I'm looking to evaluate whether the following is true or false:

For rotation of a rigid body about an arbitrary axis, the angular momentum always points along the axis of rotation.


$$r = x \hat i + y \hat j + z \hat k$$ and

$$ v = v_x \hat i + v_y \hat j + v_z \hat k$$

Then, the angular momentum is defined as:

$$\vec L = \vec r \times m\vec v$$

$$r \times v = (yv_z-zv_y)\hat i - (xv_z-zv_x)\hat j + (xv_y-yv_x) \hat k$$

$$L = m \left((yv_z-zv_y)\hat i - (xv_z-zv_x)\hat j + (xv_y-yv_x) \hat k\right)$$

Here, $L$ is not along a principle axis, but has $3$ components. If the particles rotating exhibit $v_x,v_y$ or $v_z = 0$, it is plain to see $L$ is now on one principle axis. However, it isn't clear to me that if $L$ has $3$ components that it doesn't still point along the axis of rotation. Maybe the axis of rotation has $3$ components as well?

Basically, I think my confusion boils down to whether angular momentum's vector determines the axis of rotation (which I think would make sense since it's proportional to $\omega$) which would make the statement true, however, it doesn't seem impossible to me that the axis of rotation and angular momentum can not necessarily be parallel.


marked as duplicate by Emilio Pisanty, Qmechanic Nov 19 '17 at 6:01

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