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Momento angular

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Momento angular
Momento
angular
Angular momentum of a particle
The instantaneous angular momen
defined by the cross product of the p
instantaneous linear momentum p:
L
This allows us to write Equation 11.
!
z
L = r × p
O
r
m
y
p
φ
Javier Junquera
x
Active Figure 11.4 The angular
momentum L of a particle of mass
m and linear momentum p located
at the vector position r is a vector
given by L ! r ! p. The value of L
depends on the origin about which
it is measured and is a vector perpendicular to both r and p.
At the Active Figures link
at http://www.pse6.com, you
can change the position vector
r and the momentum vector p
which is the rotational analog of Newt
causes the angular momentum L to ch
change. Equation 11.11 states that the
time rate of change of the particle’s
Note that Equation 11.11 is valid on
gin. (Of course, the same origin must
thermore, the expression is valid for
The SI unit of angular momentum
and the direction of L depend on the
we see that the direction of L is perp
Figure 11.4, r and p are in the xy pla
p ! mv, the magnitude of L is
L
where # is the angle between r and p
p (# ! 0 or 180°). In other words, wh
line that passes through the origin, t
respect to the origin. On the other h
Bibliografía
FUENTE PRINCIPAL
Física, Volumen 1, 3° edición
Raymod A. Serway y John W. Jewett, Jr.
Ed. Thomson
ISBN: 84-9732-168-5
Capítulo 10
Física para Ciencias e Ingeniería, Volumen 1, 7° edición
Raymod A. Serway y John W. Jewett, Jr.
Cengage Learning
ISBN 978-970-686-822-0
Capítulo 11
Tips on Physics
R. P. Feynman, R. B. Leighton, y M. Sands
Ed. Pearson Addison Wesley
ISBN: 0-8053-9063-4
Capítulo 3-3 y siguientes
Definición de momento angular o cinético
The instantaneous angular momen
defined by the cross product of the p
instantaneous linear momentum p:
L
Angular momentum of a particle
Consideremos una partícula de masa m, con un vector de posición
This allows us to write Equation 11.
y que se mueve con una cantidad de movimiento
!
z
which is the rotational analog of Newt
causes the angular momentum L to ch
change. Equation 11.11 states that the
time rate of change of the particle’s
O
y
Note that Equation 11.11 is valid on
m p
r
gin. (Of course, the same origin must
φ
thermore, the expression is valid for
x
The SI unit of angular momentum
Active Figure 11.4 The angular
and the direction of L depend on the
momentum L of a particle of mass
we see that the direction of L is perp
m and linear momentum p located
at the vector position r is a vector
Figure 11.4, r and p are in the xy pla
El momento angular instantáneogiven by
deLla
relativo
semagnitude
define como
! partícula
r ! p. The value
of L al origen
p ! mv, O
the
of L isel
L = r × p
on the
origin about which
producto vectorial de su vectordepends
posición
instantáneo
y del momento lineal instantáneo
it is measured and is a vector perpendicular to both r and p.
At the Active Figures link
at http://www.pse6.com, you
can change the position vector
r and the momentum vector p
L
where # is the angle between r and p
p (# ! 0 or 180°). In other words, wh
line that passes through the origin, t
respect to the origin. On the other h
Definición de momento angular o cinético
The instantaneous angular momen
defined by the cross product of the p
instantaneous linear momentum p:
L
Angular momentum of a particle
Consideremos una partícula de masa m, con un vector de posición
This allows us to write Equation 11.
y que se mueve con una cantidad de movimiento
!
z
L = r × p
Tanto el módulo, la dirección como
el sentido del momento angular
dependen del origen que se elija
O
r
m
y
p
φ
x
Active Figure 11.4 The angular
momentum L of a particle of mass
m and linear momentum p located
at the vector position r is a vector
given by Lal!plano
r ! p. The
value of L
Dirección: perpendicular
formado
por
depends on the origin about which
Sentido: regla de lait ismano
derecha
measured
and is a vector perpendicular to both r and p.
Módulo:
At the Active Figures link
2/s
SI: kg • myou
atUnidades
http://www.pse6.com,
can change the position vector
r and the momentum vector p
which is the rotational analog of Newt
causes the angular momentum L to ch
change. Equation 11.11 states that the
time rate of change of the particle’s
Note that Equation 11.11 is valid on
gin. (Of course, the same origin must
thermore, the expression is valid for
The SI unit of angular momentum
and the direction of L depend on the
we see that the direction of L is perp
Figure 11.4, r and p are in the xy pla
py
! mv, the magnitude of L is
L
where # is the angle between r and p
p (# ! 0 or 180°). In other words, wh
line that passes through the origin, t
respect to the origin. On the other h
Momento angular o cinético:
Casos particulares
cuando
es paralelo a
. Es decir, cuando la partícula se mueve a lo largo de una línea
recta que pasa por el origen tiene un momento angular nulo con respecto a ese origen
máxima cuando
es perpendicular a
. En ese momento la partícula se mueve
exactamente igual que si estuviera en el borde de una rueda que gira alrededor del origen
en el plano definido por
y
(movimiento circular).
Módulo
Dirección y sentido
Conservación del momento angular
En general, si sobre la partícula actuase más de una fuerza
Ecuación análoga para las rotaciones de las segunda ley de Newton para las traslaciones
Esta ecuación es válida:
- sólo si los momentos de todas las fuerzas involucradas y el momento angular se
miden con respecto al mismo origen.
-válida para cualquier origen fijo en un sistema de referencia inercial.
Conservación del momento angular
Si
Esto se verifica si:
La fuerza se anula
(caso, por ejemplo, de la partícula libre)
La fuerza es paralela a la posición
(fuerzas centrales)
(ley de Gravitación Universal)
Analogías entre rotaciones y traslaciones
Traslaciones
Rotaciones
Una fuerza neta sobre una partícula
produce un cambio en el momento
lineal de la misma
Un torque neto sobre una partícula
produce un cambio en el momento
angular de la misma
Una fuerza neta actuando sobre una
partícula es igual a la razón de cambio
temporal del momento lineal de la partícula
Una torque neto actuando sobre una partícula
es igual a la razón de cambio temporal del
momento angular de la partícula
tance a from the pole? (a) zero (b) mvd (c) mva (d) impossible to determine
tum about any axis displaced
from the path of the particle.
Momento
de ofuna
partícula
en un
Example 11.3angular
Angular Momentum
a Particle
in Circular Motion
movimiento
circular
Solution The linear momentum of the particle is always
A particle moves in the
xy plane in a circular path of radius
changing (in direction, not magnitude). You might be
tempted, therefore, to conclude that the angular momeneltum
plano
en un
movimiento
dehowradio r.
of the xy
particle
is always
changing. In circular
this situation,
ever, this iscon
not respecto
the case—letal
us origen
see why. O
From
Equation
momento angular
si su
velocidad
11.12, the magnitude of L is given by
r, as shown in Figure 11.5. Find the magnitude and direction of its angular momentum relative to O when its linear
Supongamos
una partícula que se mueve en
velocity is v.
Hallar la magnitud y dirección de su
lineal es
L ! mvr sin 90% ! mvr
y
v
r
O
m
x
Figure 11.5 (Example 11.3) A particle moving in a circle of radius r has an angular momentum
about O that has magnitude
Magnitud
mvr. The vector L ! r " p points out of the diagram.
Como
el used
momento
linealv de
la partícula
where
we have
$ ! 90° because
is perpendicular
to r.está en
This value
of L is constant
because(en
all three
factors on no
the en
constante
cambio
dirección,
right are constant.
magnitud),
pensar
que el
The direction ofpodríamos
L also is constant,
even though
themomento
diangular
de
la
partícula
también
cambia
rection of p ! m v keeps changing. You can visualize this by de
applying the right-hand
to find the
direction
of L !
manera rule
contínua
con
el tiempo
r " p ! m r " v in Figure 11.5. Your thumb points upward
and away from the page; this is the direction of L Hence, we
can write the vector expression L ! (mvr)k̂. If the particle
embargo
es eland
caso
were to move Sin
clockwise,
L wouldeste
point no
downward
into
the page. A particle in uniform circular motion has a
constant angular momentum about an axis through the
Dirección
center of its path.
Perpendicular al plano de la pantalla y saliendo
hacia fuera (regla de la mano derecha)
Angular Momentum of a System of Particles
Una partícula en un movimiento circular uniforme tiene un momento angular
In Section 9.6, constante
we showed thatcon
Newton’s
second a
lawun
foreje
a particle
could be
extended
to
respecto
que pase
por
el centro
de la trayectoria
a system of particles, resulting in:
Momento angular total de un sistema de partículas
El momento angular total de un sistema de partículas con respecto a un determinado
punto se define como la suma vectorial de los momento angulares de las partículas
individuales con respecto a ese punto.
En un sistema continuo habría que reemplazar la suma por una integral
Momento angular total de un sistema de partículas
A priori, para cada partícula i tendríamos que calcular el torque asociado con:
- fuerzas internas entre las partículas que componen el sistema
- fuerzas externas
Sin embargo, debido al principio de acción y reacción, el torque neto
debido a las fuerzas internas se anula.
Se puede concluir que el momento angular total de un sistema de
partículas puede variar con el tiempo si y sólo si existe un torque neto
debido a las fuerzas externas que actúan sobre el sistema
Momento angular total de un sistema de partículas
El torque neto (con respecto a un eje que pase por un origen en un
sistema de referencia inercial) debido a las fuerzas externas que actúan
sobre un sistema es igual al ritmo de variación del momento angular
total del sistema con respecto a dicho origen
Momento angular de un sólido rígido en rotación
Consideremos una placa que rota alrededor de un eje perpendicular y
que coincide con el eje z de un sistema de coordenadas
Cada partícula del objeto rota en el plano xy
alrededor del eje z con una celeridad angular
El momento angular de una partícula de masa
que rota en torno al eje z es
Y el momento angular del sistema angular (que en este
caso particular sólo tiene componente a lo largo de z)
Momento angular de un sólido rígido en rotación
Y el momento angular del sistema angular (que en este
caso particular sólo tiene componente a lo largo de z)
Donde se ha definido el momento de inercia del objeto
con respecto al eje z como
En este caso particular, el momento angular tiene la misma dirección que la velocidad angular
Momento angular de un sólido rígido en rotación
En general, la expresión
no siempre es válida.
