# Angular momentum in rotating non-inertial reference frame

I know in inertial reference frame rigid body we can write $$L=I\omega$$ where $$L$$ is the angular momentum of the rigid body, $$I$$ is the moment of inertia and $$\omega$$ is the angular velocity of the rigid body.

In a non inertial reference frame what will be the angular velocity of the rigid body? Will the relation $$L=I \omega$$ still hold?

I tried by starting with $$V_{\text{o in r frame}} = V_{\text{r frame}} + rw$$ where $$r$$ is position vector w.r.t $$r$$ frame .

And angular velocity $$= r\times mv$$, Substituting $$v$$ and proceeding. I am not ending anywhere.

Any help is appreciated.

It won't hold.

Consider an inertial system $$S$$ with origin $$O$$ and anotther system $$S'$$ with origin $$O'$$. The vectors of the basis of $$S'$$ rotate wrt the vectors of the basis of $$S$$ with angular velocity $$\mathbb{\Omega}$$. Then, $$$$\mathbf{L}^{O} = \mathbf{L}^{O'} + \mathbf{r_{O'}}^{O} \times \mathbf{P}^{O} + M\mathbf{R}^{O} \times \mathbf{v_{O'}}^{O} - M\mathbf{r_{O'}}^{O} \times \mathbf{v_{O'}}^{O}$$$$ where the superscripts refers to the origin from where it's considered and the capitalized magnitudes refer to the center of mass position and velocity.

We can write the first term $$$$\mathbf{L}^{O'} = I^{O'}\mathbf{\Omega}$$$$ getting $$$$\mathbf{L}^{O} = I^{O'}\mathbf{\Omega} + \mathbf{r_{O'}}^{O} \times \mathbf{P}^{O} + M\mathbf{R}^{O} \times \mathbf{v_{O'}}^{O} - M\mathbf{r_{O'}}^{O} \times \mathbf{v_{O'}}^{O}$$$$

There are 2 particular cases in which this is symplified a lot. Take $$O'$$ as

• the center of mass. $$$$\mathbf{L}^{O} = I^{CM}\mathbf{\Omega} + \mathbf{R}^{O} \times \mathbf{P}^{O}$$$$ The angular momentum can be seen as a sum of the angular momentum of the body wrt to the center of mass (spin) and the angular momentum of the center of mass wrt to $$O$$ (orbital).

• a point with no speed wrt to $$O$$. $$$$\mathbf{L}^{O} = I^{O'}\mathbf{\Omega} + \mathbf{r_{O'}}^{O} \times \mathbf{P}^{O}$$$$ but if $$O'$$ is fixed, $$S'$$ is inertial and we might as well consider $$O = O'$$, in which case $$$$\mathbf{L}^{O} = I^{O}\mathbf{\Omega}$$$$

Let's assume that the rigid body rotates about its center of mass (COM), and the origin of the inertial frame $$F$$ is taken at the COM of the rigid body. Let $$F'$$ be the rotating frame and $$\Omega$$ the angular velocity of $$F'$$ relative to $$F$$. Assuming further that the origin of $$F'$$ coincides with $$F$$, then the angular velocity $$\omega'$$ of the rigid body in $$F'$$ is simply $$\omega'=\omega-\Omega$$. When $$\omega'>0,$$ it means that the rigid body is rotating in the same direction w.r.t $$F'$$ as to $$F$$. and vice versa.

Now, consider the origin of the $$F'$$ is not at the center of mass of the rigid body, there is an orbital motion of the COM w.r.t $$F'$$. The orbital velocity $$v'$$ of the COM relative to $$F'$$ can be found by using the velocity transformation law $$v=V+\Omega\,\times r'+v',$$ where $$v$$ is the velocity vector of the COM relative to $$F$$, $$V$$ is the velocity vector of the origin of $$F'$$ relative to $$F$$, $$r'$$ is the COM's position vector relative to $$F'$$ and $$v'$$ is the velocity vector of COM relative to $$F'$$. In this case, $$v$$ is zero since the COM is certainly not moving w.r.t to $$F$$ and $$V$$ vanishes if we assume there is no translational velocity of the origin of $$F'$$ relative to $$F$$. So the orbital angular momentum of the COM relative to $$F'$$ is $$v'=-\Omega\,\times r',$$ and the orbital angular momentum of COM is $$L'_{orb}=Mr'\times v',$$ where $$M$$ is the mass of the rigid body. The total angular momentum of the rigid body relative to $$F'$$ is therefore $$L'=L'_{orb}+I\omega.$$ The second term remains as the rotational motion of the rigid body about its COM is unchanged unless the origin is fixed at the COM of the rigid body.

• Okay, so it will not hold ...what if we consider non inertial frame , but not rotating? Commented Aug 16, 2021 at 4:00
• Or will that be a completely different question? Commented Aug 16, 2021 at 4:01
• @Rover It does not hold in general since there is an additional motion of COM relative to the frame. For example, in a translational accelerating frame, unless the radius vector of the COM is parallel with its velocity, there is always an extra angular momentum induced by the acceleration. Commented Aug 16, 2021 at 10:53