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# 3.12: Torque, Angular Momentum and a Moving Point

[ "article:topic", "authorname:tatumj", "showtoc:no" ]

In Figure III.7 I draw the particle $$m_{i}$$, which is just one of $$n$$ particles, $$n-1$$ of which I haven’t drawn and are scattered around in 3-space. I draw an arbitrary origin O, the centre of mass C of the system, and another point Q, which may (or may not) be moving with respect to O. The question I am going to ask is: Does the equation $$\dot{\bf L } = \boldsymbol\tau$$ apply to the point Q? It obviously does if Q is stationary, just as it applies to O. But what if Q is moving? If it does not apply, just what is the appropriate relation?

The theorem that we shall prove – and interpret - is

$\dot{\bf L }_{Q} = {\boldsymbol\tau}_{Q} + M{\bf r }_{Q}^{\prime} \times\ddot{\bf r}_{Q}. \label{eq:3.12.1}$

We start:

${\bf L }_{Q} =\sum ({\bf r }_{i} -{\bf r }_{Q})\times [m_{i}({\bf v }_{i}-{\bf v }_{Q})] \label{eq:3.12.2}$

$\therefore \quad \dot{\bf L }_{Q} =\sum ({\bf r }_{i} -{\bf r }_{Q})\times m_{i}(\dot{\bf v }_{i} - \dot{\bf v }_Q)+\sum (\dot{\bf r }_{i} -\dot{\bf r }_{Q})\times m_i({\bf v }_{i} - {\bf v }_Q). \label{eq:3.12.3}$

The second term is zero, because $$\dot{\bf r}={\bf v}$$

Continue:

$\dot{\bf L } =\sum ({\bf r }_{i} -{\bf r }_{Q})\times m_{i}\dot{\bf v }_{i} - \sum m_{i}{\bf r }_{i}\times \dot{\bf v}_{Q} + \sum m_{i}{\bf r }_{Q}\times \dot{\bf v }_{Q} \label{eq:3.12.4}$

Now $$m_{i} \dot{\bf v }_{i} = {\bf F }_{i}$$ , so that the first term is just $${\boldsymbol\tau}_{Q}$$

Continue:

$$\dot{\bf L } = {\boldsymbol\tau}_{Q} - \sum m_{i} {\bf r }_{i} \times dot{\bf v}_{Q} + \sum M_{i} {\bf r }_{Q} \times dot{\bf v }_{Q}$$

$$= {\boldsymbol\tau}_{Q} -M\overline{\bf r }\times \ddot{\bf r }_{Q} +M{\bf r }_{Q}\times \ddot{\bf r }_{Q}$$

$$= {\boldsymbol\tau}_{Q} + M({\bf r }_{Q} - \overline{\bf {\bf r } })\times \ddot{\bf {\bf r } }_{Q}$$

$\therefore \dot{\bf L}_{Q} = {\boldsymbol\tau}_{Q} + M{\bf r }^{\prime}_{Q}\times \ddot{r}_{Q} \qquad Q.E. D \label{eq:3.12.5}$

Thus in general, $$\dot{\bf L}_{Q} \neq {\boldsymbol\tau}_{Q}$$ , but $$\dot{\bf L}_{Q} = {\boldsymbol\tau}_{Q}$$ under any of the following three circumstances:

1. $${\bf r }^{\prime}_{Q} = 0$$ - that is, Q coincides with C.
2. $$\ddot{\bf r }_{Q} = 0$$ - that is, Q is not accelerating.
3. $$\ddot{\bf r }_{Q}$$ and $${\bf r }^{\prime}_{Q}$$ are parallel, which would happen, for example, if O were a centre of attraction or repulsion and Q were accelerating towards or away from O.

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