Torque/Proofs

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A proof that torque is equal to the time-derivative of angular momentum can be stated as follows:

The definition of angular momentum for a single particle is:

$\mathbf{L} = \mathbf{r} \times \mathbf{p}$

where "×" indicates the vector cross product. The time-derivative of this is:

$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}$

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv, we can see that:

$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}$

But the cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma, (Newton's 2nd law) we obtain:

$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}$

And by definition, torque τ = r×F. Note that there is a hidden assumption that mass is constant — this is quite valid in non-relativistic mechanics. Also, total (summed) forces and torques have been used — it perhaps would have been more rigorous to write: