dm)\vec {r}'\right)\times \vec V_{cm}=0.$$ 4. ##~(dm)\vec {r}'\times (\vec{\omega}\times \vec {r}')##
Triple-product rule once more
$$\vec {r}'\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec {r}'\cdot \vec {r}')-\vec {r}'(\vec{\omega}\cdot \vec {r}').$$The second term on the RHS vanishes because the vectors are orthogonal. Integrating the first term, $$\vec L_4=\int \vec{\omega}(\vec {r}'\cdot \vec {r}')=\vec{\omega}\int (dm){r'}^2=I_{cm}~\vec{\omega}.$$This is called the angular momentum
about the center of mass. Thus, the total angular momentum about point A of the translating and rotating rigid body is $$\begin{align} & \vec L=\vec L_1+\vec L_4=M\vec R \times \vec V_{cm}+I_{cm}~\vec{\omega}.\end{align}$$ The first term in this expression, angular momentum
of the center of mass, depends on the choice of reference point A. The second term, angular momentum
about the center of mass, does not.
The bottom line is that one can always write the angular momentum of a translating and rotating rigid body in the form of Equation (1).
Note that a simplification occurs if point A is chosen anywhere on the straight line path of the CM. Then ##\vec {R}## and ##\vec V_{cm}## are parallel and the angular momentum
of the CM vanishes. This is why in the asteroid problem and all problems like it, it is convenient to choose the CM as a point of reference for the angular momentum and immediately write the angular momentum conservation equation in the form $$m~v~d=I_{cm}~\omega$$ where ##d## is the vertical distance between the path of the asteroid and the CM of the combined object. This equation says that the initial orbital angular momentum about the CM of the asteroid before the collision is converted to spin angular momentum of the combined object after the collision. Furthermore, as we have seen, choosing point A away from the path of the CM adds equal amounts of (orbital) angular momentum of the CM on both sides of the momentum conservation equation which cancel out. (See post #178.)