Transcript Angular Momentum

Physics 111: Mechanics
Lecture 11
Dale Gary
NJIT Physics Department
Angular Momentum




Vectors
Cross Product
Torque using vectors
Angular Momentum
March 25, 2016
Vector Basics



We will be using vectors a lot in this
course.
Remember that vectors have both
magnitude and direction e.g. a, q
You should know how to find the
components of a vector from its
magnitude and direction
a x  a cos q
Ways of writing vector notation
a y  a sin q

You should know how to find a vector’s
magnitude and direction from its
components
a  ax2  a y2
F  ma


F  ma
F  ma
y

a
ay
q
ax
q  tan 1 a y / ax
March 25, 2016
x
Projection of a Vector in Three
Dimensions




Any vector in three dimensions
can be projected onto the x-y
plane.
The vector projection then
makes an angle f from the x
axis.
Now project the vector onto
the z axis, along the direction
of the earlier projection.
The original vector a makes an
angle q from the z axis.
z

a
q
f
x
March 25, 2016
y
Vector Basics (Spherical Coords)


You should know how to generalize the
case of a 2-d vector to three dimensions,
e.g. 1 magnitude and 2 directions a, q , f
Conversion to x, y, z components
a x  a sin q cos f
z

a
a y  a sin q sin f

a  a a a
2
x
2
y
f
2
z
q  cos 1 a z / a
f  tan 1 a y / ax

q
a z  a cos q
Conversion from x, y, z components
a sin q
x
Unit vector notation:

a  axiˆ  a y ˆj  az kˆ
March 25, 2016
y
A Note About Right-Hand
Coordinate Systems


A three-dimensional
coordinate system MUST obey
the right-hand rule.
Curl the fingers of your RIGHT
HAND so they go from x to y.
Your thumb will point in the z
direction.
z
y
x
March 25, 2016
Cross Product
 
C  A B




B

B sin q
The cross product of two vectors says
something about how perpendicular they are.

Magnitude:
 

A

A sin q
q
C  A  B  AB sin q




q is smaller angle between the vectors
Cross product of any parallel vectors = zero
Cross product is maximum for perpendicular
vectors
Cross products of Cartesian unit vectors:
iˆ  ˆj  kˆ; iˆ  kˆ   ˆj; ˆj  kˆ  iˆ
iˆ  iˆ  0; ˆj  ˆj  0; kˆ  kˆ  0
y
j
i
x
k
z
i
j
March 25, 2016
k
Cross Product

Direction: C perpendicular to
both A and B (right-hand rule)






Place A and B tail to tail
Right hand, not left hand
Four fingers are pointed along
the first vector A
“sweep” from first vector A
into second vector B through
the smaller angle between
them
Your outstretched thumb
points the direction of C
First practice
   
A B  B  A ?
A B  B  A
   
A B  B  A ?
March 25, 2016
More about Cross Product



The quantity ABsinq is the area of the
parallelogram formed by A and B
The direction of C is perpendicular to
the plane formed by A and B
Cross product is not commutative
A B  B  A



The distributive law
  
   
A (B  C)  A B  A C
The derivative of cross product
obeys the chain rule
Calculate cross product


d   dA   dB
A B 
 B  A
dt
dt
dt


 
A  B  ( Ay Bz  Az By )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax By  Ay Bx )kˆ
March 25, 2016
Derivation



 
How do we show that A  B  ( Ay Bz  Az By )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax By  Ay Bx )kˆ ?

