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Chapter 11Chapter 11
Angular MomentumAngular Momentum
Schedule 2+ Weeks left!Schedule 2+ Weeks left!
10-Apr
Ch 11: Angular Mom.
Ch 11: Angular Mom.+ Chapt 12. Ch 12: Statics
17-Apr Ch 12: Statics Ch 15: Oscillations Ch 15: Oscillations
24-Apr Ch 15: Oscillations No Class No Class
1-May FINAL EXAM No Class No Class
So…So…
This weekThis week– Angular MomentumAngular Momentum– Statics (?)Statics (?)– Quiz on FridayQuiz on Friday
Next WeekNext Week– Statics & OscillationsStatics & Oscillations
Following WeekFollowing Week– One ClassOne Class
Then … the final examinationThen … the final examination– VERY CLOSE!VERY CLOSE!
The ExamThe Exam
You had seen most of it.You had seen most of it. It is not graded yet.It is not graded yet. You probably did ok!You probably did ok!
– Or not.Or not.
Let’s just make a few comments on the Let’s just make a few comments on the examexam
As shown in the figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Have you seen this one before??
A rod of length 30.0 cm has linear mass density (mass-per-length) that varies with position. It is given by = 50.0 g/m + 20.0 x g/m2,where x is the distance from one end, measured in meters. (a) What is the mass of the rod? (b) How far from the x = 0 end is its center of mass? This one is new.
L
CM
L
xdmM
X
dmM
0
0
1
This one is in your notes
So is the other oneSo is the other one
The Vector Product (Brief Review) There are instances where the
product of two vectors is another vector A X B
The vector product of two vectors is also called the cross product
You can learn more about this in the textbook and in the notes on the website.
The Vector Product and Torque The torque vector
lies in a direction perpendicular to the plane formed by the position vector and the force vector
= r x F The torque is the
vector (or cross) product of the position vector and the force vector
The Vector Product Defined Given two vectors, A and B The vector (cross) product of A
and B is defined as a third vector, C
C is read as “A cross B” The magnitude of C is AB sin
is the angle between A and B
More About the Vector Product
The quantity AB sin is equal to the area of the parallelogram formed by A and B
The direction of C is perpendicular to the plane formed by A and B
The best way to determine this direction is to use the right-hand rule
Properties of the Vector Product The vector product is not
commutative. The order in which the vectors are multiplied is important
To account for order, rememberA x B = - B x A
If A is parallel to B ( = 0o or 180o), then A x B = 0 Therefore A x A = 0
More Properties of the Vector Product If A is perpendicular to B, then |A x
B| = AB The vector product obeys the
distributive law A x (B + C) = A x B + A x C
Final Properties of the Vector Product The derivative of the cross product
with respect to some variable such as t is
where it is important to preserve the multiplicative order of A and B
d d d
dt dt dt
A BA B B A
Vector Products of Unit Vectors
ˆ ˆ ˆ ˆ ˆ ˆ 0
ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆˆ ˆ
ˆ ˆ ˆˆ ˆ
i i j j k k
i j j i k
j k k j i
k i i k j
Vector Products of Unit Vectors, cont Signs are interchangeable in cross
products A x (-B) = - A x B jiji ˆˆˆˆ
Using Determinants The cross product can be
expressed as
Expanding the determinants gives
ˆ ˆ ˆ
ˆ ˆ ˆy z x yx zx y z
y z x yx zx y z
A A A AA AA A A
B B B BB BB B B
i j k
A B i j k
ˆ ˆ ˆy 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
Torque Vector Example Given the force
= ?
m)ˆ00.5ˆ00.4(
N)ˆ00.3ˆ00.2(
jir
jiF
ˆ ˆ ˆ ˆ [(4.00 5.00 )N] [(2.00 3.00 )m]
ˆ ˆ ˆ ˆ[(4.00)(2.00) (4.00)(3.00)
ˆ ˆ ˆ ˆ(5.00)(2.00) (5.00)(3.00)
ˆ2.0 N m
r F i j i j
i i i j
j i i j
k
)(0.2)5234(
032
054 Nmxx kk
kji
Fr
Using the determinant …
New Definition
Angular Momentum: L
p=mv
Angular Momentum Consider a particle of mass m
located at the vector position r and moving with linear momentum p
Adding the term
( )
d
dtd
dtd
dt
pr F r
rp
r p
This is equal to zerobecause v=(dr/dt) isparallel to p
New Rules The torque applied to a mass m is equal
to the time rate of change of its angular momentum.
