sobre momento angular | Rotation Around A Fixed Axis | 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 ˆ If the particle can write the vector expression L ! (mvr)k. 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