Start with
A  Axiˆ  Ay ˆj  Az kˆ

B  Bxiˆ  By ˆj  Bz kˆ
Then  
A  B  ( Axiˆ  Ay ˆj  Az kˆ)  ( Bxiˆ  By ˆj  Bz kˆ)
 Axiˆ  ( Bxiˆ  By ˆj  Bz kˆ)  Ay ˆj  ( Bxiˆ  By ˆj  Bz kˆ)  Az kˆ  ( Bxiˆ  By ˆj  Bz kˆ)


But
So
iˆ  ˆj  kˆ; iˆ  kˆ   ˆj; ˆj  kˆ  iˆ
iˆ  iˆ  0; ˆj  ˆj  0; kˆ  kˆ  0
 
A  B  Axiˆ  By ˆj  Axiˆ  Bz kˆ  Ay ˆj  Bxiˆ  Ay ˆj  Bz kˆ
 Az kˆ  Bxiˆ  Az kˆ  By ˆj
iˆ
 
A  B  Ax
Bx
March 25, 2016
ˆj
Ay
By
kˆ
Az
Bz
Torque as a Cross Product
 
  r F


The torque is the cross product of a force
vector with the position vector to its point
of application
  rF sin q  r F  r F


The torque vector is perpendicular to the
plane formed by the position vector and
the force vector (e.g., imagine drawing
them tail-to-tail)
Right Hand Rule: curl fingers from r to F,
thumb points along torque.
Superposition:
 


 net   i   ri  Fi (vector sum)
all i
all i
Can have multiple forces applied at multiple
points.
 Direction of net is angular acceleration axis

March 25, 2016
Calculating Cross Products
 
Find: A  B
Where:

A  2iˆ  3 ˆj

B  iˆ  2 ˆj
 
Solution: A  B  (2iˆ  3 ˆj )  (iˆ  2 ˆj )
 2iˆ  (iˆ)  2iˆ  2 ˆj  3 ˆj  (iˆ)  3 ˆj  2 ˆj
 0  4iˆ  ˆj  3 ˆj  iˆ  0  4kˆ  3kˆ  7kˆ
i
j
k
Calculate torque given a force and its location


F  (2iˆ  3 ˆj ) N
r  (4iˆ  5 ˆj )m
 
Solution:   r  F  (4iˆ  5 ˆj )  (2iˆ  3 ˆj )
 4iˆ  2iˆ  4iˆ  3 ˆj  5 ˆj  2iˆ  5 ˆj  3 ˆj

iˆ ˆj kˆ
A B  4 5 0
2 3 0
 0  4iˆ  3 ˆj  5 ˆj  2iˆ  0  12kˆ  10kˆ  2kˆ (Nm)
March 25, 2016
Net torque example: multiple forces at a single point
3 forces applied at point r :
r  r cos q ˆi  0 ˆj  r sin q kˆ
F1  2 ˆi; F2  2 kˆ ; F3  2 ˆj; r  3; q  30o

F1
z
F2

F3

r
q
y
Find the net torque about the origin:
τ net  r  Fnet  r  ( F1  F2  F3 )
 (r ˆi  r kˆ )  (2ˆi  2ˆj  2kˆ )
x
z
x
set
rx  rsin( q)  3sin(30 o )  1.5
rz  rcos(q)  3cos(30o )  2.6
 2rx ˆi × ˆi  2rx ˆi × ˆj  2rx ˆi ×kˆ  2rz kˆ × ˆi  2rz kˆ × ˆj  2rzkˆ ×kˆ
τ net  0  2rxkˆ  2rx ()ˆj  2rz ˆj  2rz ()ˆi  0
oblique rotation axis
ˆ
ˆ
ˆ
 τ  3i  2.2 j  5.2k
through origin
i
net
Here all forces were applied at the same point.
For forces applied at different points, first calculate
the individual torques, then add them as vectors,
 


i.e., use:
net 
 i   ri  Fi
all i
all i
(vector sum)
j
March 25, 2016
k
Angular Momentum

Same basic techniques that were used in linear
motion can be applied to rotational motion.





