This is a physical ppt

1. Physics 111: Mechanics
Lecture 11
Dale Gary
NJIT Physics Department

2. May 15, 2023
Angular Momentum
 Vectors
 Cross Product
 Torque using vectors
 Angular Momentum

3. May 15, 2023
Vector Basics
 We will be using vectors a lot in this
course.
 Remember that vectors have both
magnitude and direction e.g.
 You should know how to find the
components of a vector from its
magnitude and direction
 You should know how to find a vector’s
magnitude and direction from its
components
a
m
F
a
m
F
m





a
F
Ways of writing vector notation
q
y
x
a

q
q
sin
cos
a
a
a
a
y
x


x
y
y
x
a
a
a
a
a
/
tan 1
2
2




q
q
,
a
ax
ay

4. May 15, 2023
q
y
x
a

z
f
q
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.
y
x
a

z
f

5. May 15, 2023
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
 Conversion to x, y, z components
 Conversion from x, y, z components
 Unit vector notation:
q
y
x
a

z
f
f
q,
,
a
q
f
q
f
q
cos
sin
sin
cos
sin
a
a
a
a
a
a
z
y
x



x
y
z
z
y
x
a
a
a
a
a
a
a
a
/
tan
/
cos
1
1
2
2
2







f
q
k
a
j
a
i
a
a z
y
x
ˆ
ˆ
ˆ 



sin
a q

6. May 15, 2023
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.
y
x
z

7. May 15, 2023
Cross Product
 The cross product of two vectors says
something about how perpendicular they are.
 Magnitude:
 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:
q
sin
AB
B
A
C 


 

B
A
C





A

B

q
q
sin
A

q
sin
B

0
ˆ
ˆ
;
0
ˆ
ˆ
;
0
ˆ
ˆ
ˆ
ˆ
ˆ
;
ˆ
ˆ
ˆ
;
ˆ
ˆ
ˆ













k
k
j
j
i
i
i
k
j
j
k
i
k
j
i
y
x
z
i
j
k
i
k
j

8. May 15, 2023
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
   

9. May 15, 2023
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
 The distributive law
 The derivative of cross product
obeys the chain rule
 Calculate cross product
  dt
B
d
A
B
dt
A
d
B
A
dt
d











C
A
B
A
C
B
A












 )
(
A B B A
   
k
B
A
B
A
j
B
A
B
A
i
B
A
B
A
B
A x
y
y
x
z
x
x
z
y
z
z
y
ˆ
)
(
ˆ
)
(
ˆ
)
( 









10. May 15, 2023
Derivation
 How do we show that ?
 Start with
 Then
 But
 So
k
B
j
B
i
B
B
k
A
j
A
i
A
A
z
y
x
z
y
x
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ








)
ˆ
ˆ
ˆ
(
ˆ
)
ˆ
ˆ
ˆ
(
ˆ
)
ˆ
ˆ
ˆ
(
ˆ
)
ˆ
ˆ
ˆ
(
)
ˆ
ˆ
ˆ
(
k
B
j
B
i
B
k
A
k
B
j
B
i
B
j
A
k
B
j
B
i
B
i
A
k
B
j
B
i
B
k
A
j
A
i
A
B
A
z
y
x
z
z
y
x
y
z
y
x
x
z
y
x
z
y
x





















0
ˆ
ˆ
;
0
ˆ
ˆ
;
0
ˆ
ˆ
ˆ
ˆ
ˆ
;
ˆ
ˆ
ˆ
;
ˆ
ˆ
ˆ













k
k
j
j
i
i
i
k
j
j
k
i
k
j
i
j
B
k
A
i
B
k
A
k
B
j
A
i
B
j
A
k
B
i
A
j
B
i
A
B
A
y
z
x
z
z
y
x
y
z
x
y
x
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ















k
B
A
B
A
j
B
A
B
A
i
B
A
B
A
B
A x
y
y
x
z
x
x
z
y
z
z
y
ˆ
)
(
ˆ
)
(
ˆ
)
( 








z
y
x
z
y
x
B
B
B
A
A
A
k
j
i
B
A
ˆ
ˆ
ˆ





11. May 15, 2023
 The torque is the cross product of a force
vector with the position vector to its point
of application
 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.
Torque as a Cross Product
F
r







