A solid sphere is spinning about z axis. I know that its angular momentum about the z axis will be L=Iw. Where I is the moment of inertia about its central axis and w is its angular speed about z axis. But, what will be the angular momentum about x or y axis or about any arbitrary axis other than z axis ? Is it zero or non zero?

  • 1
    $\begingroup$ Does this answer your question? Angular momentum of a purely rotating body about any axis $\endgroup$
    – Rishi
    Commented Apr 28, 2023 at 5:11
  • 1
    $\begingroup$ @Rishi the Q&A in the linked pages are irrelevant to this question. $\endgroup$ Commented Apr 28, 2023 at 6:45
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 no, you are wrong, Rishi's linked page answers it correctly. $\endgroup$ Commented Apr 28, 2023 at 7:52
  • $\begingroup$ @naturallyInconsistent If I am wrong, please, let me know where in the question or on the answers there is mention of an axis not parallel to the rotational axis. $\endgroup$ Commented Apr 28, 2023 at 9:32
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 although the answers there don't exactly address the issue here, the linked question seems to be just a broader version of this one. It asks about the projection of the angular momentum about an arbitrary point in an arbitrary direction (while this question is just about the direction). $\endgroup$
    – Rishi
    Commented Apr 28, 2023 at 18:17

1 Answer 1


We'll use the classical treatment for angular momentum, where $\mathbf{L}=I\boldsymbol{\omega}$. Here, $I$ is the moment of inertia tensor. Since a sphere is isotropic, its moment of inertia tensor is proportional to the identity matrix, and hence its angular momentum is always parallel to its angular velocity.

When the angular velocity is along the $z$-axis, the only nonzero component of $\mathbf{L}$ is about the $z$-axis. The angular momentum about another axis, given by unit vector $\mathbf{\hat{n}}$, is the projection of $\mathbf{L}$ onto that axis, $\mathbf{L}\boldsymbol{\cdot}\mathbf{\hat{n}}$. In particular, the projection of $\mathbf{L}$ onto the $x$ and $y$ axes are $0$.

  • $\begingroup$ Probably your answer could be more complete if you would add a few words to clarify the relation between "angular momentum around an axis" and "cartesian component of the angular momentum". $\endgroup$ Commented Apr 28, 2023 at 6:48
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 Made an edit! $\endgroup$
    – DanDan面
    Commented Apr 28, 2023 at 7:11
  • $\begingroup$ I have to say, I am utterly confused. W.r.t where are you taking the angular momentum ? Isn't it dependent on where you take it around $\endgroup$
    – Babu
    Commented Apr 29, 2023 at 4:29
  • $\begingroup$ @TrystwithFreedom it's encoded in $I$, here I have chosen the center of mass of the sphere $\endgroup$
    – DanDan面
    Commented Apr 29, 2023 at 4:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.