Transcript Chapter 11 - UCF Physics

Chapter 11
Angular Momentum
The Vector Product and Torque
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The torque vector lies in
a direction perpendicular
to the plane formed by
the position vector and
the force vector
  Fr
The torque is the vector
(or cross) product of the
position vector and the
force vector
Torque Vector Example
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Given the force and location
F  (2.00 ˆi  3.00 ˆj) N
r  (4.00 ˆi  5.00 ˆj) m

Find the torque produced
  r  F  [(4.00ˆi  5.00ˆj)N]  [(2.00ˆi  3.00ˆj)m]
 [(4.00)(2.00)ˆi  ˆi  (4.00)(3.00)ˆi  ˆj
(5.00)(2.00)ˆj  ˆi  (5.00)(3.00)ˆi  ˆj
 2.0 kˆ N  m
Angular Momentum
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Consider a particle of mass m located at the
vector position r and moving with linear
momentum p
Find the net torque
dp
r   F    r 
dt
dr
Add the term
 p  sinceit  0 
dt
d (r  p )
  dt
Angular Momentum
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The instantaneous angular
momentum L of a particle
relative to the origin O is
defined as the cross product
of the particle’s instantaneous
position vector r and its
instantaneous linear
momentum p

L  r p
Torque and Angular Momentum
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The torque is related to the angular momentum
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Similar to the way force is related to linear momentum
dL
  dt
The torque acting on a particle is equal to the
time rate of change of the particle’s angular
momentum
This is the rotational analog of Newton’s Second
Law
   and L
must be measured about the same origin
Angular Momentum
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The SI units of angular momentum are
(kg.m2)/ s
Both the magnitude and direction of the
angular momentum depend on the choice of
origin
The magnitude is L = mvr sinf
 f is the angle between p and r
The direction of L is perpendicular to the
plane formed by r and p
Angular Momentum of a Particle,
Example
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The vector L = r  p is
pointed out of the diagram
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The magnitude is
L = mvr sin 90o = mvr
sin 90o is used since v
is perpendicular to r
A particle in uniform
circular motion has a
constant angular
momentum about an axis
through the center of its
path
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Angular Momentum of a System of
Particles
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The total angular momentum of a system of
particles is defined as the vector sum of the
angular momenta of the individual particles

Ltot  L1  L2 
 Ln   Li
i

Differentiating with respect to time
dLtot
dLi

  i
dt
dt
i
i
Angular Momentum of a System of
Particles
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Any torques associated with the internal forces
acting in a system of particles are zero
Therefore,


ext
dL tot

dt
The net external torque acting on a system about
some axis passing through an origin in an inertial
frame equals the time rate of change of the total
angular momentum of the system about that origin
Angular Momentum of a System of
Particles
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The resultant torque acting on a system
about an axis through the center of mass
equals the time rate of change of angular
momentum of the system regardless of the
motion of the center of mass
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This applies even if the center of mass is
accelerating, provided  and L are evaluated
relative to the center of mass
System of Objects, Example
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The masses are
connected by a light
cord that passes over
a pulley; find the
linear acceleration
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Conceptualize
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The sphere falls, the
pulley rotates and the
block slides
Use angular
momentum approach
System of Objects, Example
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Categorize
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The block, pulley and sphere are a nonisolated
system
Use an axis that corresponds to the axle of the
pulley
Analyze
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At any instant of time, the sphere and the block
have a common velocity v
Write expressions for the total angular momentum
and the net external torque
System of Objects, Example
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Solve the expression for the linear
acceleration
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The system as a whole was analyzed so that
internal forces could be ignored
Only external forces are needed
Angular Momentum of a Rotating
Rigid Object
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Each particle of the object
rotates in the xy plane
about the z axis with an
angular speed of w
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The angular momentum of
an individual particle is
Li = mi ri2 w

L and w are directed
along the z axis
Angular Momentum of a Rotating
Rigid Object
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To find the angular momentum of the entire
object, add the angular momenta of all the
individual particles