Si un objeto rígido rota alrededor de un eje arbitrario, el momento angular y la velocidad angular
podrían apuntar en direcciones diferentes.
En este caso, el momento de inercia no puede ser tratado como un escalar.
Estrictamente hablando,
se aplica sólo en el caso de un sólido rígido de cualquier forma
que rota con respecto a uno de los tres ejes mutuamente perpendiculares (denominados ejes
principales de inercia) y que pasan por su centro de masa.
Ecuación del movimiento para la rotación
de un sólido rígido
Supongamos que el eje de rotación del sólido coincide con uno de sus ejes principales,
de modo que el momento angular tiene la misma dirección que la velocidad angular
Derivando esta expresión con respecto al tiempo
Si asumimos que el momento de inercia no cambia con el tiempo
(esto ocurre para un cuerpo rígido)
El torque externo neto que actúa sobre un sólido rígido que rota alrededor de
un eje fijo es igual al momento de inercia con respecto al eje de rotación
multiplicado por la aceleración angular del objeto con respecto a ese eje
Ecuación del movimiento para la rotación
de un sólido rígido
Supongamos que el eje de rotación del sólido no coincide con uno de sus ejes principales,
de modo que el momento angular tiene la misma dirección que la velocidad angular
Pero como el momento angular ya no es paralelo a la velocidad angular,
ésta no tiene por qué ser constante
Conservación del momento angular
El momento angular total de un sistema es contante, tanto en dirección como en
módulo si el torque resultante debido a las fuerzas externas se anula
Tercera ley de conservación: en un sistema aislado se conserva:
- energía total
- el momento lineal
- el momento angular
El principio de conservación del momento angular es un resultado general que se
puede aplicar a cualquier sistema aislado.
El momento angular de un sistema aislado se conserva tanto si el sistema es un
cuerpo rígido como si no lo es.
Conservación del momento angular
El momento angular total de un sistema es contante, tanto en dirección como en
módulo si el torque resultante debido a las fuerzas externas se anula
Para un sistema aislado consistente en un conjunto de partículas, la ley de
conservación se escribe como
Conservación del momento angular
Si la masa de un sistema aislado que rota sufre un redistribución,
el momento de inercia cambia
Como la magnitud del momento angular del sistema es
La ley de conservación del momento angular requiere que el
producto de I por ω permanezca constante
Es decir, para un sistema aislado, un cambio en I requiere un cambio en ω
Esta expresión es válida para:
- una rotación en torno a un eje fijo.
- una rotación alrededor de un eje que pase por el centro de masas de un
sistema que rota.
Lo único que se requiere es que el torque neto de la fuerza externa se anule
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
elástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
pi ! pf
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
Como6.0
elkg#m/s
disco%y(2.0
la kg)v
barra
forman un sistema aislado
df ! (1.0 kg)vs
y la colisión es elástica:
we apply the law
of conservation
of angular mo-  Now
Se conserva
la energía
total
mentum, using the initial position of the center of the stick
-  our
Sereference
conserva
el momento
lineal of anas
point. We
know that the component
-  Se
conserva
momento
angular
gular
momentum
of theeldisk
along the axis
perpendicular
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
to the plane of the ice is negative. (The right-hand rule
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
Tenemos tres incógnitas
and move to the right.
Li ! Lf
y tres leyes de conservación
% rm d vdi ! %rm d vdf & I"
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
elástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
pi ! pf
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
Como6.0
elkg#m/s
disco%y(2.0
la kg)v
barra
forman un sistema aislado
df ! (1.0 kg)vs
y la colisión es elástica:
Now we apply
the law of conservation
of angular moConservación
del momento
lineal
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
mentum, using the initial position of the center of the stick
as our reference point. We know that the component of angular momentum of the disk along the axis perpendicular
to the plane of the ice is negative. (The right-hand rule
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Li ! Lf
% rm d vdi ! %rm d vdf & I"
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
elástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
pi ! pf
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
Como6.0
elkg#m/s
disco%y(2.0
la kg)v
barra
forman un sistema aislado
df ! (1.0 kg)vs
y la colisión es elástica:
Now we apply
the law of conservation
of angularangular
moConservación
del momento
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
mentum, using the initial position of the center of the stick
as our reference point. We know that the component of anLa componente
delthe
momento
angular del
gular momentum
of the disk along
axis perpendicular
largo(The
de right-hand
la dirección
to the plane of disco
the ice a
is lo
negative.
rule
shows that
Ld points into theal
ice.)