The force applied to a mass is still equal to the time rate of change of its linear momentum.
From the definition, if there is no torque applied to a mass, its angular momentum will remain constant!
So .. no force, no torque, no change in angular momentum!
r sin()
Again ….. the definition of L 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 x p
Torque and Angular Momentum The torque is related to the angular
momentum Similar to the way force is related to linear
momentum
This is the rotational analog of Newton’s Second Law and L must be measured about the same origin This is valid for any origin fixed in an inertial frame
d
dt L
New Rules … Add the following associations to our previous list
).()(
)()(
momentumangularLmomentump
torqueforceF
More About Angular Momentum The SI units of angular momentum
are (kg.m2)/ s Both the magnitude and direction
of L depend on the choice of origin The magnitude of L = mvr sin
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 Executing Circular Motion about a point. The vector L = r x p is
pointed out of the diagram
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
Angular Momentum of a System of Particles 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
Differentiating with respect to timetot i
ii i
d d
dt dt L L
Angular Momentum of a System of Particles, cont Any torques associated with the internal
forces acting in a system of particles are zero
Therefore,
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
totext
d
dt L
Angular Momentum of a System of Particles, final 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 This applies even if the center of mass
is accelerating, provided and L are evaluated relative to the center of mass
Angular Momentum of a Rotating Rigid Object Each particle of the
object rotates in the xy plane about the z axis with an angular speed of
The angular momentum of an individual particle is Li = mi ri
2 L and are directed
along the z axis
=Ii
Angular Momentum of a Rotating Rigid Object, cont To find the angular momentum of
the entire object, add the angular momenta of all the individual particles
This also gives the rotational form of Newton’s Second Law
2z i i i
i i
L L m r I
extzdL dI I
dt dt
Angular Momentum of a Rotating Rigid Object, final If a symmetrical object rotates
about a fixed axis passing through its center of mass, the vector form holds: L = I where L is the total angular
momentum measured with respect to the axis of rotation
Angular Momentum of a Bowling Ball The momentum of
inertia of the ball is 2/5MR 2
The angular momentum of the ball is Lz = I
The direction of the angular momentum is in the positive z direction
Conservation of Angular Momentum 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 Net torque = 0 -> means that the system is
isolated Ltot = constant or Li = Lf
For a system of particles, Ltot = Ln = constant
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
Conservation of Angular Momentum:The Merry-Go-Round 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
As the person moves toward the center of the rotating platform, the angular speed will increase
To keep L constant
Motion of a Top The only external
forces acting on the top are the normal force n and the gravitational force M g
The direction of the angular momentum L is along the axis of symmetry
The right-hand rule indicates that = r F = r M g is in the xy plane
Motion of a Top, cont The direction of d L is parallel to that
of in part. The fact that Lf = d L + Li indicates that the top precesses about the z axis. The precessional motion is the motion
of the symmetry axis about the vertical The precession is usually slow relative
to the spinning motion of the top
Gyroscope A gyroscope can be
used to illustrate precessional motion
The gravitational force Mg produces a torque about the pivot, and this torque is perpendicular to the axle
The normal force produces no torque
Gyroscope, cont The torque results in a
change in angular momentum d L in a direction perpendicular to the axle. The axle sweeps out an angle d in a time interval dt.
The direction, not the magnitude, of L is changing
The gyroscope experiences precessional motion
Precessional Frequency Analyzing the previous vector
triangle, the rate at which the axle rotates about the vertical axis can be found
p is the precessional frequency
p
d Mgh
dt I
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