# 11.2 Angular momentum  (Page 3/8)

 Page 3 / 8

## Significance

Since the meteor is accelerating downward toward Earth, its radius and velocity vector are changing. Therefore, since $\stackrel{\to }{l}=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p}$ , the angular momentum is changing as a function of time. The torque on the meteor about the origin, however, is constant, because the lever arm ${\stackrel{\to }{r}}_{\perp }$ and the force on the meteor are constants. This example is important in that it illustrates that the angular momentum depends on the choice of origin about which it is calculated. The methods used in this example are also important in developing angular momentum for a system of particles and for a rigid body.

Check Your Understanding A proton spiraling around a magnetic field executes circular motion in the plane of the paper, as shown below. The circular path has a radius of 0.4 m and the proton has velocity $4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}$ . What is the angular momentum of the proton about the origin? From the figure, we see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have
$\stackrel{\to }{l}=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p}=\left(0.4\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{i}\text{)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(1.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\phantom{\rule{0.2em}{0ex}}\text{kg}\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\right)\stackrel{^}{j}\right)=2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-21}\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\stackrel{^}{k}$

## Angular momentum of a system of particles

The angular momentum of a system of particles is important in many scientific disciplines, one being astronomy. Consider a spiral galaxy, a rotating island of stars like our own Milky Way. The individual stars can be treated as point particles, each of which has its own angular momentum. The vector sum of the individual angular momenta give the total angular momentum of the galaxy. In this section, we develop the tools with which we can calculate the total angular momentum of a system of particles.

In the preceding section, we introduced the angular momentum of a single particle about a designated origin. The expression for this angular momentum is $\stackrel{\to }{l}=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p},$ where the vector $\stackrel{\to }{r}$ is from the origin to the particle, and $\stackrel{\to }{p}$ is the particle’s linear momentum. If we have a system of N particles, each with position vector from the origin given by ${\stackrel{\to }{r}}_{i}$ and each having momentum ${\stackrel{\to }{p}}_{i},$ then the total angular momentum of the system of particles about the origin is the vector sum of the individual angular momenta about the origin. That is,

$\stackrel{\to }{L}={\stackrel{\to }{l}}_{1}+{\stackrel{\to }{l}}_{2}+\cdots +{\stackrel{\to }{l}}_{N}.$

Similarly, if particle i is subject to a net torque ${\stackrel{\to }{\tau }}_{i}$ about the origin, then we can find the net torque about the origin due to the system of particles by differentiating [link] :

$\frac{d\stackrel{\to }{L}}{dt}=\sum _{i}\frac{d{\stackrel{\to }{l}}_{i}}{dt}=\sum _{i}{\stackrel{\to }{\tau }}_{i}.$

The sum of the individual torques produces a net external torque on the system, which we designate $\sum \stackrel{\to }{\tau }.$ Thus,

$\frac{d\stackrel{\to }{L}}{dt}=\sum \stackrel{\to }{\tau }.$

[link] states that the rate of change of the total angular momentum of a system is equal to the net external torque acting on the system when both quantities are measured with respect to a given origin. [link] can be applied to any system that has net angular momentum, including rigid bodies, as discussed in the next section.

## Angular momentum of three particles

Referring to [link] (a), determine the total angular momentum due to the three particles about the origin. (b) What is the rate of change of the angular momentum? Three particles in the xy- plane with different position and momentum vectors.

## Strategy

Write down the position and momentum vectors for the three particles. Calculate the individual angular momenta and add them as vectors to find the total angular momentum. Then do the same for the torques.

## Solution

1. Particle 1: ${\stackrel{\to }{r}}_{1}=-2.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{i}+1.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{j},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{p}}_{1}=2.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(4.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\stackrel{^}{j}\right)=8.0\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m}\text{/}\text{s}\stackrel{^}{j},$
${\stackrel{\to }{l}}_{1}={\stackrel{\to }{r}}_{1}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{p}}_{1}=-16.0\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\stackrel{^}{k}.$

Particle 2: ${\stackrel{\to }{r}}_{2}=4.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{i}+1.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{j},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{p}}_{2}=4.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(5.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\stackrel{^}{i}\right)=20.0\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m}\text{/}\text{s}\stackrel{^}{i}$ ,
${\stackrel{\to }{l}}_{2}={\stackrel{\to }{r}}_{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{p}}_{2}=-20.0\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\stackrel{^}{k}.$

Particle 3: ${\stackrel{\to }{r}}_{3}=2.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{i}-2.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{j},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{p}}_{3}=1.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(3.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\stackrel{^}{i}\right)=3.0\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m}\text{/}\text{s}\stackrel{^}{i}$ ,
${\stackrel{\to }{l}}_{3}={\stackrel{\to }{r}}_{3}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{p}}_{3}=6.0\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\stackrel{^}{k}.$