Applying
perpendicular
plano
delconservation
hielo esofnegativa
angular momentum to the system gives
(regla de la mano derecha)
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
Li ! Lf
% rm d vdi ! %rm d vdf & I"
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
elástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
pi ! pf
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
Como6.0
elkg#m/s
disco%y(2.0
la kg)v
barra
forman un sistema aislado
df ! (1.0 kg)vs
y la colisión es elástica:
Now weConservación
apply the law of conservation
of angular
mode la energía
mecánica
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
mentum, using the initial position of the center of the stick
as our reference point. We know that the component of anSolo tenemos
(tanto en
gular momentum
of the diskenergía
along the cinética
axis perpendicular
translacional
como
rotacional
to the planeforma
of the ice
is negative. (The
right-hand
rule
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Li ! Lf
% rm d vdi ! %rm d vdf & I"
su
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
elástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
Resolvemos el sistema de las tres
pi ! pf
ecuaciones
con tres incógnitas
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
6.0 kg#m/s % (2.0 kg)vdf ! (1.0 kg)vs
Now we apply the law of conservation of angular momentum, using the initial position of the center of the stick
as our
reference point. variables
We know that en
the component
of anDespejando
la primera
y segunda
gular momentum of the disk along the axis perpendicular
ecuación, y sustituyendo en la tercera
to the plane of the ice is negative. (The right-hand rule
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Li ! Lf
% rm d vdi ! %rm d vdf & I"
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
i
student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0Lm/s
golpea una
barra de 1 kg y longitud
4.0 m11.10
que se
apoya
sobre una superficie de hielo sin rozamiento. Asumimos
que la colisión es
Interactive
Example
Disk
and Stick
elástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg
disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de laarecolisión
lar momentum
all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a problem
in which
all three conservation laws might
and(b) 
that
disk does notde
deviate
from its original
of después
Latheceleridad
traslación
de laline
barra
de la
colisión
play a part. To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c)  La velocidad angular de la barra después
de la colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
momento
de inercia
barra
respectoThe
a su
dethemasas
de 1.33 kg m2
multaneously.
first centro
comes from
law of thees
conservaafter the El
collision.
The moment
of inertia ofde
thela
stick
about con
its center of mass is 1.33 kg · m2.
tion of linear momentum:
Despejando variables en la primera y segunda
pi ! pf y sustituyendo en la tercera
ecuación,
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right. (La otra solución carece de
6.0 kg#m/s % (2.0 kg)vdf ! (1.0 kg)vs
Now we apply the law of conservation of angular momentum, using the initial position of the center of the stick
as our reference point. We know that the component of angular momentum
of the disk
along the
axis perpendicular
Sustituyendo
datos
y resolviendo
la
to the plane of the ice is negative. (The right-hand rule
segundo grado
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Li ! Lf
sentido físico)
% rm d vdi ! %rm d vdf & I"
ecuación de
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
perfectamente inelástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
¿Qué pasaría si la colisión fuera perfectamente
pi ! pf inelástica?
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
En este
caso, el disco se adhiere a la barra después
6.0 kg#m/s % (2.0 kg)vdf ! (1.0 kg)vs
de la colisión
Now we apply
the law of conservation
of angular moConservación
del momento
lineal
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
mentum, using the initial position of the center of the stick
as our reference point. We know that the component of angular momentum of the disk along the axis perpendicular
to the plane of the ice is negative. (The right-hand rule
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Li ! Lf
% rm d vdi ! %rm d vdf & I"
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
perfectamente inelástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
¿Qué pasaría si la colisión fuera perfectamente
pi ! pf inelástica?
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf & m svs
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
Cálculo del centro de masas
(necesario para la parte rotacional)
6.0 kg#m/s % (2.0 kg)vdf ! (1.0 kg)vs
Now we apply the law of conservation of angular momentum,
using the initial
position ofde
thela
center
of the
stick origen
Tomamos
el centro
barra
como
as our reference point. We know that the component of anJusto en el instante de la colisión, la posición
gular momentum of the disk along the axis perpendicular
centro
de masas
estarárule
en
to the plane of del
the ice
is negative.
(The right-hand
shows that Ld points into the ice.) Applying conservation of
angular momentum to the system gives
Li ! Lf
% rm d vdi ! %rm d vdf & I"
Es decir, a 0,67 m del borde superior de la barra
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
perfectamente inelástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
¿Qué pasaría si la colisión fuera perfectamente
pi ! pf inelástica?
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf &
m svmomento
s
Conservación
del
angular
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
6.0 kg#m/s % (2.0 kg)vdf ! (1.0 kg)vs
Now we apply the law of conservation of angular momentum, using the initial position of the center of the stick
ahora es
la We
distancia
delcomponent
disco al
(0.67 m)
as our reference
point.
know that the
of CDM
angular momentum of the disk along the axis perpendicular
El plane
sistema
con
respecto
to the
of theva
ice ais rotar
negative.
(The
right-hand al
rulecentro de
shows
that Ld así
points
into tenemos
the ice.) Applying
masas,
que
queconservation
calcular oflos nuevos
angular momentum
to the system
momentos
de inercia
degives
la barra (teorema de Steiner)
Li ! Lf
% rm d vdi ! %rm d vdf & I"
other part of the system is the student plus the stool she is
sitting on. So, we can now state that
Lf ! Li ! L student&stool % Li
Problema de conservación del momento angular
Figure 11.12 (Example 11.9) The wheel is initially spinning
when the student is at rest. What happens when the wheel is
inverted?
2L
L student&stool !
