Question

The angular momentum of a particle m is defined by $\mathbf{L}\mathbf{=}\mathbf{mr}\mathbf{\times }\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$(see end of Section 3). Show that$\frac{\mathbf{dL}}{\mathbf{dt}}\mathbf{=}\mathbf{mr}\mathbf{\times }\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{r}}{{\mathbf{dt}}^{\mathbf{2}}}$

Short answer

The proof is given as follows.

$\frac{\mathrm{dL}}{\mathrm{dt}}=\mathrm{mr}\times \frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}$

Step-by-step solution

Given equation

The angular momentum of a particle m is defined by $\mathrm{L}=\mathrm{mr}\times \left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$.

Definition of Angular Momentum

The angular momentum L of m about point O is defined by the equation$\mathbf{L}\mathbf{=}\mathbf{r}\mathbf{\times }\left(\mathrm{mv}\right)\mathbf{=}\mathbf{mr}\mathbf{\times }\mathbf{v}$

Proof of the statement

To prove the given condition, Consider,

$\mathrm{L}=\mathrm{mr}\times \left(\frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}\right)$,then:

$\frac{\mathrm{dL}}{\mathrm{dt}}=\mathrm{m}\frac{\mathrm{dr}}{\mathrm{dt}}\times \frac{\mathrm{dr}}{\mathrm{dt}}+\mathrm{mr}\times \frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}$

But since the cross product of a vector with itself results in the zero vector, then:

$\frac{\mathrm{dL}}{\mathrm{dt}}=\mathrm{mr}\frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}$

Hence, the result is $\frac{\mathrm{dL}}{\mathrm{dt}}=\mathrm{m}\frac{\mathrm{dr}}{\mathrm{dt}}\times \frac{\mathrm{dr}}{\mathrm{dt}}+\mathrm{mr}\times \frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}$