91Ó°ÊÓ

The rotation between angular momentum and angular velocity of the bar can be

$\stackrel{→}{L}=I\stackrel{→}{\mathrm{Ó\neg }}$

Here, $l$is the moment of inertia of the bar and $\mathrm{Ó\neg }$is the angular velocity of the bar, and the bar is rotating counter clockwise the angular velocity is in the$\widehat{k}$direction.

The rotating bar is as shown in the below figure:

The mass of bar is $500g$convert the units of mass from $gtokg$

$M=500g$

$=\left(500\mathrm{g}\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)$

$=0.5\mathrm{kg}$

The angular velocity of the bar is $120rpm.$Convert the units of bar from $rpmandrads$

$\stackrel{→}{\mathrm{Ó\neg }}=\left(120\frac{\mathrm{rev}}{\min }\right)\left(\frac{2\mathrm{Ï€}\mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)$

$=12.6\widehat{k}\frac{\mathrm{rad}}{\mathrm{s}}$

The moment of inertia $l$can be found using the equation for a thin rod around the center:

$I=\frac{1}{12}M{m}^2$

Substitute $0.5kg$for $2.0m$for $l$in the above frmula t solve the mooemnt of inertia of the bar

$I=\frac{1}{12}\left(0.5\mathrm{kg}\right)(2.0\mathrm{m}{)}^2$

$=0.167\mathrm{kg}â‹\dots {\mathrm{m}}^2$

Substitute $0.167\mathrm{kg}â‹\dots {\mathrm{m}}^2\text{for}/\text{and}12.6\widehat{k}\frac{\mathrm{rad}}{\mathrm{s}}\text{for}\stackrel{→}{\mathrm{Ó\neg }}$in equation $\stackrel{→}{L}=I\stackrel{→}{\mathrm{Ó\neg }}$to solve the angular momentum of the rotating bar.

$\stackrel{→}{L}=I\stackrel{→}{\mathrm{Ó\neg }}$

$\begin{array}{r}=\left(0.167\mathrm{kg}â‹\dots {\mathrm{m}}^2\right)\left(12.6\widehat{k}\frac{\mathrm{rad}}{\mathrm{s}}\right)\\ \end{array}$

$=2.1\widehat{k}\mathrm{kg}â‹\dots \frac{{\mathrm{m}}^2}{\mathrm{s}}$

The angular momentum is out of the page (counterclockwise rotation )