Un disco de 2.0 kg que vuela con una celeridad de 3.0 m/s
golpea una i barra de 1 kg y longitud
4.0
m que
se apoya
que la colisión es
Example
11.10
Disk andsobre
Stick una superficie de hielo sin rozamiento. AsumimosInteractive
perfectamente inelástica y que el disco no se desvía de su trayectoria original.
stick. Because the disk and stick form an isolated system, we
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of
Encontrar:
can assume that total energy, linear momentum, and angulength 4.0 m that is lying flat on nearly frictionless ice, as
(a) 
La
celeridad
de
traslación
del
disco
después
de la colisión
lar momentum
are all conserved. Thus, we can categorize
shown in Figure 11.13. Assume that the collision is elastic
this as a de
problem
in which all three conservation laws might
and
the disk does
nottraslación
deviate from its
line ofdespués
(b) 
Lathatceleridad
de
deoriginal
la barra
la colisión
To analyze the problem, first note that we have
motion. Find the translational speed of the disk, the transla(c) 
La velocidad angular de la barra después play
de alapart.
colisión
three unknowns, and so we need three equations to solve sitional speed of the stick, and the angular speed of the stick
inercia
deoflathebarra
con respecto
a The
su centro
masas
esthede
1.33 kg m2
multaneously.
first comesde
from
the law of
conservaafterEl
themomento
collision. The de
moment
of inertia
stick about
its center of mass is 1.33 kg · m2.
tion of linear momentum:
¿Qué pasaría si la colisión fuera perfectamente
pi ! pf inelástica?
Solution Conceptualize the situation by considering Figure
11.13 and imagining what happens after the disk hits the
m dvdi ! m dvdf &
m svmomento
s
Conservación
del
angular
(2.0 kg)(3.0 m/s) ! (2.0 kg)vdf & (1.0 kg)vs
Before
After
(1)
vdi = 3.0 m/s
vdf
2.0 m
ω
vs
Figure 11.13 (Example 11.10) Overhead view of a disk striking
a stick in an elastic collision, which causes the stick to rotate
and move to the right.
6.0 kg#m/s % (2.0 kg)vdf ! (1.0 kg)vs
Now we apply the law of conservation of angular momentum, using the initial position of the center of the stick
as our reference point. We know that the component of angular momentum of the disk along the axis perpendicular
Despejando
la ice
velocidad
y sustituyendo
to
the plane of the
is negative. angular
(The right-hand
rule
shows that Ld points into valores
the ice.) Applying
conservation of
anteriores
angular momentum to the system gives
Li ! Lf
% rm d vdi ! %rm d vdf & I"
los
Movimiento de precesión de los giróscopos
350
C H A P T E R 1 1 • Angular Momentum
Trompo:
cuerpo simétrico que gira alrededor de un eje de simetría mientras un punto
11.5 The
of Gyroscopes and Tops
de este eje permanece
fijoMotion
(una peonza)
caso particularAde
trompo
en el que
elmotion
punto
pasa
por
el centro
de
veryun
unusual
and fascinating
type of
youfijo
probably
have
observed
is that of
a masas
Giróscopo:
top spinning about its axis of symmetry, as shown in Figure 11.14a. If the top spins very
Supongamos el movimiento rapidly,
de una
gira rápidamente
entorno
eje
deFig.simetría
thepeonza
symmetry que
axis rotates
about the z axis, sweeping
outaa su
cone
(see
Precessional motion
La peonza actúa como un giróscopo y cabría
esperar que su orientación en el espacio
permaneciera invariable
z
L
CM
(a)
n
r
Mg
11.14b). The motion of the symmetry axis about the vertical—known as precessional
motion—is usually slow relative to the spinning motion of the top.
It is quite natural to wonder why the top does not fall over. Because the center of
mass is not directly above the pivot point O, a net torque is clearly acting on the top
about O—a torque resulting from the gravitational force Mg. The top would certainly
fall over if it were not spinning. Because it is spinning, however, it has an angular momentum L directed along its symmetry axis. We shall show that this symmetry axis
moves about the z axis (precessional motion occurs) because the torque produces a
change in the direction of the symmetry axis. This is an excellent example of the importance of the directional nature of angular momentum.
The essential features of precessional motion can be illustrated by considering the
simple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
downward gravitational force Mg and the normal force n acting upward at the pivot
point O. The normal force produces no torque about the pivot because its moment
arm through that point is zero. However, the gravitational force produces a torque
! ! r " Mg about O, where the direction of ! is perpendicular to the plane formed
by r and Mg. By necessity, the vector ! lies in a horizontal xy plane perpendicular to
the angular momentum vector. The net torque and angular momentum of the gyroscope are related through Equation 11.13:
y
O
τ
x
Sin embargo, si la peonza está inclinada, se observa que su
eje de simetría gira alrededor del eje , formando en su
desplazamiento la figura de un cono.
A este movimiento se le denomina movimiento de precesión
∆L
Li
Lf
!!
