# Confusion about angular momentum of rigid bodies

I know that angular momentum of a rigid body has a parallel and perpendicular component to the angular velocity. I also know that the perpendicular component is due to constraints that keep the axis fixed, right? Then I know that perpendicular component is zero if the axis of rotation is also symmetry axis of rigid body. Here with symmetry we mean material symmetry, so that the center of mass is always on the symmetry axis, right? Then why the angular momentum with respect to the center of mass can also has a perpendicular component? Intuitively I really don't understand why is it needed since the center of mass is on the axis of rotation, if the perpendicular component is always due to constraints that keep the axis fixed. Basically I think that I don't understand how can be possible that principal axis of inertia don't pass through the center of mass.

Then another doubt: the decomposition of angular momentum in parallel and perpendicular component and all the following consequences, hold also for a moving axis of rotation?

I am honestly confused about your question, but no worries, I will try to be as thorough as possible in the hope of helping you. At this point, I will not prove any equation from first principles (Newtow's Laws), but should you want that or any further clarification, just ask.

First let´s have a look at the equation of motion:

$$\sum \vec{M_{O_i}^{ext}} = \dot{\vec{H_O}}$$

In this equation the net external moment (also known as net external torque), $$\sum \vec{r_{i/O}} \times \vec{F_i^{ext}}$$, is equated to the time derivative of angular momentum. The assumption of central internal forces was made to avoid having to sum internal moments as well, but that's nothing to worry about since in typical mechanical systems that assumption is valid. It should be noted that point O has to either be a fixed point in space or the center of gravity (G).

$$\vec{H_G}$$ can be calculated using the following equation. The moments and products of inertia should be calculated with respect to point G.

$$\vec{H_G} = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{yx} & I_{yy} & -I_{yz} \\ -I_{zx} & -I_{zy} & I_{zz} \end{bmatrix} \cdot \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{bmatrix}$$ For any other point in space,

$$\vec{H_O} = \vec{H_G} + \vec{r_{G/O}} \times \vec{L}$$ , where $$\vec{L}$$ is the linear momentum (which is equal to $$M\cdot\vec{v_G})$$

Note: Remember to take the time derivative of the angular momentum when using the equation of motion.

Now that the fundamentals have been established, I will address some specific points in your question.

(i) angular momentum of a rigid body has a parallel and perpendicular component to the angular velocity. True. Unless the angular velocity is an eigenvector (of the inertia matrix), the angular momentum does not point in the direction of the angular velocity.

(ii) perpendicular component is zero if the axis of rotation is also symmetry axis of rigid body. Rotational symmetry about an axis, means that if the inertia tensor is rotated until that axis matches one of the coordinated axis (x,y,z), the inertia tensor will then be diagnolized and two of its moments of inertia will be the same. In such a case, an angular velocity pointing in the direction of the axis of symmetry is clearly an eigenvector, so there are no perpendicular components.

(iii) why the angular momentum with respect to the center of mass can also has a perpendicular component. That depends on the angular velocity, if it is an eigenvector, angular momentum will point in the same direction. Otherwise, it may have other components.

If you are struggling to accept what the equations are showing you, I suggest you go through the trouble of deriving yourself the equations.

The last thing I would like to mention is your comments on "constraints". The constraits in problems typically are either connected with the loading (forces and moments) or with the kinematics (such as the direction of the angular velocity or the velocity of a particular point), NOT with the direction of the angular momentum, much less, the direction of its derivative.

• First of all thanks for the very complete answer. When we say axis of symmetry, we mean the geometrical axis of symmetry or the material axis of symmetry? For example for a non homogeneous sphere they are different. So, if we always mean the material axis of symmetry, and rotations about it have perpendicular angular momentum=0, then we have to conclude, since rotations about G can have it different from 0, that G doesn't always lays on the symmetry axis, right?
– abc
Jan 13 at 16:40
• And about the physical intuition for perpendicular component of angular momentum, I talked about constraints because I thought that it is due by the constraints that keep the axis of rotation fixed. For example when we do the balancing of wheels, I thought it was because we wanted to reduce the perpendicular angular momentum when for example the wheel doesn't rotate anymore about its axis of symmetry (which is where the center of mass lies)
– abc
Jan 13 at 16:46
• Yes, we mean mass weighted average. Not all bodies have axis of symmetry, but those that do have, if homogeneous, always contain point G. But again, no matter the point in space you are calculating angular momentum with respect to, there will generally be a direction for the angular velocity for which the corresponding angular momentum will point in a direction other than that of angular velocity. However, for a homegeneous body, if angular velocity points along a axis of symmetry there will not be a perpendicular component. Jan 15 at 20:13
• Ok thanks. Take for example any rigid body, even with no geometrical symmetry or non homogeneous. It will certainly have a center of mass. If we mean that the axis of symmetry is not the geometrical one but simply the axis passing through the center of mass of the object, then how can rotations about center of mass have a non zero perpendicular component of angular momentum? Or maybe do we mean that that component is surely zero (other than in general for rotations about principal axis) for rotations about the GEOMETRICAL axis of symmetry, and not the "material" one?
– abc
Jan 16 at 17:40
• Anyway I think I quite understand now the tensor of inertia, but there's only one other thing that I'm not sure about. Take for example a door that it's opening. It's rotating not about an axis of symmetry and I'm quite sure not about a principal axis. Then it will have also a perpendicular component of angular momentum. Can we say that this component is due to the constraints of the door that in this case keep the axis fixed, because the center of mass is moving?
– abc
Jan 16 at 17:46