Proof of angular momentum

From Free net encyclopedia

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:

<math>\mathbf{L} = \mathbf{r} \times \mathbf{p}</math>

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

<math>\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}</math>

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:

<math>\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + m\mathbf{v} \times \mathbf{v}</math>

But the cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma, we obtain:

<math>\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}</math>

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: