A barbell consist of two small balls, each with mass $m=0.4kg,$at the ends of a very low mass rod of length $d=0.6m$. It is mounted on the end of a low-mass rigid rod of length$b=0.9m$(figure). The apparatus is set in motion in such a way that although the rod rotates clockwise with angular speed localid="1668604224599" ${\omega }_1=15rad/s$, the barbell maintains its vertical orientation. Calculate these vector quantities: (a) localid="1668604234930" ${\overrightarrow{L}}_{rot},$ (b) localid="1668604258387" ${\overrightarrow{L}}_{trans,B,}$(c) localid="1668604276383" ${\overrightarrow{L}}_{tot,B.}$

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 translational angular momentum of bar-bell -rod system is $\left|{\overrightarrow{L}}_{trans-barbell}\right|=\left|{\overrightarrow{r}}_{bar-bell}\times {\overrightarrow{P}}_{total}\right|$

Here, ${\overrightarrow{r}}_{bar-bell}$is the position vector of the center of mass of the bar-bell, and ${\overrightarrow{P}}_{total}$is the linear momentum vector of the bar-bell.

The expression for the magnitude of linear momentum is, $\left|{\overrightarrow{P}}_{total}\right|=M_{tot}{V}_{CM}$

Here, $M_{tot}$is the total mass of the bar-bell, and $V_{CM}$is the velocity of center of mass of the bar-bell.

The expression for the velocity of center of mass of the bar-bell is, $V_{CM}=b{\omega }_1$

Here,$b$is the distance from the edge of the center of mass of the bar-bell.

Thus,

$\left|{\overrightarrow{L}}_{trans-barbell}\right|=b\left(2m\right)\left(b{\omega }_1\right)$

$=\left(2m\right)b^2{\omega }_1$

Here, $2m$is the total mass of the bar-bell, and is the angular speed of rotational of the rod.

Substitute $0.4kg$for $m,0.9m$for $b,$and $15rad/s$for ${\omega }_1$in $\left|{\overrightarrow{L}}_{trans-barbell}\right|=\left(2m\right)b^2{\omega }_1$

$\left|{\overrightarrow{L}}_{trans-barbell}\right|=\left(2\left(0.4kg\right)\right){\left(0.9m\right)}^2\left(15rad/s\right)$

$=9.72kg·m^2/s$

As a result, the size of the bar-bell system's translational angular momentum is, $9.72kg·m^2/s$and its direction is into the page because the rod is rotating clockwise.

(c) The expression for the total angular momentum of the rod-barbell system is ${\overrightarrow{L}}_{total}={\overrightarrow{L}}_{rot}+{\overrightarrow{L}}_{trans-barbell}$

Substitute $0$for ${\overrightarrow{L}}_{rot}$, and $9.72kg·m^2/s$into the page for ${\overrightarrow{L}}_{trans-barbell}$in ${\overrightarrow{L}}_{total}={\overrightarrow{L}}_{rot}+{\overrightarrow{L}}_{trans-barbell}$

${\overrightarrow{L}}_{total}=9.72kg·m^2/s$into the page

Thus, the total angular momentum of the bar-bell rod system is,

$9.72kg·m^2/s$, into the page.