Does the spin angular momentum count?

[christang_1023]

Taking the Earth orbiting the sun as an example, when I consider the angular momentum of the Earth about the sun, should the spin angular momentum be counted?
I'm confused that if it's counted, the spin angular momentum, Lcm=Icm×ωspin, is different from other angular momentum regarding the sun as the axis.

 

[PeroK]

Taking the Earth orbiting the sun as an example, when I consider the angular momentum of the Earth about the sun, should the spin angular momentum be counted?
I'm confused that if it's counted, the spin angular momentum, Lcm=Icm×ωspin, is different from other angular momentum regarding the sun as the axis.

"Counted" in what sense?

 

[christang_1023]

"Counted" in what sense?

Sorry, my English is not that good.
I was trying to say if I calculate the total angular momentum of the Earth about the axis, do I need to add the spin angular momentum of the earth?

 

[PeroK]

Sorry, my English is not that good.
I was trying to say if I calculate the total angular momentum of the Earth about the axis, do I need to add the spin angular momentum of the earth?

About what axis?

 

[christang_1023]

About what axis?

the sun

 

[PeroK]

the sun

Okay. In general, you have a body with an angular momentum about a given point. And that body is also rotating about an axis through its centre of mass. Your question is whether the rotation about the centre of mass gives additional angular momentum about a point? Or, whether the rotational angular momentum cancels out?

What do you think the answer is?

 

[christang_1023]

Okay. In general, you have a body with an angular momentum about a given point. And that body is also rotating about an axis through its centre of mass. Your question is whether the rotation about the centre of mass gives additional angular momentum about a point? Or, whether the rotational angular momentum cancels out?

What do you think the answer is?

$$\vec{L}=\sum \vec{r_{i}}\times \vec{p_{i}}=\sum \vec{r_{i}}\times (\vec{p_{i}^{cm}}+\vec{p_{cm}})=\sum \vec{r_{i}} \times \vec{p_{i}^{cm}}+\sum \vec{r_{i}} \times \vec{p_{cm}}$$

Because in the center of mass reference, the total momentum is equal to 0, can I say the rotational angular momentum cancels out?

 

[PeroK]

$$\vec{L}=\sum \vec{r_{i}}\times \vec{p_{i}}=\sum \vec{r_{i}}\times (\vec{p_{i}^{cm}}+\vec{p_{cm}})=\sum \vec{r_{i}} \times \vec{p_{i}^{cm}}+\sum \vec{r_{i}} \times \vec{p_{cm}}$$

Because in the center of mass reference, the total momentum is equal to 0, can I say the rotational angular momentum cancels out?

You need to take $\vec{r_i} = \vec{r}_{cm} + \vec{q_i}$ and $\vec{p_i} = m_i\vec{v}_i = m_i(\vec{v}_{cm} + \vec{u_i})$.

Where $q, u$ are the displacements and velocities relative to the centre of mass.

To answer the question, think about the angular momentum about the centre of the object.

 

[christang_1023]

You need to take $\vec{r_i} = \vec{r}_{cm} + \vec{q_i}$ and $\vec{p_i} = m_i\vec{v}_i = m_i(\vec{v}_{cm} + \vec{u_i})$.

Where $q, u$ are the displacements and velocities relative to the centre of mass.

To answer the question, think about the angular momentum about the centre of the object.

Thank you for your useful hints. I get $$ \vec{L_{tot}}=\sum m_{i}\vec{q_i} \times \vec{u_i}+\sum M_{tot}\vec{r_{cm}} \times \vec{v_{cm}}=\vec{L_{spin}}+\vec{L_{orbit}} $$, so does it mean that the rotational angular momentum actually add extra angular momentum about a point instead of cancelling out?

 

[PeroK]

Thank you for your useful hints. I get $$ \vec{L_{tot}}=\sum m_{i}\vec{q_i} \times \vec{u_i}+\sum M_{tot}\vec{r_{cm}} \times \vec{v_{cm}}=\vec{L_{spin}}+\vec{L_{orbit}} $$, so does it mean that the rotational angular momentum actually add extra angular momentum about a point instead of cancelling out?

Yes, the total angular momentum is the angular momentum of the centre of mass plus the spin angular momentum. You can prove it by expanding the expression in post #7 as I suggested in post #8. Two of the terms are zero, and the other two are as required.

 

[christang_1023]

Yes, the total angular momentum is the angular momentum of the centre of mass plus the spin angular momentum. You can prove it by expanding the expression in post #7 as I suggested in post #8. Two of the terms are zero, and the other two are as required.

I get what you said, and thank you very much!