(b)
Figure 11.14 Precessional motion
dL
dt
From this expression, we see that the nonzero torque produces a change in angular
momentum d L—a change that is in the same direction as !. Therefore, like the torque
vector, d L must also be perpendicular to L. Figure 11.15b illustrates the resulting precessional motion of the symmetry axis of the gyroscope. In a time interval dt, the
change in angular momentum is d L ! L f " Li ! ! dt. Because d L is perpendicular to
L, the magnitude of L does not change (! Li ! ! ! L f !). Rather, what is changing is the
direction of L. Because the change in angular momentum d L is in the direction of !,
which lies in the xy plane, the gyroscope undergoes precessional motion.
La velocidad
del eje de simetría alrededor del eje vertical es normalmente lenta con
of a top spinningangular
about its symmetry axis. (a) The only external
respecto
lathevelocidad angular de la peonza alrededor de su eje de simetría
forces acting
on the topaare
normal force n and the gravitational force Mg. The direction of
the angular momentum L is along
Movimiento de precesión de los giróscopos
Origen del movimiento de precesión
350
C H A P T E R 1 1 • Angular Momentum
The no
Motion
of Gyroscopes
Tops
¿Por qué11.5
la peonza
mantiene
su direcciónand
de giro?
A very unusual and fascinating type of motion you probably have observed is that of a
top spinning about its axis of symmetry, as shown in Figure 11.14a. If the top spins very
rapidly, the symmetry axis rotates about the z axis, sweeping out a cone (see Fig.
11.14b). The motion of the symmetry axis about the vertical—known as precessional
motion—is usually slow relative to the spinning motion of the top.
It is quite natural to wonder why the top does not fall over. Because the center of
mass is not directly above the pivot point O, a net torque is clearly acting on the top
about O—a torque resulting from the gravitational force Mg. The top would certainly
fall over if it were not spinning. Because it is spinning, however, it has an angular momentum L directed along its symmetry axis. We shall show that this symmetry axis
moves about the z axis (precessional motion occurs) because the torque produces a
change in the direction of the symmetry axis. This is an excellent example of the importance of the directional nature of angular momentum.
The essential features of precessional motion can be illustrated by considering the
simple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
downward gravitational force Mg and the normal force n acting upward at the pivot
point O. The normal force produces no torque about the pivot because its moment
arm through that point is zero. However, the gravitational force produces a torque
! ! r " Mg about O, where the direction of ! is perpendicular to the plane formed
by r and Mg. By necessity, the vector ! lies in a horizontal xy plane perpendicular to
the angular momentum vector. The net torque and angular momentum of the gyroscope are related through Equation 11.13:
Como el centro de masas de la peonza no se encuentra
directamente sobre el punto de pivote , hay un par neto
con respecto a
que actúa sobre la peonza.
Precessional motion
z
L
El par está producido por la fuerza de la gravedad
CM
(a)
n
r
Mg
y
O
τ
x
Como está girando, la peonza tiene un momento angular
cuya dirección coincide con el eje de simetría de la peonza
∆L
Li
Si no estuviera girando, la peonza caería
Lf
!!
(b)
Figure 11.14 Precessional motion
of a top spinning about its symmetry axis. (a) The only external
forces acting on the top are the
normal force n and the gravitational force Mg. The direction of
the angular momentum L is along
dL
dt
From this expression, we see that the nonzero torque produces a change in angular
momentum d L—a change that is in the same direction as !. Therefore, like the torque
vector, d L must also be perpendicular to L. Figure 11.15b illustrates the resulting precessional motion of the symmetry axis of the gyroscope. In a time interval dt, the
change in angular momentum is d L ! L f " Li ! ! dt. Because d L is perpendicular to
L, the magnitude of L does not change (! Li ! ! ! L f !). Rather, what is changing is the
direction of L. Because the change in angular momentum d L is in the direction of !,
which lies in the xy plane, the gyroscope undergoes precessional motion.
El par provoca un cambio en la dirección del eje de simetría que a la postre es el responsable del
movimiento de este eje de simetría con respecto al eje
Movimiento de precesión de los giróscopos
Origen del movimiento de precesión
350
C H A P T E R 1 1 • Angular Momentum
The no
Motion
of Gyroscopes
Tops
¿Por qué11.5
la peonza
mantiene
su direcciónand
de giro?
Las dos fuerzas que actúan sobre la peonza son:
-  Su peso:
actúa hacia abajo
-  La normal:
actúa hacia arriba en el punto de pivote
la normal no produce ningún par alrededor del pivote
porque su brazo de palanca con respecto a dicho
punto es cero
Precessional motion
z
L
CM
(a)
n
r
Mg
y
O
τ
x
∆L
Li
A very unusual and fascinating type of motion you probably have observed is that of a
top spinning about its axis of symmetry, as shown in Figure 11.14a. If the top spins very
rapidly, the symmetry axis rotates about the z axis, sweeping out a cone (see Fig.
11.14b). The motion of the symmetry axis about the vertical—known as precessional
motion—is usually slow relative to the spinning motion of the top.
It is quite natural to wonder why the top does not fall over. Because the center of
mass is not directly above the pivot point O, a net torque is clearly acting on the top
about O—a torque resulting from the gravitational force Mg. The top would certainly
fall over if it were not spinning. Because it is spinning, however, it has an angular momentum L directed along its symmetry axis. We shall show that this symmetry axis
moves about the z axis (precessional motion occurs) because the torque produces a
change in the direction of the symmetry axis. This is an excellent example of the importance of the directional nature of angular momentum.
