Let’s compare the Momentum principle and the Angular momentum principle in a simple situation. Consider a $m$mass falling near the Earth (figure). Neglecting air resistance, the momentum principle gives $dp{}_y/dt=-mg$, yielding $dp{}_y/dt=-g$(nonrelativistic). Choose a location $A$off to the side, on the ground. Apply the angular momentum principle to find an algebraic expression for the rate of change of angular momentum of the mass about location $A$.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

The expression for the angular momentum of an object about a point is,

${\overrightarrow{L}}_r=\overrightarrow{r}\times \overrightarrow{p}$

Here,$\overrightarrow{r}$is the position vector of the object, and$\overrightarrow{p}$is the linear momentum of the object.

Compute the rate of change of angular momentum of the object.

$\frac{d{\overrightarrow{L}}_r}{dt}=\frac{d}{dt}\left(\overrightarrow{r}\times \overrightarrow{p}\right)$

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

$=\overrightarrow{v}\times m\overrightarrow{v}+\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}$

$=0+\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}$

Thus, the rate of change of angular momentum of the object is,

$\frac{d{\overrightarrow{L}}_r}{dt}=\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}$

$=0,0,-mgx$

Therefore, the rate of change of angular momentum is $-mgx$.

Thus, by angular momentum principle