91Ó°ÊÓ

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 & magnitude using $\left|\stackrel{→}{r}\right|\left|\stackrel{→}{p}\right|\sin \mathrm{θ}$From Point $D$to point $p$

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

Fig:1

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

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

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

Fig:2

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

Direction & magnitude from E to p.

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

Fig:3

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

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

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

Direction & magnitude from F to p.

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

Fig:4

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

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

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

Direction & magnitude from G to p.

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

Fig:5

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

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

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

Direction & magnitude from H to p.

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

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