F becomes 
m becomes I
a becomes 
v becomes ω
x becomes θ
Linear momentum defined as p  mv
 What if mass of center of object is not moving,
but it is rotating?
 Angular momentum
L  Iω

March 25, 2016
Angular Momentum I

Angular momentum of a rotating rigid object
L  Iω





L


L has the same direction as  *
L is positive when object rotates in CCW
L is negative when object rotates in CW
Angular momentum SI unit: kg-m2/s
Calculate L of a 10 kg disk when  = 320 rad/s, R = 9 cm = 0.09 m
L = I and I = MR2/2 for disk
L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s
*When rotation is about a principal axis
March 25, 2016
Angular Momentum II
Angular momentum of a particle
L  I  mr 2  mv r  mvr sin f  rp sin f
 Angular momentum of a particle

L  r ×p  m(r × v )




r is the particle’s instantaneous position vector
p is its instantaneous linear momentum
Only tangential momentum component contribute
Mentally place r and p tail to tail form a plane, L is
perpendicular to this plane
March 25, 2016
Angular Momentum of a Particle in
Uniform Circular Motion
Example: A particle moves in the xy plane in a circular path of
radius r. Find the magnitude and direction of its angular
momentum relative to an axis through O when its velocity is v.



The angular momentum vector
points out of the diagram
The magnitude is
L = rp sinq = mvr sin(90o) = mvr
A particle in uniform circular motion
has a constant angular momentum
about an axis through the center of
its path
O
March 25, 2016
Angular momentum III

Angular momentum of a system of particles
Lnet  L1  L2  ...  Ln   Li   ri  pi
all i


all i
angular momenta add as vectors
be careful of sign of each angular momentum
for this case:
Lnet  L1  L2  r1  p1  r2  p 2
Lnet  r1 p1  r2 p2
March 25, 2016
Example: calculating angular momentum for particles
Two objects are moving as shown in the figure .
What is their total angular momentum about point
O?
m2
Lnet  L1  L2  r1  p1  r2  p 2
Lnet  r1mv1 sin q1  r2 mv2 sin q 2
 r1mv1  r2 mv2
 2.8  3.1 3.6  1.5  6.5  2.2
 31.25  21.45  9.8 kg m 2 /s
m1
Direction of L is out of screen.
March 25, 2016
Angular Momentum for a Car

What would the angular momentum about point “P” be if
the car leaves the track at “A” and ends up at point “B”
with the same velocity ?
A) 5.0  102
B) 5.0 
A
106
B
C) 2.5  104
D) 2.5 
106
E) 5.0  103
P
L  r  p  p r  pr sin(q)
March 25, 2016
Recall: Linear Momentum and
Force



Linear motion: apply force to a mass
The force causes the linear momentum to change
The net force acting on a body is the time rate of
change of its linear momentum
Fnet
dv dp
 F  ma  m

dt dt
p  mv
IL Fnet t  p
 
t
March 25, 2016
Angular Momentum and Torque



Rotational motion: apply torque to a rigid body
The torque causes the angular momentum to change
The net torque acting on a body is the time rate of
change of its angular momentum
Fnet
dp
 F 
dt
τ net
dL
 τ 
dt
and L are to be measured about the same origin
The origin must not be accelerating (must be an
inertial frame)
 τ

March 25, 2016
Demonstration





 dp

 dL
Fnet  F 
 net   
dt
dt

Start from dL  d (r  p )  m d (r  v )
dt dt
dt
Expand using derivative chain rule



   
dL
d  
 dr   dv 
 m (r  v )  m  v  r    mv  v  r  a 
dt
dt
dt 
 dt

   
  
  

dL
 mv  v  r  a   mr  a  r  (ma )  r  Fnet   net
dt
March 25, 2016
What about SYSTEMS of Rigid
Bodies?

 dL i
Rotational 2 law for a single body : i 
dt
 • individual
Total angular momentum 
angular momenta Li
Lsys   L i • all about same
of a system of bodies:
origin


dLsys
• i = net torque on particle “i”

dLi


  i
• internal torque pairs are
dt
dt
included in sum
i
nd
BUT… internal torques in the sum cancel in Newton 3rd law
pairs. Only External Torques contribute to Lsys

dLsys
dt


  i ,ext   net
net external torque on the system
i
Nonisolated System: If a system interacts with its environment in the
sense that there is an external torque on the system, the net external
torque acting on a system is equal to the time rate of change of its
angular momentum.
March 25, 2016
Example: A Non-isolated System

