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## Main Question or Discussion Point

The angular momentum of a system of particles α = 1,2,3,...n can be written as the sum of the two angular momenta:

[tex]\vec{L} = \vec{R} \times \vec{P} + \sum_{\alpha} \vec{r_{\alpha}}' x \vec{\rho_{\alpha}}'[/tex]

where the first term is the angular momentum of the center of mass with all mass M = Ʃ m

If we differentiate with respect to time, we get the torque:

[tex]\vec{N} = \frac{d}{dt} \vec{L} = \frac{d}{dt} \vec{R} \times \vec{P} + \frac{d}{dt} \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{\rho_{\alpha}}' = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}'[/tex]

where

We can write the effective force in terms of the real force by:

[tex]\vec{F_{\alpha}}' = \vec{F_{\alpha}} - m_{\alpha} \frac{d^2}{dt^2} \vec{R}[/tex]

And so the above becomes:

[tex]\vec{N} = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}} - \sum{m_{\alpha} \vec{r_{\alpha}}' \times \frac{d^2}{dt^2} \vec{R}}[/tex]

But the second sum is zero because of how the center of mass is defined and so:

[tex]\vec{N} = \vec{N} = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}[/tex]

This says that:

[tex]\vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}' = \vec{N} = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}[/tex]

So eiher I have an inconsistency in my derivation or the torque can be measured by looking at the observed accelerations in both the inertial frame and the non-inertial frame so as long as the non-inertial frame has the motion characteristics as the center of mass. Is this true?

[tex]\vec{L} = \vec{R} \times \vec{P} + \sum_{\alpha} \vec{r_{\alpha}}' x \vec{\rho_{\alpha}}'[/tex]

where the first term is the angular momentum of the center of mass with all mass M = Ʃ m

_{α}, and the second term is the total angular momentum of the system with respect to the center of mass.If we differentiate with respect to time, we get the torque:

[tex]\vec{N} = \frac{d}{dt} \vec{L} = \frac{d}{dt} \vec{R} \times \vec{P} + \frac{d}{dt} \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{\rho_{\alpha}}' = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}'[/tex]

where

**F**is the net external force on the system and**F**' is the "effective" force as seen with respect to the center of mass._{α}We can write the effective force in terms of the real force by:

[tex]\vec{F_{\alpha}}' = \vec{F_{\alpha}} - m_{\alpha} \frac{d^2}{dt^2} \vec{R}[/tex]

And so the above becomes:

[tex]\vec{N} = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}} - \sum{m_{\alpha} \vec{r_{\alpha}}' \times \frac{d^2}{dt^2} \vec{R}}[/tex]

But the second sum is zero because of how the center of mass is defined and so:

[tex]\vec{N} = \vec{N} = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}[/tex]

This says that:

[tex]\vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}' = \vec{N} = \vec{R} \times \vec{F} + \sum_{\alpha} \vec{r_{\alpha}}' \times \vec{F_{\alpha}}[/tex]

So eiher I have an inconsistency in my derivation or the torque can be measured by looking at the observed accelerations in both the inertial frame and the non-inertial frame so as long as the non-inertial frame has the motion characteristics as the center of mass. Is this true?