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It's proved in my K&K mechanics textbook that in pure rotation about an axis passing through the body it's angular momentum is $I\omega$.

What about if I want to find the angular momentum about an axis outside of it? Is it going to be the same?

In the special case of the body rotating about it's center of mass ,will it's angular momentum be the same for all axis?

My attempt to prove that for the above special case the angular momentum will be the same

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Let the angular momentum about the centre of mass be $L_0$ and about another axis be $L$

Then we can write$\begin{aligned} \vec{L} &=\sum R_{i} \times m_{i} v_{i} \\ &=\sum\left(R_{0}+r_{i}\right) \times m_{i} v_{i} \\ &=\sum R_{0} \times m_{i} v_{i}+\sum r_{i} \times m_{i} v_{i} \\ &=\left(R_{0} \times \sum m_{i} v_{i}\right)+L_{0} \end{aligned}$

We know that the velocity of centre of mass is zero therefore $\sum m_{i} v_{i}=0$

$\Rightarrow \quad \vec{L}=\vec{L}_{0}$

Where am I going wrong then?

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  • $\begingroup$ Note: Angular momentum is always measured about a point rather than an axis $\endgroup$ – DatBoi Jan 28 at 15:41
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No, it won't be the same. Always remember to specify the axis with respect to which you are evaluating the moment of inertia. Usually we calculate the moment of inertia with respect to the center of mass which is the smallest possible of your rigid body. So it would be more correct to write $I_0$ for the moment of inertia or $I_{CM}$. There is a theorem that can answer to your question.
The Huygens-Steiner theorem, or parallel axis theorem, allows to calculate the moment of inertia of a solid with respect to an axis parallel to the one passing through the center of mass avoiding in many cases (where there is a symmetrical structure) the laborious direct calculation.
The moment of inertia with respect to an axis $a$, parallel to another $c$ passing through the center of mass, is obtained by adding to the initial moment of inertia with respect to $c$ the product between the mass of the body itself and the square of the distance between the axes $c$ and from $a$. $$I_z=I_{cm}+Md^2$$ So if you have different moment of inertia, at the same angular velocity, you will have different angular momentum.
With this formula you can also see that the smallest moment of inertia is when $d=0$ so the moment of inertia respect to the axes through the CM: $I_{CM}$.
Hope this can help you.

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  • $\begingroup$ If the body is rotating about it's center of mass will it's angular momentum be the same for all axis? $\endgroup$ – Kashmiri Jan 10 at 12:04
  • $\begingroup$ No, it won't be because moment of inertia depends also on which axes you took. The are same cases where it's true but there must be particular symmetries. For example a disc: the moment of inertia through the CM and parallel to the radius is the same for every axes that is parallel to the radius and that crosses the CM. But the moment of inertia through the CM and perpendicular the radius is different. $\endgroup$ – Hitman Reborn Jan 10 at 12:14
  • $\begingroup$ Another example is the sphere: moment of inertia is the same for every axes that crosses the CM. This is due by the particular symmetry of the sphere. $\endgroup$ – Hitman Reborn Jan 10 at 12:21
  • $\begingroup$ Please have a look now $\endgroup$ – Kashmiri Jan 10 at 12:55
  • $\begingroup$ Now I am going to show that the moment of inertia is the same: let be $I_{CM}$ and $I_a$ the moment of inertia respect another axes. For definition: $I_a=\sum_im_iR^2_{i_a}$ but since $R_a=R_{CM}-d$ where d is the distance between the 2 axes we have: $$I_a=\sum_im_iR^2_{i_{CM}}+\sum_im_id^2-2d\sum_im_iR_{i_{CM}}$$ the last term is 0. So we have $$I_a=I_{CM}+Md^2$$ So for sure the moment of inertia is different. If angular momentum is conserved as you wrote and moment of inertia is different it implies that the angular velocity is changed. What you wrote is right but angular velocity changes. $\endgroup$ – Hitman Reborn Jan 10 at 13:20
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No, it wouldn't not be same as Moment of Inertia will change .

$\omega$ is same for every particle in a rigid body .

Hence , $ L_{axis}=I_{axis}\omega$ for a rigid body in pure rotation.

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  • $\begingroup$ If the body is rotating about it's center of mass will it's angular momentum be the same for all axis? $\endgroup$ – Kashmiri Jan 10 at 12:04
  • $\begingroup$ @YasirSadiq You wrote $\Sigma m_iv_i=0$. What if i replace $v_i$ with $\omega r_i$? $\endgroup$ – Bhavay Jan 10 at 14:30
  • $\begingroup$ You've use $\vec{\omega}$cross $r$ $\endgroup$ – Kashmiri Jan 10 at 14:50
  • $\begingroup$ @YasirSadiq Yes. Now, I can't seem to find the mistake in your work. $\endgroup$ – Bhavay Jan 10 at 15:13
  • $\begingroup$ Thank you for giving your time to go through it. :) $\endgroup$ – Kashmiri Jan 10 at 15:17

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