A sphere mass m1 and a block
of mass m2 are connected by a
light cord that passes over a
pulley. The radius of the pulley
is R, and the mass of the thin
rim is M. The spokes of the
pulley have negligible mass.
The block slides on a
frictionless, horizontal surface.
Find an expression for the
linear acceleration of the two
objects.
a

a
 ext  m1 gR
March 25, 2016
Masses are connected by a light cord. Find the
linear acceleration a.
• Use angular momentum approach
• No friction between m2 and table
• Treat block, pulley and sphere as a non
isolated system rotating about pulley axis.
a
As sphere falls, pulley rotates, block slides
• Constraints:
Equal v's and a's for block and sphere
v  ωR for pulley α  d / dt
a  αR  dv / dt

a
I
• Ignore internal forces, consider external forces only
• Net external torque on system:  net  m1 gR about
• Angular momentum of system:
(not constant)
Lsys
center of wheel
 m1vR  m2vR  Iω  m1vR  m2vR  MR 2ω
dLsys
 m1aR  m2 aR  MR 2α  (m1R  m2 R  MR)a  τ net  m1 gR
dt
m1 g
same result followed from earlier
a 
method using 3 FBD’s & 2nd law
M  m1  m2
March 25, 2016
Isolated System

Isolated system: net external torque acting on
a system is ZERO


no external forces
net external force acting on a system is ZERO
 τ ext
Ltot  constant
dLtot

0
dt
or
Li  L f
March 25, 2016
Angular Momentum Conservation
Ltot  constant
or
Li  L f
Here i denotes initial state, f is the final state
 L is conserved separately for x, y, z direction
 For an isolated system consisting of particles,

Ltot   Ln  L1  L2  L3 

 constant
For an isolated system that is deformable
I ii  I f  f  constant
March 25, 2016
First Example



A puck of mass m = 0.5 kg is
attached to a taut cord passing
through a small hole in a
frictionless, horizontal surface. The
puck is initially orbiting with speed
vi = 2 m/s in a circle of radius ri =
0.2 m. The cord is then slowly
pulled from below, decreasing the
radius of the circle to r = 0.1 m.
What is the puck’s speed at the
smaller radius?
Find the tension in the cord at the
smaller radius.
March 25, 2016
Angular Momentum Conservation
m = 0.5 kg, vi = 2 m/s, ri = 0.2 m,
rf = 0.1 m, vf = ?
 Isolated system?
 Tension force on m exert zero
torque about hole, why?

Li  L f
L  r  p  r  (mv )
Li  mri vi sin 90  mri vi L f  mrf v f sin 90  mrf v f
ri
0.2
v f  vi 
2  4 m/s
rf
0.1
v 2f
42
T  mac  m  0.5
 80 N
rf
0.1
March 25, 2016
Isolated
System

τ net  0

L
about z - axis

 L  constant
I ω  I
i
initial
i
final
f
ωf
Moment of inertia
changes
March 25, 2016
Controlling spin () by changing I (moment of inertia)
In the air, net = 0
L is constant
L  I i i  I f  f
Change I by curling up or stretching out
- spin rate  must adjust
Moment of inertia changes
March 25, 2016
Example: A merry-go-round problem
A 40-kg child running at 4.0
m/s jumps tangentially onto a
stationary circular merry-goround platform whose radius is
2.0 m and whose moment of
inertia is 20 kg-m2. There is
no friction.
Find the angular velocity of
the platform after the child
has jumped on.
March 25, 2016
The Merry-Go-Round
The moment of inertia of the
system = 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 moment of inertia
decreases.
 The angular speed must
increase since the angular
momentum is constant.

March 25, 2016
Solution: A merry-go-round problem
Ltot   Iiωi 
I
f
ωf
Li  I ii  I0  mc vT r  mc vT r
L f  I f ω f  ( I  mc r 2 )ω f
I = 20 kg.m2
VT = 4.0 m/s
mc = 40 kg
r = 2.0 m
0 = 0
( I  mc r 2 )ω f  mc vT r
ωf 
mc vT r
40  4  2

 1.78 rad/s
2
2
I  mc r
10  40  2
March 25, 2016