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Torque as measured from center of mass.

  1. Oct 10, 2013 #1
    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 mα, 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 Fα' is the "effective" force as seen with respect to the center of mass.

    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?
  2. jcsd
  3. Oct 11, 2013 #2


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    As you say, the total angular momentum separates into two parts: the one of the center of mass (orbital), and the one with respect to the center of mass (spin).
    The time derivative of the total angular momentum will equal the net external torque on the system, which separates again into a term on the center of mass and a term with respect to the center of mass.

    Some of the torque will contribute to the change in orbital angular momentum, and this is expressed just as the cross product of the center of mass vector and the net force on the total system.

    The rest of the torque will contribute to a change in the spin angular momentum, and can be nonzero even if the net force is zero.
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