Question

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" ${\overrightarrow{L}}_{tot,B.}$

Short answer

The translational angular momentum of the bar-bell system is-$9.72kg·m^2/s$.

The total angular momentum of the bar bell rod system. $9.72kg·m^2/s$into the page.

Step-by-step solution

Definition of Angular Momentum.

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.

Find the rotational angular momentum of bar bell.

(a) Solve for the rotational angular momentum of the bar-bell rod system using the formula for rotational angular momentum.

(b) The expression for the rotational angular momentum of bar-bell rod system is, ${\overrightarrow{L}}_{rot}=I{\omega }_{bar-bell}$

Here,is the moment of inertia of bar-bell

The rotating angular momentum of the bar-bell is 0 since its angular velocity is zero.

Find the magnitude of translational angular momentum of the bar-bell system.

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.

Find the total angular momentum of the bar bell rod system.

(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.