Question

Write the relation between angular momentum and torque.

Answer 1

Relation between angular momentum and torque:

The angular momentum of a particle is given by

$\stackrel{\rightharpoonup }{\mathrm{L}}=\stackrel{\rightharpoonup }{\mathrm{r}}\times \stackrel{\rightharpoonup }{\mathrm{p}}$

Differentiating both sides of the above equation with respect to t,

$\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{L}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\stackrel{\rightharpoonup }{\mathrm{r}}\mathrm{X}\stackrel{\rightharpoonup }{\mathrm{p}}\right)$

$=\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{r}}}{\mathrm{dt}}\times \stackrel{\rightharpoonup }{\mathrm{p}}+\stackrel{\rightharpoonup }{\mathrm{r}}\times \frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{p}}}{\mathrm{dt}}$

Now $\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{r}}}{\mathrm{dt}}=\stackrel{\rightharpoonup }{\mathrm{v}}$ and, $\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{p}}}{\mathrm{dt}}=\stackrel{\rightharpoonup }{\mathrm{F}}$

Therefore equation (A) becomes

$\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{L}}}{\mathrm{dt}}=\stackrel{\rightharpoonup }{\mathrm{v}}\times \stackrel{\rightharpoonup }{\mathrm{p}}+\stackrel{\rightharpoonup }{\mathrm{r}}\times \stackrel{\rightharpoonup }{\mathrm{F}}$…(equation B)

since $\stackrel{\rightharpoonup }{\mathrm{v}}$and $\stackrel{\rightharpoonup }{\mathrm{p}}$are parallel vectors,$\stackrel{\rightharpoonup }{\mathrm{v}}\times \stackrel{\rightharpoonup }{\mathrm{p}}=0$

Also $\stackrel{\rightharpoonup }{\mathrm{r}}\times \stackrel{\rightharpoonup }{\mathrm{F}}=\stackrel{\rightharpoonup }{\mathrm{\tau }}$, the torque on the particle.

Therefore equation (B) gives

$\stackrel{\rightharpoonup }{\mathrm{\tau }}=\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{L}}}{\mathrm{dt}}$

Therefore, torque on a particle is equal to the time rate of change of its angular momentum.