15. Dinamica Del Movimiento Rotacional

7/29/2019 15. Dinamica Del Movimiento Rotacional

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A N G U L A R M O M E N T U M

CHAPTER 11


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PRODUCTO VECTORIAL

Existe instancias donde el producto de vectoresresultar ser otro vector. Anteriormente se observo que el producto de dos vectores

tambin puede ser un escalar.

Ello es denominado producto punto y es aplicado acircunstancias donde los vectores son paralelos.

El producto vectorial de dos vectores esdenominado producto cruz y es aplicado acircunstancias donde los vectores sonperpendiculares.


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PRODUCTO VECTORIAL Y TORQUE

El vector torque est enuna direccinperpendicular al plano

formado por losvectores posicin yfuerza.

t = rx F

El torque es el productovectorial (o cruz) delvector posicin y delvector fuerza.


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DEFINICIN DE PRODUCTOVECTORIAL

Dado para dos vectores, A y B

El producto vectorial (cruz) de A y B es definidocomo un tercer vector, C.

C es ledo como A cruz B

La magnitud de C esAB sen q Donde, q es el ngulo entre A y B.


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EL PRODUCTO VECTORIAL

El mduloAB senq esigual al rea delparalelogramo formado

por los vectores A y B. La direccin de C es

perpendicular al planoformado por los vectoresA y B.

La mejor forma dedeterminar la direccines usando la regla de lamano derecha.


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PROPIEDADES DEL PRODUCTOVECTORIAL 1

El producto vectorial no es conmutativo. El orden enla determinacin de la multiplicacin es importante.

Tomando en cuenta el orden, recordar:

A x B = - B x A

Si A es paralelo a B (q = 0o or 180o), luego, A x B = 0 De la misma manera: A x A = 0

Si A es perpendicular a B, luego |A x B| = AB


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PROPIEDADES DEL PRODUCTOVECTORIAL 2

El producto vectorial obedece a la ley distributiva: A x (B + C) = A x B + A x C

La derivada del producto cruz con respecto adeterminadas variables tales como t es:

donde es importante para ejecutar lamultiplicacin el orden de A y B.

d d d

dt dt dt

A B

A B B A


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PRODUCTO VECTORIAL CONVECTORES UNITARIOS

0

i i j j k k

i j j i k

j k k j i

k i i k j

jiji

Los signos son intercambiables en el producto cruz. A x (-B) = - A x B


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USANDO DETERMINANTES

El producto cruz tambin puede ser expresadocomo:

Desarrollando cada determinante se obtiene:

y z x yx z

x y z

y z x yx z

x y z

A A A AA AA A A

B B B BB BB B B

i j k

A B i j k

y z z y x z z x x y y xA B A B A B A B A B A B A B i j k


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VECTOR TORQUE, EJEMPLO

Dados los datos:

Determine t = ?m)00.500.4(

N)00.300.2(

jir

jiF

[(4.00 5.00 )N] [(2.00 3.00 )m] [(4.00)(2.00) (4.00)(3.00)

(5.00)(2.00) (5.00)(3.00)

2.0 N m

t

r F i j i j

i i i j

j i i j

k


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MOMENTO ANGULAR 1

Considerando una partcula de masa mdeterminado por el vector posicin r y movindosecon cantidad de movimientop.

Adicionando el trmino

( )

d

dt

d

dt

d

dt

t

t

pr F r

r

p

r p


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MOMENTO ANGULAR 2

El momento angularinstantneo L de unapartcula relativa alorigen O es definidocomo el producto cruzde del vector posicinr y su cantidad de

movimientoinstantneop de lapartcula. L = r xp


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TORQUE Y MOMENTO ANGULAR

El torque es relativo al momento angular. De manera idntica como la fuerza se comporta

en la cantidad de movimiento.

Esto es la rotacin anlogo a la segunda ley

de Newton. St y L deben ser medidos respecto a un mismo

origen.

Valido para sistemas de referencia inerciales.

d

dtt

L


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MOMENTO ANGULAR 3

The SI units of angular momentum are (kg.m2)/ s

Both the magnitude and direction of L depend onthe choice of origin

The magnitude of L = mvrsin f f is the angle between p and r The direction of L is perpendicular to the plane

formed by r and p


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ANGULAR MOMENTUM OF A

PARTICLE, EXAMPLE The vectorL = r x p is

pointed out of thediagram

The magnitude isL = mvrsin 90o = mvr sin 90o is used since v is

perpendicular tor

A particle in uniformcircular motion has aconstant angularmomentum about anaxis through the

center of its path


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ANGULAR MOMENTUM OF A SYSTEM

OF PARTICLES

The total angular momentum of a system ofparticles is defined as the vector sum of the angularmomenta of the individual particles

Ltot = L1 + L2+ + Ln =S

Li Differentiating with respect to time

tot ii

i i

d d

dt dt t

L L


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OF PARTICLES, CONT

Any torques associated with the internalforces acting in a system of particles arezero

Therefore,

The net external torque acting on a system about

some axis passing through an origin in an inertialframe equals the time rate of change of the totalangular momentum of the system about thatorigin

totext

ddt

t L


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OF PARTICLES, FINAL

The resultant torque acting on a system about anaxis through the center of mass equals the time rateof change of angular momentum of the systemregardless of the motion of the center of mass This applies even if the center of mass is accelerating,

provided tand L are evaluated relative to the center ofmass


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ROTATING RIGID OBJECT Each particle of the

object rotates in thexy plane about thez

axis with an angularspeed of w

The angularmomentum of an

individual particle is Li= mi ri

2w

L and w are directedalong thez axis


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ROTATING RIGID OBJECT, CONT

To find the angular momentum of the entire object,add the angular momenta of all the individualparticles

