15. Dinamica Del Movimiento Rotacional
-
Upload
crizthian-ames -
Category
Documents
-
view
231 -
download
1
Transcript of 15. Dinamica Del Movimiento Rotacional
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
1/39
A N G U L A R M O M E N T U M
CHAPTER 11
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
2/39
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.
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
3/39
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.
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
4/39
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.
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
5/39
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.
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
6/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
7/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
8/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
9/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
10/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
11/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
12/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
13/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
14/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
15/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
16/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
17/39
ANGULAR MOMENTUM OF A SYSTEM
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
18/39
ANGULAR MOMENTUM OF A SYSTEM
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
19/39
ANGULAR MOMENTUM OF A
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
20/39
ANGULAR MOMENTUM OF A
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
21/39
ANGULAR MOMENTUM OF A
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
22/39
ANGULAR MOMENTUM OF A
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
23/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
24/39
CONSERVATION OF ANGULAR
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
25/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
26/39
CONSERVATION OF ANGULAR
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
27/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
28/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
29/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
30/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
31/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
32/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
33/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
34/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
35/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
36/39
ANGULAR MOMENTUM OF A
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
37/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
38/39
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
-
7/29/2019 15. Dinamica Del Movimiento Rotacional
39/39
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