Question

Determine both the direction and magnitude of the angular momentum of the particle in Figure 11.13, relative to the locations D, E,F, G, and H. We've already analyzed the angular momentum relative toA, B, and C in the example given above. Notice how the magnitude and direction of the angular momentum relative to the different locations differ in magnitude and direction.

Short answer

Direction and magnitude of the angular momentum of particle p relative to location D is in negative z- direction and magnitude$50kgm/s^2$

Direction and magnitude of the angular momentum of particle p relative to location E is in negative z- direction and magnitude$50kgm/s^2$

Direction and magnitude of the angular momentum of particle p relative to location F is in negative z- direction and magnitude$50kgm/s^2$

Direction and magnitude of the angular momentum of particle p relative to location G is in negative z- direction and magnitude$40kgm/s^2$

Direction and magnitude of the angular momentum of particle p relative to location H is undefined.

Step-by-step solution

Definition of Angular momentum

The measure of rotational motion is angular momentum.

Translational (or "orbital") angular momentum is used to describe motions like the Earth's orbit around the Sun. The movement of the Earth around its own axis is an example of rotational (or "spin") angular momentum.

The Angular Momentum Principle ties a change in angular momentum to the net torque, or twist, given to a system, while the Momentum Principle relates a change in momentum to the net force on a system.

Direction and magnitude of the angular momentum of the particle relative to location D.

Direction & magnitude using $\left|\overrightarrow{r}\right|\left|\overrightarrow{p}\right|\sin \theta$From Point $D$to point $p$

$\begin{array}{rcl}\overrightarrow{r_B}& =& (4,5,0)m\\ \overrightarrow{p}& =& (10,0,0)kg-m/s\\ & & \end{array}$

Fig:1

In fig.1, place $\left|\overrightarrow{r_D}\right|$& $\overrightarrow{p}$tail to tail. Vector $\overrightarrow{r_D}$& $\overrightarrow{p}$define $xy$plane. Now use right hand rule, the thumb point into the page $(\otimes ),$so the angular momentum is innegative zdirection.

$\begin{array}{rcl}\left|\overrightarrow{L_D}\right|& =& \left|\overrightarrow{r_D}\right|\left|\overrightarrow{p}\right|\sin \theta \\ \left|\overrightarrow{r_D}\right|& =& \sqrt{4^2+5^2}=6.40m\\ \left|\overrightarrow{p}\right|& =& 10kgm/s\\ & & \end{array}$

From fig.2, $\sin \theta =\frac{5}{6.40}$

Fig:2

$\begin{array}{rcl}\left|\overrightarrow{L_D}\right|& =& (6.40)\left(10\right)\left(\frac{5}{6.40}\right)=50kgm/s^2\\ \overrightarrow{L_D}& =& (0,0,-50)kgm/s^2\\ & & \end{array}$

Step 3: Direction and magnitude of the angular momentum of the particle relative to location E.

Direction & magnitude from E to p.

$\begin{array}{rcl}\overrightarrow{r_E}& =& (4,5,0)\\ \overrightarrow{p}& =& (10,0,0)kgm/s\\ & & \end{array}$

Fig:3

In fig.3, place $\overrightarrow{r_E}and\overrightarrow{p}$tail to tail. Vector $\overrightarrow{r_E}and\overrightarrow{p}$define $xy$plane. Using right hand rule, thumb point into page $(\otimes )$so angular momentum is in negative z direction.

$\begin{array}{rcl}\left|\overrightarrow{L_E}\right|& =& \left|\overrightarrow{r_E}\right|\left|\overrightarrow{p}\right|\sin \theta \\ \left|\overrightarrow{r_E}\right|& =& \sqrt{4^2+5^2}=6.40m\\ \left|\overrightarrow{p}\right|& =& 10kgm/s\\ & & \end{array}$

$\begin{array}{rcl}\sin \theta & =& \frac{5}{6.40}\\ \left|\overrightarrow{L_E}\right|& =& (6.40)\left(10\right)\left(\frac{5}{6.40}\right)=50kgm/s^2\\ \left|\overrightarrow{L_E}\right|& =& (0,0,-50)kgm/s^2\\ & & \end{array}$

Step 4: Direction and magnitude of the angular momentum of the particle relative to location F.

Direction & magnitude from F to p.

$\begin{array}{rcl}\left|\overrightarrow{r_F}\right|& =& (-4,5,0)m\\ \overrightarrow{p}& =& (10,0,0)kgm/s\\ & & \end{array}$

Fig:4

In fig.4 $\overrightarrow{r_F}and\overrightarrow{p}$,are placed tail to tail. $\overrightarrow{r_F}and\overrightarrow{p}$defined in $xy$plane. Using right hand rule, thumb point into the pages $(\otimes ),$o the angular momentum is innegative zdirection.

$\begin{array}{rcl}\left|\overrightarrow{L_F}\right|& =& \left|\overrightarrow{r_F}\right|\left|\overrightarrow{p}\right|\sin \theta \\ \left|\overrightarrow{L_F}\right|& =& \sqrt{4^2+6^2}\\ \left|\overrightarrow{p}\right|& =& 10kgm/s\\ & & \end{array}$

$\begin{array}{rcl}\sin \theta & =& \frac{5}{6.40}\\ \left|\overrightarrow{L_F}\right|& =& (6.40)\left(10\right)\left(\frac{5}{6.40}\right)=50kgm/s^2\\ \left|\overrightarrow{L_F}\right|& =& (0,0,-50)kgm/s^2\\ & & \end{array}$

Step 5: Direction and magnitude of the angular momentum of the particle relative to location G.

Direction & magnitude from G to p.

$$\begin{gathered} \overrightarrow{r_G}=(-4,-3,0)m \\ \overrightarrow{p}=(10,0,0)kgm/s \end{gathered}$$

Fig:5

In fig.5,$\overrightarrow{r_G}\&\overrightarrow{p}$are placed tail to tail. $\overrightarrow{r_G}\&\overrightarrow{p}$defines in $xy$plane. Using right hand rule, thumb point into the page $(\otimes ).$So, the angular momentum is innegative zdirection.

$\begin{array}{rcl}\left|\overrightarrow{L_G}\right|& =& \left|\overrightarrow{r_H}\right|\left|\overrightarrow{p}\right|\sin \theta \\ \left|\overrightarrow{r_G}\right|& =& \sqrt{3^2+4^2}=5\\ \left|\overrightarrow{p}\right|& =& 10kgm/s\\ & & \end{array}$

$\begin{array}{rcl}\sin \theta & =& \frac{4}{5}\\ \left|\overrightarrow{L_G}\right|& =& \left(5\right)\left(10\right)\left(\frac{4}{5}\right)=40kgm/s^2\\ \overrightarrow{L_G}& =& (0,0,-40)kgm/s^2\\ & & \end{array}$

Step 6: Direction and magnitude of the angular momentum of the particle relative to location H.

Direction & magnitude from H to p.

About location$H,r_{\perp }$ is zero because the momentum points straight toward H (the impact parameter is zero).

Therefore $\left|\overrightarrow{L_H}\right|=0$& the direction of $\overrightarrow{L_H}$is undefined.