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 $b=0.9m$. The apparatus is set in motion in such a way that it again rotates clockwise with angular speed ${\omega }_1=15\text{rad/s,}$but in addition, the barbell rotates clockwise about its center, with an angular speed ${\omega }_2=20\text{rad/s,}$(figure). Calculate these vector quantities: (a) ${\overrightarrow{L}}_{rot},$(b) ${\overrightarrow{L}}_{trans,B,}$(c)${\overrightarrow{L}}_{tot,B.}$

Short answer

The rotational angular momentum of bar-bel is$1.44kg·m^2/s$ and its direction is into the page.

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

and its direction is into the page.

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

To solve for the rotating angular momentum of the bar-bell rod system, use the formula for rotational angular momentum.

$\left|{\overrightarrow{L}}_{rot}\right|=I{\omega }_2$

Here,$I$is the moment of inertia of bar-bell, androle="math" localid="1668599691435" ${\omega }_2$is the angular speed of the bar-bell.

The expression for the moment of inertia of the bar-bell is,

$I=m{\left(\frac{d}{2}\right)}^2+m{\left(\frac{d}{2}\right)}^2$

Here, $m$is the mass of each point mass, and $d$is the distance between the two point masses of bar-bell.

Substitute $m{\left(\frac{d}{2}\right)}^2+m{\left(\frac{d}{2}\right)}^2$for $I$in $\left|{\overrightarrow{L}}_{rot}\right|=I{\omega }_2$to write the final expression for the rotational angular momentum.

Thus, $\left|{\overrightarrow{L}}_{rot}\right|=2\left(m{\left(\frac{d}{2}\right)}^2\right){\omega }_2$

Substitute $0.4kg$for $m$, $0.6m$for $d$, and $20rad/s$for ${\omega }_2$in $\left|{\overrightarrow{L}}_{rot}\right|=2\left(m{\left(\frac{d}{2}\right)}^2\right){\omega }_2.$

$\left|{\overrightarrow{L}}_{rot}\right|=2\left(\left(0.4kg\right){\left(0.3m\right)}^2\right)\left(20rad/s\right)$

$=1.44kg·m^2/s$

Thus, the rotational angular momentum of bar-bel is$1.44kg·m^2/s$ and its direction is into the page.

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 rod to the center of mass of the bar-bell.

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

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

Here, $2m$is the total mass of the bar-bell, and ${\omega }_1$is the angular speed of rotation of the rod.

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

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

$=9.72kg·m^2/s$

Therefore, the magnitude of translational angular momentum of the bar-bell system is, $9.72kg·m^2/s$and its direction is into the page because the rod in rotating clockwise.

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

The expression for the total angular momentum of the rod-barbell system is

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

Substitute $1.44kg·m^2/s$into the page for ${\overrightarrow{L}}_{rot}$, and $9.72kg·m^2/s,$into the page for ${\overrightarrow{L}}_{trans-barbell}$in ${\overrightarrow{L}}_{total}={\overrightarrow{L}}_{rot}+{\overrightarrow{L}}_{trans-barbell}$

${\overrightarrow{L}}_{total}=1.44kg·m^2/s+9.72kg·m^2/s,$into the page $=11.16kg·m^2/s$into the page

Thus, the total angular momentum of the bar-bell rod system is, $9.72kg·m^2/s$, into the page

The vector-form of total angular momentum is