torque

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Let the force acting on the p^{th}\, particle be \mathbf{F}_p\, which is broken up into an external component and a component due to forces within the body,

\mathbf{F}_p = \mathbf{F}_{p,ext} + \sum_{q} \mathbf{F}_{p,q}\,.

Then

m_p \frac{d^2}{dt^2} \left( \mathbf{R} + \mathbf{r}_p \right) =\mathbf{F}_{p,ext} + \sum_{q} \mathbf{F}_{p,q}\,.

Summing over all particles gives

\sum_p m_p \frac{d^2}{dt^2} \left( \mathbf{R} + \mathbf{r}_p \right)\, = \sum_p \mathbf{F}_{p,ext} + \sum_p \sum_{q} \mathbf{F}_{p,q}\,,
M \frac{d^2}{dt^2}\mathbf{R}\, = \sum_p \mathbf{F}_{p,ext}\,,
M \frac{d\mathbf{V}}{dt}\, = \mathbf{F}_{ext}\,,

since \mathbf{F}_{p,q} = -\mathbf{F}_{q,p}\,. For an individual particle, then

m_p \frac{d}{dt} \left( \mathbf{V} + \mathbf{v}_p \right) =\mathbf{F}_{p}\,, or
m_p \frac{d\mathbf{v}_p}{dt} = \mathbf{F}_{p} - \frac{m_p}{M}\mathbf{F}_{ext}\,.

Thus \mathbf{f}_p \equiv \mathbf{F}_{p} - \frac{m_p}{M}\mathbf{F}_{ext}\, is a force that does not contribute to the acceleration of the center of mass. To see this, note that \sum_p \mathbf{f}_p = 0\,. These forces only give rise to rotations. We define the moment acting on the p^{th}\, particle to be

\boldsymbol{\tau}_p = \mathbf{r}_p \times \mathbf{f}_p\,.

Furthermore, we can separate the moment into internal and external contributions:

\boldsymbol{\tau}_p = \boldsymbol{\tau}_{p,ext} + \sum_{q} \boldsymbol{\tau}_{p,q}\,,

where

\boldsymbol{\tau}_{p,ext} = \mathbf{r}_p \times \left( \mathbf{F}_{p,ext} - \frac{m_p}{M}\mathbf{F}_{ext} \right)\,,
\boldsymbol{\tau}_{p,q} = \mathbf{r}_p \times \mathbf{F}_{p,q}\,.

We define the torque to be the sum of all the moments acting on the body:

\boldsymbol{\tau}_{ext} = \sum_p \boldsymbol{\tau}_{p,ext}\,.

Note then that

\sum_{p,q} \boldsymbol{\tau}_{p,q}\, =\frac{1}{2} \sum_{p,q} \left( \mathbf{r}_p + \mathbf{r}_p\right)\times \mathbf{F}_{p,q}\,,
= \frac{1}{2} \sum_{p,q} \left( \mathbf{r}_p - \mathbf{r}_q\right)\times \mathbf{F}_{p,q}\,, since \mathbf{F}_{pq} = -\mathbf{F}_{qp}\,,
= 0\,,

since \mathbf{F}_{pq}\, is directed along \mathbf{r}_p - \mathbf{r}_q\,.

Torque and angular momentum

The change in angular momentum of the rigid body is

\frac{d\mathbf{L}_{total}}{dt}\, = \sum_p \left(\mathbf{R}+\mathbf{r}_p\right) \times \frac{d}{dt}\left(\mathbf{V}+\mathbf{v}_p\right)\,
= M \mathbf{R} \times \frac{d\mathbf{V}}{dt} + \sum_p m_p \mathbf{r}_p\times \frac{d\mathbf{v}_p}{dt}\,,

On the other hand, since \boldsymbol{\omega} = \hat{\mathbf{r}}_p \times \frac{\mathbf{v}_p}{r_p}\, and \mathbf{v}_p = \boldsymbol{\omega}\times\mathbf{r}_p\,,

\frac{d\mathbf{v}_p}{dt} = \dot{\boldsymbol{\omega}} \times \mathbf{r}_p\,,

so that

\frac{d\mathbf{L}_{total}}{dt} = M \mathbf{R} \times \frac{d\mathbf{V}}{dt} + \sum_p m_p \mathbf{r}_p\times \dot{\boldsymbol{\omega}} \times \mathbf{r}_p\,,

which, following the derivation of the moment of inertia tensor, is

\frac{d\mathbf{L}_{total}}{dt}= M \mathbf{R} \times \frac{d\mathbf{V}}{dt} + \mathbf{I} \dot{\boldsymbol{\omega}}\,.

Furthermore, since

\sum_p m_p \mathbf{r}_p\times \frac{d\mathbf{v}_p}{dt} = \sum_p \boldsymbol{\tau}_p = \sum_p \boldsymbol{\tau}_{p,ext} = \boldsymbol{\tau}_{ext}\,,

we obtain

\frac{d\mathbf{L}_{total}}{dt} = \mathbf{R}\times\mathbf{F}_{ext} + \boldsymbol{\tau}_{ext}\,.

Summary

M \frac{d\mathbf{V}}{dt}\, = \mathbf{F}_{ext}\,,
\mathbf{I} \dot{\boldsymbol{\omega}}\, = \boldsymbol{\tau}_{ext}\,,
\frac{d\mathbf{L}_{total}}{dt}\, = \mathbf{R}\times\mathbf{F}_{ext} + \boldsymbol{\tau}_{ext}\,.
back to moment of inertia tensor
on to Euler's equations
Retrieved from "http://www.physics.thetangentbundle.net/wiki/Classical_mechanics/rigid_body_dynamics/torque"
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