The essential features of precessional motion can be illustrated by considering the
simple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
downward gravitational force Mg and the normal force n acting upward at the pivot
point O. The normal force produces no torque about the pivot because its moment
arm through that point is zero. However, the gravitational force produces a torque
Dirección
! ! r " Mg about O, where the direction of ! is perpendicular to the plane formed
perpendicular
perpendicular toal
by r and Mg. By necessity, the vector ! lies in a horizontal xy plane
the angular momentum vector. The net torque and angular momentum
of the gyroformado
por
scope are related through Equation 11.13:
Par con respecto a
debido a la gravedad
dL
Lf
(b)
Figure 11.14 Precessional motion
of a top spinning about its symmetry axis. (a) The only external
forces acting on the top are the
normal force n and the gravitational force Mg. The direction of
the angular momentum L is along
plano
y
Obligatoriamente el vector ! !sedtencuentra en el plano horizontal
(perpendicular
alwe
peso)
perpendicular
vectora change
momento
angular
From this expression,
see that
the nonzero torqueal
produces
in angular
like masas
the torquecon
momentum
d L—a
change that is de
in the
direction del
as !. Therefore,
(que lleva
la dirección
lasame
posición
centro de
vector, d L must also be perpendicular to L. Figure 11.15b illustrates the resulting prea ) In a time interval dt, the
cessional motion of the symmetry respecto
axis of the gyroscope.
change in angular momentum is d L ! L f " Li ! ! dt. Because d L is perpendicular to
L, the magnitude of L does not change (! Li ! ! ! L f !). Rather, what is changing is the
direction of L. Because the change in angular momentum d L is in the direction of !,
which lies in the xy plane, the gyroscope undergoes precessional motion.
Movimiento de precesión de los giróscopos
Origen del movimiento de precesión
350
C H A P T E R 1 1 • Angular Momentum
The no
Motion
of Gyroscopes
Tops
¿Por qué11.5
la peonza
mantiene
su direcciónand
de giro?
A very unusual and fascinating type of motion you probably have observed is that of a
El par neto y el momento angular están relacionados por
top spinning about its axis of symmetry, as shown in Figure 11.14a. If the top spins very
Precessional motion
z
rapidly, the symmetry axis rotates about the z axis, sweeping out a cone (see Fig.
11.14b). The motion of the symmetry axis about the vertical—known as precessional
motion—is usually slow relative to the spinning motion of the top.
It is quite natural to wonder why the top does not fall over. Because the center of
mass is not directly above the pivot point O, a net torque is clearly acting on the top
about O—a torque resulting from the gravitational force Mg. The top would certainly
fall over if it were not spinning. Because it is spinning, however, it has an angular momentum L directed along its symmetry axis. We shall show that this symmetry axis
moves about the z axis (precessional motion occurs) because the torque produces a
change in the direction of the symmetry axis. This is an excellent example of the importance of the directional nature of angular momentum.
The essential features of precessional motion can be illustrated by considering the
simple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
downward gravitational force Mg and the normal force n acting upward at the pivot
point O. The normal force produces no torque about the pivot because its moment
arm through that point is zero. However, the gravitational force produces a torque
! ! r " Mg about O, where the direction of ! is perpendicular to the plane formed
by r and Mg. By necessity, the vector ! lies in a horizontal xy plane perpendicular to
the angular momentum vector. The net torque and angular momentum of the gyroscope are related through Equation 11.13:
El cambio en el vector momento angular
producido por el par
va en la misma dirección del par.
Por ello
también tiene que ser perpendicular a
L
CM
(a)
n
r
Mg
y
O
τ
x
Dado que
∆L
Li
En un periodo de tiempo determinado
momento angular es
Lf
(b)
es perpendicular
dL a
!!
el cambio en el
el módulo de
no cambia
dt
From this expression, we see that the nonzero torque produces a change in angular
momentum d L—a change that is in the same direction as !. Therefore, like the torque
vector, d L must also be perpendicular to L. Figure 11.15b illustrates the resulting precessional motion of the symmetry axis of the gyroscope. In a time interval dt, the
change in angular momentum is d L ! L f " Li ! ! dt. Because d L is perpendicular to
L, the magnitude of L does not change (! Li ! ! ! L f !). Rather, what is changing is the
direction of L. Because the change in angular momentum d L is in the direction of !,
which lies in the xy plane, the gyroscope undergoes precessional motion.
Lo que cambia es la dirección de .
Puesto que el cambio en el momento angular
va en la dirección de
(situado en el plano
la peonza experimenta un movimiento de precesión
Figure 11.14 Precessional motion
of a top spinning about its symmetry axis. (a) The only external
forces acting on the top are the
normal force n and the gravitational force Mg. The direction of
the angular momentum L is along
),
Movimiento de precesión de una peonza
Descripción cuantitativa
350
C H A P T E R 1 1 • Angular Momentum
11.5
The Motion of Gyroscopes and Tops
En el intervalo de tiempo
el vector momento angular
rota un ángulo
que es también el ángulo que rota el eje.
