Question

A barbell spins around a pivot at A its center at (Figure). The barbell consists of two small balls, each with mass $\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{\text{kg}}$at the ends of a very low mass road of length $\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{\text{m}}$. The barbell spins clockwise with angular speed${\mathit{\omega }}_{\mathbf{0}}\mathbf{=}\mathbf{20}$ radians/s. (a) Consider the two balls separately, and calculate${\overrightarrow{\mathbf{L}}}_{\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{s}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{A}}$ and ${\overrightarrow{\mathbf{L}}}_{\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{s}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{A}}$(both direction and magnitude in each case). (b) Calculate ${\overrightarrow{\mathbf{L}}}_{\mathbf{tot}\mathbf{,}\mathbf{A}}\mathbf{=}{\overrightarrow{\mathbf{L}}}_{\mathbf{trans}\mathbf{,}\mathbf{I}\mathbf{,}\mathbf{A}}\mathbf{+}{\overrightarrow{\mathbf{L}}}_{\mathbf{trans}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{A}}$(both direction and magnitude). (c) Next, consider the two balls together and calculate for the barbell. (d) What is the direction of the angular velocity ${\overrightarrow{\mathbf{\omega }}}_{\mathbf{0}}$? (e) Calculate ${\overrightarrow{\mathbf{L}}}_{\mathbf{r}\mathbf{o}\mathbf{t}}\mathbf{=}\mathit{I}{\overrightarrow{\mathbf{\omega }}}_{\mathbf{0}}$ (both direction and magnitude). (f) How does ${\overrightarrow{\mathbf{L}}}_{\mathbf{r}\mathbf{o}\mathbf{t}}$compare to ${\overrightarrow{\mathbf{L}}}_{\mathbf{r}\mathbf{o}\mathbf{t}\mathbf{,}\mathbf{A}}$? The point is the form $\mathit{I}\mathit{\omega }$ is just a convenient way of calculating the (rotational) angular momentum of multiparticle system. In principle one can always calculate the angular momentum simply by adding up the individual angular momentum of all the particles. (g) calculate ${\mathit{K}}_{\mathbf{r}\mathbf{o}\mathbf{t}}$.

Short answer

1. The magnitude of the translational angular momentum of the barbell at location 1 about the point A is$0.72kg.m^2/s$.
2. Total translational angular momentum of the barbell is$1.44kg.m^2/s$.
3. The moment of the inertia of the barbell rotating on its axis is role="math" localid="1668421674840" $I=0.072kg.m^2$.
4. The direction of the angular momentum of the barbell point into the page because the barbell rotates about the pivot in the clockwise direction. Hence, the direction of the angular speed $\overrightarrow{{\omega }_0}$of the barbell point into the page.
5. The direction of the angular momentum of the barbell point to the page because the barbell rotates about the pivot point in the clockwise direction.
6. The total angular momentum of the barbell is the same as the rotational angular momentum of the barbell.
7. The rotational kinetic energy is$14.4\text{J}$ .

Step-by-step solution

Definition of Angular Momentum:

In physics, angular momentum (rarely angular momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the total angular momentum of a closed system remains constant. Angular momentum has both direction and magnitude, and both are conserved.

Formula to find the translational angular momentum is as follows.

$L=m\overrightarrow{v}\overrightarrow{r}$

Here, $m$is the mass, $\overrightarrow{v}$ is the velocity and $\overrightarrow{r}$is the perpendicular distance.

A barbell spins around a pivot at its center at is A shown in the following figure.

(a) Find the magnitude of the translational angular momentum of the barbell at location 1 about the point  :

Let be the separation between the two balls. From the figure, the distance from each ball to its pivot point $\left(A\right)$is,

$\overrightarrow{r}=\frac{d}{2}$

The translational angular momentum of the barbell at location 1 about the point $A$is calculated as follows,

$L_{trans,1,A}=m\overrightarrow{v}\overrightarrow{r}$

$=m\left(\overrightarrow{r}{\overrightarrow{\omega }}_0\right)\overrightarrow{r}$

$=m\overrightarrow{\omega }{}_0{\overrightarrow{r}}^2$

Substitute $\frac{d}{2}$for $\overrightarrow{r}$in the above equation.