This also gives the rotational form of NewtonsSecond Law

2

z i i i

i i

L L m r Iw w

extzdL dI I

dt dt

wt


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ROTATING RIGID OBJECT, FINAL

The rotational form of Newtons Second Lawis also valid for a rigid object rotating abouta moving axis provided the moving axis:

(1) passes through the center of mass(2) is a symmetry axis

If a symmetrical object rotates about a fixedaxis passing through its center of mass, the

vector form holds: L = Iw where L is the total angular momentum measured

with respect to the axis of rotation


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BOWLING BALL The momentum of

inertia of the ball is2/5MR 2

The angularmomentum of theball is Lz = Iw

The direction ofthe angularmomentum is inthe positivezdirection


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CONSERVATION OF ANGULAR

MOMENTUM

The total angular momentum of a system isconstant in both magnitude and direction ifthe resultant external torque acting on the

system is zero Net torque = 0 -> means that the system is

isolated

Ltot

= constant orLi

= Lf

For a system of particles, Ltot = SLn = constant


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MOMENTUM, CONT

If the mass of an isolated systemundergoes redistribution, the moment

of inertia changes The conservation of angular momentum

requires a compensating change in theangular velocity

Iiwi = Ifwf This holds for rotation about a fixed axis and forrotation about an axis through the center of mass of amoving system

The net torque must be zero in any case


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CONSERVATION LAW SUMMARY

For an isolated system -

(1) Conservation of Energy:

Ei = Ef

(2) Conservation of Linear Momentum: pi = pf

(3) Conservation of Angular Momentum:

Li = Lf


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MOMENTUM:

THE MERRY-GO-ROUND The moment of inertia

of the system is themoment of inertia of

the platform plus themoment of inertia ofthe person Assume the person can

be treated as a particle

As the person movestoward the center ofthe rotating platform,the angular speed willincrease To keep L constant


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MOTION OF A TOP

The only externalforces acting on thetop are the normal

force n and thegravitational forceMg

The direction of theangular momentum L

is along the axis ofsymmetry

The right-hand ruleindicates that t = r F= r

M g is in thexyplane


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MOTION OF A TOP, CONT

The direction of d L is parallel to that of t in part. Thefact that Lf= d L + Li indicates that the top precessesabout thez axis. The precessional motion is the motion of the symmetry axis

about the vertical

The precession is usually slow relative to the spinning motionof the top


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GYROSCOPE

A gyroscope can beused to illustrateprecessional motion

The gravitational forceMg produces a torqueabout the pivot, andthis torque is

perpendicular to theaxle

The normal forceproduces no torque


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GYROSCOPE, CONT

The torque results in achange in angularmomentum d L in a

direction perpendicularto the axle. The axlesweeps out an angle dfin a time interval dt.

The direction, not the

magnitude, of L ischanging

The gyroscopeexperiences

precessional motion


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GYROSCOPE, FINAL

To simplify, assume the angular momentum due tothe motion of the center of mass about the pivot iszero Therefore, the total angular momentum is L = Iwdue to its

spin

This is a good approximation when wis large


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PRECESSIONAL FREQUENCY

Analyzing the previous vector triangle, the rate atwhich the axle rotates about the vertical axis canbe found

wp is the precessional frequencyp

d Mgh

dt I

fw

w


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GYROSCOPE IN A SPACECRAFT

The angular momentumof the spacecraft aboutits center of mass is zero

A gyroscope is set intorotation, giving it anonzero angularmomentum

The spacecraft rotates inthe direction opposite tothat of the gyroscope

So the total momentumof the system remains

zero


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ANGULAR MOMENTUM AS A

FUNDAMENTAL QUANTITY

The concept of angular momentum is alsovalid on a submicroscopic scale

Angular momentum has been used in the

development of modern theories of atomic,molecular and nuclear physics

In these systems, the angular momentumhas been found to be a fundamental

quantity Fundamental here means that it is an intrinsic

property of these objects

It is a part of their nature


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FUNDAMENTAL ANGULAR

MOMENTUM

Angular momentum has discrete values

These discrete values are multiples of afundamental unit of angular momentum

The fundamental unit of angular momentum is h-bar

Where his called Plancks constant2

34 kg m1.054 102 sh


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MOLECULE Consider the

molecule as a rigid

rotor, with the twoatoms separated by afixed distance

The rotation occurs

about the center ofmass in the plane ofthe page with aspeed of

CMI

w


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CLASSICAL IDEAS IN SUBATOMIC

SYSTEMS

Certain classical concepts and models areuseful in describing some features of atomic andmolecular systems

Proper modifications must be made A wide variety of subatomic phenomena can be

explained by assuming discrete values of theangular momentum associated with a particular

type of motion


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NIELS BOHR

Niels Bohr was a Danish physicist

He adopted the (then radical) idea of discreteangular momentum values in developing his theory

of the hydrogen atom Classical models were unsuccessful in describing many

aspects of the atom


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BOHRS HYDROGEN ATOM

The electron could occupy only those circular orbitsfor which the orbital angular momentum was equalto n

where n is an integer

This means that orbital angular momentum isquantized

15. Dinamica Del Movimiento Rotacional

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