Physics 111: Mechanics
Lecture 11
Dale Gary
NJIT Physics Department
Angular Momentum




Vectors
Cross Product
Torque using vectors
Angular Momentum
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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 ?
April 12, 2020
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
April 12, 2020
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
 
A  B  Ax
ˆj
kˆ
Ay
Az
Bx
By
Bz
iˆ
April 12, 2020
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

April 12, 2020
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
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

k
iˆ
ˆj kˆ
A B  4 5 0
2 3 0
 0  4iˆ  3 ˆj  5 ˆj  2iˆ  0  12kˆ  10kˆ  2kˆ (Nm)
April 12, 2020
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  2rz kˆ ×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
April 12, 2020
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

April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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.
April 12, 2020
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)
April 12, 2020
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
dv dp
Fnet  F  ma  m

dt dt
p  mv
IL Fnet t  p
 
t
April 12, 2020
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

April 12, 2020
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
April 12, 2020
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.
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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
April 12, 2020
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.
April 12, 2020
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
April 12, 2020
Isolated
System

τ net  0

L
about z - axis

 L  constant
I ω  I
i
initial
i
final
f
ωf
Moment of inertia
changes
April 12, 2020
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
April 12, 2020
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.
April 12, 2020
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.
April 12, 2020
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
April 12, 2020