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

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

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

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, $V_{CM}=b{\mathrm{Ó\neg }}_1$

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

Thus,

$\left|{\stackrel{→}{L}}_{trans-barbell}\right|=b\left(2m\right)\left(b{\mathrm{Ó\neg }}_1\right)$

$=\left(2m\right)b^2{\mathrm{Ó\neg }}_1$

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 ${\mathrm{Ó\neg }}_1$in $\left|{\stackrel{→}{L}}_{trans-barbell}\right|=\left(2m\right)b^2{\mathrm{Ó\neg }}_1$

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

$=9.72kg·m^2/s$

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 ${\stackrel{→}{L}}_{total}={\stackrel{→}{L}}_{rot}+{\stackrel{→}{L}}_{trans-barbell}$

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

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

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

$9.72kg·m^2/s$, into the page.