A partir del triángulo que define
en la figura (b)
Precessional motion
z
A very unusual and fascinating type of motion you probably have observed is that of a
top spinning about its axis of symmetry, as shown in Figure 11.14a. If the top spins very
rapidly, the symmetry axis rotates about the z axis, sweeping out a cone (see Fig.
11.14b). The motion of the symmetry axis about the vertical—known as precessional
motion—is usually slow relative to the spinning motion of the top.
It is quite natural to wonder why the top does not fall over. Because the center of
mass is not directly above the pivot point O, a net torque is clearly acting on the top
about O—a torque resulting from the gravitational force Mg. The top would certainly
fall over if it were not spinning. Because it is spinning, however, it has an angular momentum L directed along its symmetry axis. We shall show that this symmetry axis
moves about the z axis (precessional motion occurs) because the torque produces a
change in the direction of the symmetry axis. This is an excellent example of the importance of the directional nature of angular momentum.
The essential features of precessional motion can be illustrated by considering the
simple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
downward gravitational force Mg and the normal force n acting upward at the pivot
point O. The normal force produces no torque about the pivot because its moment
arm through that point is zero. However, the gravitational force produces a torque
! ! r " Mg about O, where the direction of ! is perpendicular to the plane formed
by r and Mg. By necessity, the vector ! lies in a horizontal xy plane perpendicular to
the angular momentum vector. The net torque and angular momentum of the gyroscope are related through Equation 11.13:
Por otra parte, el módulo del momento del peso viene definido por
L
CM
(a)
n
r
Mg
y
O
τ
x
Definiendo la velocidad angular
de precesión como
dL
∆L
Li
Como
Lf
!!
(b)
Figure 11.14 Precessional motion
of a top spinning about its symmetry axis. (a) The only external
forces acting on the top are the
normal force n and the gravitational force Mg. The direction of
the angular momentum L is along
dt
From this expression, we see that the nonzero torque produces a change in angular
Independiente del
momentum d L—a change that is in the same direction as !. Therefore, like the torque
ángulo
inclinación
vector, d L must also be perpendicular to L. Figure 11.15b illustrates the
resultingde
precessional motion of the symmetry axis of the gyroscope. In a time interval dt, the
change in angular momentum is d L ! L f " Li ! ! dt. Because d L is perpendicular to
L, the magnitude of L does not change (! Li ! ! ! L f !). Rather, what is changing is the
direction of L. Because the change in angular momentum d L is in the direction of !,
which lies in the xy plane, the gyroscope undergoes precessional motion.
El resultado es válido siempre que
out O—a torque resulting from the gravitational force Mg. The top would certainly
over if it were not spinning. Because it is spinning, however, it has an angular montum L directed along its symmetry axis. We shall show that this symmetry axis
ves about the z axis (precessional motion occurs) because the torque produces a
nge in the direction of the symmetry axis. This is an excellent example of the imporce of the directional nature of angular momentum.
The essential features of precessional motion can be illustrated by considering the
ple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
wnward gravitational force Mg and the normal force n acting upward at the pivot
nt O. The normal force produces no torque about the pivot because its moment
m through that point is zero. However, the gravitational force produces a torque
! r " Mg about O, where the direction of ! is perpendicular to the plane formed
r and Mg. By necessity, the vector ! lies in a horizontal xy plane perpendicular to
angular momentum vector. The net torque and angular momentum of the gyrope are related through Equation 11.13:
Movimiento de precesión de los giróscopos
Descripción cuantitativa
Repitiendo el proceso anterior para el caso de un giróscopo
!!
dL
dt
m this expression, we see that the nonzero torque produces a change in angular
mentum d L—a change that is in the same direction as !. Therefore, like the torque
tor, d L must also be perpendicular to L. Figure 11.15b illustrates the resulting presional motion of the symmetry axis of the gyroscope. In a time interval dt, the
nge in angular momentum is d L ! L f " Li ! ! dt. Because d L is perpendicular to
the magnitude of L does not change (! Li ! ! ! L f !). Rather, what is changing is the
ction of L. Because the change in angular momentum d L is in the direction of !,
ch lies in the xy plane, the gyroscope undergoes precessional motion.
h
En el intervalo de tiempo
el vector momento angular
rota un ángulo
que es también el ángulo que rota el eje
Del triángulo formado por los vectores
,
,y
n
O
τ
Li
Lf
Donde hemos utilizado que para ángulos pequeños
Li
Mg
dφ
φ
Lf
dL
(a)
(b)
ure 11.15 (a) The motion of a simple gyroscope pivoted a distance h from its center
mass. The gravitational force Mg produces a torque about the pivot, and this torque
erpendicular to the axle. (b) Overhead view of the initial and final angular momenvectors. The torque results in a change in angular momentum d L in a direction
pendicular to the axle. The axle sweeps out an angle d # in a time interval dt.
Dividiendo entre
y utilizando la expresión
A esta velocidad angular se la conoce como frecuancia de precesión.
El resultado es válido siempre que
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