1
$\begingroup$

We know that angular momentum of a body is defined about a point in space. Let us consider a solid cylinder whose radius is R and mass is M. It has a moment of inertia defined around the axis of rotation as $I=\frac{1}{2}MR^2$. Now if it moves with angular velocity $\omega$.
Compare the equation $L=I\omega$ with previous, the magnitude of angular momentum is $L=\frac{1}{2}MR^2\omega$. We know that angular momentum is defined about a point and NOT an axis. So my question is: Which point on the axis is our angular momentum defined about? enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ I think you can use the Parallel axis theorem to convince yourself that in this particular case it doesn't matter which point along the axis of rotation you calculate the angular momentum from. $\endgroup$
    – Amit
    May 9, 2023 at 13:31
  • 1
    $\begingroup$ $L = I \omega$ is correct, not $L = I \omega^2$. You confusing angular momentum with kinetic energy here. $\endgroup$ May 9, 2023 at 20:31
  • $\begingroup$ Oh sorry, I corrected it. $\endgroup$ May 10, 2023 at 6:34

2 Answers 2

5
$\begingroup$

Consider any two points $\vec R$ and $\vec R'$. The angular momentum about $\vec R'$ is $$\vec L' = \int \vec R'\times \vec v\ dm$$ $$= \int \vec R\times \vec v\ dm + \int (\vec R'-\vec R)\times \vec v\ dm $$ $$= \vec L +(\vec R' -\vec R)\times \vec P,$$ where $\vec L$ is the angular momentum about $\vec R$ and $\vec P$ is the linear momentum of the system. If you are in a reference frame in which $\vec P = 0$, the second term is zero, meaning the angular momentum about any point is the same. It is given by $\vec L = \mathbf I\vec \omega$, $\mathbf I$ being the moment of inertia tensor about a point on the axis of rotation.

In your case the system, a uniform cylinder spinning about a central axis, does have have $\vec P = 0$, so the angular momentum about any point is the same.

Note: angular momentum (in such a case of high symmetry) is $I\omega$, not $I\omega^2$.

$\endgroup$
1
  • $\begingroup$ Yep, thanks for reminding. I corrected $\omega^2$. $\endgroup$ May 10, 2023 at 6:36
1
$\begingroup$
  1. Angular momentum is defined about an axis, and that includes all points along the axis. This axis is parallel to the momentum vector and goes through some point in space.

  2. Mass moment of inertia tensor is defined about a point, as it represents all the MMOI values for the pencil of lines that go through that point.

  3. To move angular momentum you use the transformation rule

    $$\boldsymbol{L}_A = \boldsymbol{L}_B + (\boldsymbol{r}_A-\boldsymbol{r}_B) \times \boldsymbol{p} $$

    where $\boldsymbol{r}_A$ and $\boldsymbol{r}_B$ are the position vectors of the two points, $\boldsymbol{p}$ is the linear momentum vector and $\boldsymbol{L}_A$ and $\boldsymbol{L}_B$ are the angular momentum vectors.

    As you can see if you move either point along the direction of $\boldsymbol{p}$ then the result does not change due to the cross product $\times$ operation. This means that parallel movements along the direction of linear momentum do not change angular momentum and thus angular moment is defined about an axis.

  4. Angular momentum at point A can be defined in terms of a mass moment of inertia tensor ${\rm I}_A$ if the body is purely rotating about this point

    $$\boldsymbol{L}_B = {\rm I}_B\, \boldsymbol{\omega}$$

    $$\boldsymbol{L}_A = {\rm I}_B\, \boldsymbol{\omega} + (\boldsymbol{r}_A-\boldsymbol{r}_B) \times m (\boldsymbol{\omega} \times (\boldsymbol{r}_A-\boldsymbol{r}_B)) = {\rm I}_A \boldsymbol{\omega}$$

    since the velocity of point B that is orbiting about A is $\boldsymbol{v}_B = \boldsymbol{\omega} \times (\boldsymbol{r}_A-\boldsymbol{r}_B)$

    The above is known as the parallel axis theorem

  5. In 2D the above is simply $I_A = I_B + m d^2$ where $d$ is the perpendicular distance between the points

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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