$L_{trans,1,A}=m\overrightarrow{{\omega }_0}{\overrightarrow{r}}^2$

$$\begin{gathered} L_{trans,1,A}=m\overrightarrow{{\omega }_0}{\left(\frac{d}{2}\right)}^2 \\ =\frac{1}{4}m\overrightarrow{{\omega }_0}{d}^2 \end{gathered}$$

Thus, the magnitude of the translational angular momentum of the barbell at location 1 about the point $A$is calculated as follows,

$$\begin{gathered} \left|L_{trans,1,A}\right|=\frac{1}{4}m\overrightarrow{{\omega }_0}{d}^2 \\ =\frac{1}{4}\left(0.4kg\right)\left(20radians/s\right){\left(0.6m\right)}^2 \\ =0.72kg.m^2/s \end{gathered}$$

Similarly, the magnitude of the translational angular momentum of the barbell at location 2 about the point A is,

$$\begin{gathered} \left|L_{trans,2,A}\right|=\frac{1}{4}m\overrightarrow{{\omega }_0}{d}^2 \\ =\frac{1}{4}\left(0.4kg\right)\left(20radians/s\right){\left(0.6m\right)}^2 \\ =0.72kg.m^2/s \end{gathered}$$

Given that, the barbell spins clockwise direction with an angular speed ${\omega }_0$.

According to the right-hand rule, if the rotational motion is clockwise, the unit vector (and your right thumb) point into the plane. The direction of the thumb represents the direction of the angular momentum of the barbell.

The direction of the angular momentum of the barbell point into the page because the barbell rotates about the pivot $\left(A\right)$is in the clockwise direction.

Hence, at both locations the angular momentum of the barbell is directed into the page.

(b)Find The direction of the total angular momentum of the barbell:

Total translational angular momentum of the barbell is calculated as follows,

${\overrightarrow{L}}_{tot}={\overrightarrow{L}}_{trans,1,A}+{\overrightarrow{L}}_{trans,2,A}$

$$\begin{gathered} {\overrightarrow{L}}_{tot}=0.72kg.m^2/s+0.72kg.m^2/s \\ =1.44kg.m^2/s \end{gathered}$$

Hence, the direction of the total angular momentum of the barbell point into the page because the barbell rotates about the pivot in the clockwise direction.

(c) Find the moment of inertia of the barbell system:

The moment of the inertia of the barbell rotating on its axis is,

$$\begin{gathered} I=m{r}^2+m{r}^2 \\ =m{\left(\frac{d}{2}\right)}^2+m{\left(\frac{d}{2}\right)}^2 \\ =\frac{1}{4}m{d}^2+\frac{1}{4}m{d}^2 \end{gathered}$$

Therefore,

$I=\frac{1}{2}m{d}^2$ ...........(1)

Use the equation (1) to the find moment of inertia of the barbell rotating on its axis,

$$\begin{gathered} I=\frac{1}{2}m{d}^2 \\ =\frac{1}{2}\left(0.4\text{kg}\right)\left(0.6\text{m}\right){}^2 \\ =0.072\text{kg}·{\text{m}}^2 \end{gathered}$$

Hence, the moment of the inertia of the barbell rotating on its axis is $0.072kg.m^2$.

(d) Find the rotational angular momentum of the barbell:

The rotational angular momentum of the barbell is,

${\overrightarrow{L}}_{rot}=I\overrightarrow{{\omega }_0}$

The direction of the angular momentum of the barbell point into the page because the barbell rotates about the pivot in the clockwise direction. Hence, the direction of the angular speed $\overrightarrow{{\omega }_0}$of the barbell point into the page.

(e)The rotational angular momentum of the barbell:

The rotational angular momentum of the barbell is,

${\overrightarrow{L}}_{rot}=I\overrightarrow{{\omega }_0}$

Thus, the magnitude of the rotational angular momentum of the barbell is calculated as follows,

role="math" $$\begin{gathered} \left|{\overrightarrow{L}}_{rot}\right|=I{\omega }_0 \\ =(0.072kg.m^2)\left(20rad/s\right) \\ =1.44kg.m^2/s \end{gathered}$$

Hence the direction of the angular momentum of the barbell point into the page because the barbell rotates about the pivot point in the clockwise direction.

(f) The rotational angular momentum of the barbell: 

From part (b) and Part (e), you can conclude that the total angular momentum of the barbell is same as the rotational angular momentum of the barbell.

(g) Find the rotational kinetic energy of the barbell system:

The rotational kinetic energy of the barbell is calculated as follows,

$$\begin{gathered} K{}_{rot}=\frac{1}{2}I{\omega }_0^2 \\ =\frac{1}{2}\left(0.072kg.m^2\right){\left(20rad/s\right)}^2 \\ =14.4J \end{gathered}$$

Hence, the rotational kinetic energy is $14.4\text{J}$.