We add the individual angular momenta to find the total about the origin:
${\stackrel{\to }{l}}_{T}={\stackrel{\to }{l}}_{1}+{\stackrel{\to }{l}}_{2}+{\stackrel{\to }{l}}_{3}=-30\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\stackrel{^}{k}.$
2. The individual forces and lever arms are
$\begin{array}{c}{\stackrel{\to }{r}}_{1\perp }=1.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{j},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{F}}_{1}=-6.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{i},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{\tau }}_{1}=6.0\text{N}·\text{m}\stackrel{^}{k}\hfill \\ {\stackrel{\to }{r}}_{2\perp }=4.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{i},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{F}}_{2}=10.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{\tau }}_{2}=40.0\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\stackrel{^}{k}\hfill \\ {\stackrel{\to }{r}}_{3\perp }=2.0\phantom{\rule{0.2em}{0ex}}\text{m}\stackrel{^}{i},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{F}}_{3}=-8.0\phantom{\rule{0.2em}{0ex}}\text{N}\stackrel{^}{j},\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{\tau }}_{3}=-16.0\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\stackrel{^}{k}.\hfill \end{array}$

Therefore:
$\sum _{i}{\stackrel{\to }{\tau }}_{i}={\stackrel{\to }{\tau }}_{1}+{\stackrel{\to }{\tau }}_{2}+{\stackrel{\to }{\tau }}_{3}=30\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\stackrel{^}{k}.$

## Significance

This example illustrates the superposition principle for angular momentum and torque of a system of particles. Care must be taken when evaluating the radius vectors ${\stackrel{\to }{r}}_{i}$ of the particles to calculate the angular momenta, and the lever arms, ${\stackrel{\to }{r}}_{i\perp }$ to calculate the torques, as they are completely different quantities.

what is units?
units as in how
praise
What is th formular for force
F = m x a
Santos
State newton's second law of motion
can u tell me I cant remember
Indigo
force is equal to mass times acceleration
Santos
The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board?
Consider a wave produced on a stretched spring by holding one end and shaking it up and down. Does the wavelength depend on the distance you move your hand up and down?
how can one calculate the value of a given quantity
means?
Manorama
To determine the exact value of a percent of a given quantity we need to express the given percent as fraction and multiply it by the given number.
AMIT
meaning
Winford
briefly discuss rocket in physics
ok let's discuss
Jay
What is physics
physics is the study of natural phenomena with concern with matter and energy and relationships between them
Ibrahim
a potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inductance of the inductor that should be connected to the capacitor for the circuit to oscillate at 1125Hza potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inducta
L= 0.002H
NNAEMEKA
how did you get it?
Favour
is the magnetic field of earth changing
what is thought to be the energy density of multiverse and is the space between universes really space
tibebeab
can you explain it
Guhan
Energy can not either created nor destroyed .therefore who created? and how did it come to existence?
this greatly depend on the kind of energy. for gravitational energy, it is result of the shattering effect violent collision of two black holes on the space-time which caused space time to be disturbed. this is according to recent study on gravitons and gravitational ripple. and many other studies
tibebeab
and not every thing have to pop into existence. and it could have always been there . and some scientists think that energy might have been the only entity in the euclidean(imaginary time T=it) which is time undergone wick rotation.
tibebeab
What is projectile?
An object that is launched from a device
Grant
2 dimensional motion under constant acceleration due to gravity
Awais
Not always 2D Awais
Grant
Awais
why not? a bullet is a projectile, so is a rock I throw
Grant
bullet travel in x and y comment same as rock which is 2 dimensional
Awais
components
Awais
no all pf you are wrong. projectile is any object propelled through space by excretion of a force which cease after launch
tibebeab
for awais, there is no such thing as constant acceleration due to gravity, because gravity change from place to place and from different height
tibebeab
it is the object not the motion or its components
tibebeab
where are body center of mass on present.
on the mid point
Suzana
is the magnetic field of the earth changing?
tibebeab
does shock waves come to effect when in earth's inner atmosphere or can it have an effect on the thermosphere or ionosphere?
tibebeab
and for the question from bal want do you mean human body or just any object in space
tibebeab
A stone is dropped into a well of 19.6m deep and the impact of sound heared after 2.056 second ,find the velocity of sound in air.
9.53 m/s ?
Kyla
In this case, the velocity of sound is 350 m/s.
Zahangir
why?
Kyla
some calculations is need. then you will get exact result.
Zahangir
i mean how? isn't it just a d over t?
Kyla
calculate the time it takes the stone to hit the ground then minus the stone's time to the total time... then divide the total distance by the difference of the time
Snuggly
awit lenard. Hahahah ari ga to!
Kyla
Why do we use mochromatic light? If white light is used, how would the pattern change By Brooke Delaney By OpenStax By Yasser Ibrahim By OpenStax By OpenStax By Megan Earhart By Jazzycazz Jackson By OpenStax By OpenStax By