91Ó°ÊÓ

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,

${\stackrel{→}{L}}_r=\stackrel{→}{r}×\stackrel{→}{p}$

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

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

$\frac{d{\stackrel{→}{L}}_r}{dt}=\frac{d}{dt}\left(\stackrel{→}{r}×\stackrel{→}{p}\right)$

$=\frac{d\stackrel{→}{r}}{dt}×\stackrel{→}{p}+\stackrel{→}{r}×\frac{d\stackrel{→}{p}}{dt}$

$=\stackrel{→}{v}×m\stackrel{→}{v}+\stackrel{→}{r}×\frac{d\stackrel{→}{p}}{dt}$

$=0+\stackrel{→}{r}×\frac{d\stackrel{→}{p}}{dt}$

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

$\frac{d{\stackrel{→}{L}}_r}{dt}=\stackrel{→}{r}×\frac{d\stackrel{→}{p}}{dt}$

$=0,0,-mgx$

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

Thus, by angular momentum principle