Lz   Li   mi ri w  Iw
i

2
i
This also gives the rotational form of
Newton’s Second Law

ext
dLz
dw

I
 I
dt
dt
Angular Momentum of a Rotating
Rigid Object
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The rotational form of Newton’s Second Law is
also valid for a rigid object rotating about a
moving axis provided the moving axis:
(1) passes through the center of mass
(2) is a symmetry axis
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If a symmetrical object rotates about a fixed axis
passing through its center of mass, the vector form
holds:
L  Iw
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where L is the total angular momentum measured
with respect to the axis of rotation
Angular Momentum of a Bowling
Ball
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The momentum of
inertia of the ball is
2/5MR 2
The angular
momentum of the ball
is Lz = Iw
The direction of the
angular momentum is
in the positive z
direction
Conservation of Angular
Momentum
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The total angular momentum of a system is
constant in both magnitude and direction if the
resultant external torque acting on the system is
zero
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Net torque = 0 -> means that the system is isolated
Ltot = constant or Li = Lf
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For a system of particles,
Ltot =
L
n
= constant
Conservation of Angular
Momentum
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If the mass of an isolated system undergoes
redistribution, the moment of inertia changes
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The conservation of angular momentum requires
a compensating change in the angular velocity
Ii wi = If wf = constant
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This holds for rotation about a fixed axis and for
rotation about an axis through the center of mass of
a moving system
The net torque must be zero in any case
Conservation Law Summary
For an isolated system (1) Conservation of Energy:
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Ei = Ef
(2) Conservation of Linear Momentum:
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p i  pf
(3) Conservation of Angular Momentum:
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L i  Lf
A solid cylinder of radius 15 cm and mass 3.0 kg
rolls without slipping at a constant speed of 1.6 m/s.
(a) What is its angular momentum about its
symmetry axis? (b) What is its rotational kinetic
energy? (c) What is its total kinetic
1
energy? ( I cylinder= MR 2 )
2
A light rigid rod 1.00 m in
length joins two particles, with
masses 4.00 kg and 3.00 kg,
at its ends. The combination
rotates in the xy plane about a
pivot through the center of the
rod. Determine the angular
momentum of the system
about the origin when the
speed of each particle is 5.00
m/s.
A conical pendulum consists of a
bob of mass m in motion in a circular
path in a horizontal plane as shown.
During the motion, the supporting
wire of length  maintains the
constant angle with the vertical.
Show that the magnitude of the
angular momentum of the bob about
the center of the circle is
 m g sin f 
L

cosf


2
3
4
1/ 2
The position vector of a particle of mass 2.00 kg is
given as a function of time by r  6.00ˆi  5.00t ˆj m
Determine the angular momentum of the particle
about the origin, as a function of time.
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A particle of mass m is shot with an initial velocity vi making an
angle with the horizontal as shown. The particle moves in the
gravitational field of the Earth. Find the angular momentum of
the particle about the origin when the particle is (a) at the origin,
(b) at the highest point of its trajectory, and (c) just before it hits
the ground. (d) What torque causes its angular momentum to
change?
A wad of sticky clay with mass m and velocity vi is fired at a
solid cylinder of mass M and radius R. The cylinder is initially
at rest, and is mounted on a fixed horizontal axle that runs
through its center of mass. The line of motion of the projectile
is perpendicular to the axle and at a distance d < R from the
center. (a) Find the angular speed of the system just after the
clay strikes and sticks to the surface of the cylinder. (b) Is
mechanical energy of the clay-cylinder system conserved in this
process? Explain your answer.
Two astronauts, each having a mass of 75.0 kg, are connected
by a 10.0-m rope of negligible mass. They are isolated in
space, orbiting their center of mass at speeds of 5.00 m/s.
Treating the astronauts as particles, calculate (a) the magnitude
of the angular momentum of the system and (b) the rotational
energy of the system. By pulling on the rope, one of the
astronauts shortens the distance between them to 5.00 m. (c)
What is the new angular momentum of the system? (d) What
are the astronauts’ new speeds? (e) What is the new rotational
energy of the system? (f) How much work does the astronaut
do in shortening the rope?
A uniform solid disk is set into rotation with an angular speed
ωi about an axis through its center. While still rotating at this
speed, the disk is placed into contact with a horizontal
surface and released as in Figure. (a) What is the angular
speed of the disk once pure rolling takes place? (b) Find the
fractional loss in kinetic energy from the time the disk is
released until pure rolling occurs. (Hint: Consider torques
about the center of mass.)
Conservation of Angular Momentum:
The Merry-Go-Round
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The moment of inertia of the
system is the moment of inertia
of the platform plus the
moment of inertia of the person
 Assume the person can be
treated as a particle
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As the person moves toward
the center of the rotating
platform, the angular speed will
increase
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To keep the angular
momentum constant