 

 F
r
F
r
rF sinq

sum)
(vector
r
i
i
i
i
net i
all
all
F






 


Superposition:
 Can have multiple forces applied at multiple
points.
 Direction of net is angular acceleration axis

12. May 15, 2023
Calculating Cross Products
m
j
i
r
N
j
i
F )
ˆ
5
ˆ
4
(
)
ˆ
3
ˆ
2
( 





B
A



Solution: i
k
j
j
i
B
j
i
A ˆ
2
ˆ
ˆ
3
ˆ
2 






k
k
k
i
j
j
i
j
j
i
j
j
i
i
i
j
i
j
i
B
A
ˆ
7
ˆ
3
ˆ
4
0
ˆ
ˆ
3
ˆ
ˆ
4
0
ˆ
2
ˆ
3
)
ˆ
(
ˆ
3
ˆ
2
ˆ
2
)
ˆ
(
ˆ
2
)
ˆ
2
ˆ
(
)
ˆ
3
ˆ
2
(



























Calculate torque given a force and its location
(Nm)
ˆ
2
ˆ
10
ˆ
12
0
ˆ
2
ˆ
5
ˆ
3
ˆ
4
0
ˆ
3
ˆ
5
ˆ
2
ˆ
5
ˆ
3
ˆ
4
ˆ
2
ˆ
4
)
ˆ
3
ˆ
2
(
)
ˆ
5
ˆ
4
(
k
k
k
i
j
j
i
j
j
i
j
j
i
i
i
j
i
j
i
F
r



























Solution:
Find: Where:
ˆ
ˆ ˆ
4 5 0
2 3 0
i j k
A B
 

13. May 15, 2023
i
k
j
Net torque example: multiple forces at a single point
x
y
z
q
r

1
F

3
F

2
F

3 forces applied at point r :
ˆ ˆ ˆ
cos 0 sin
r r
q q
  
r i j k
o
ˆ ˆ
ˆ
2 ; 2 ; 2 ; 3; 30
r q
    
1 2 3
F i F k F j
Find the net torque about the origin:
net net 1 2 3
( )
ˆ ˆ ˆ
ˆ ˆ
( ) (2 2 2 )
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ
2 2 2 2 2 2
x z
x x x z z z
r r
r r r r r r
     
    
     
τ r F r F F F
i k i j k
i×i i× j i×k k ×i k × j k ×k
.
)
3cos(30
)
rcos(
r
.
)
3sin(30
)
rsin(
r
o
z
o
x
set
6
2
5
1


q



q

net
ˆ ˆ ˆ
ˆ
0 2 2 ( ) 2 2 ( ) 0
x x z z
r r r r
       
τ k j j i
net
ˆ ˆ ˆ
3 2.2 5.2
    
τ i j k
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:
sum)
(vector
F
r i
i
all
i
i
all
i
net








 

oblique rotation axis
through origin

14. May 15, 2023
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
 What if mass of center of object is not moving,
but it is rotating?
 Angular momentum
m

p v
I

L ω

15. May 15, 2023
Angular Momentum I
 Angular momentum of a rotating rigid object
 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
I

L ω 

L


16. May 15, 2023
Angular Momentum II
 Angular momentum of a particle
 Angular momentum of a particle
 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
( )
m
 
L r×p r× v
f
f

 sin
sin
2
rp
mvr
r
mv
mr
I
L 



 

17. May 15, 2023
Angular Momentum of a Particle in
Uniform Circular Motion
 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
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.

18. May 15, 2023
Angular momentum III
 Angular momentum of a system of particles
 angular momenta add as vectors
 be careful of sign of each angular momentum
net 1 2 ... n i i i
all i all i
      
 
L L L L L r p
net 1 1 2 2
r r
 
  
L p p
for this case:
net 1 2 1 1 2 2
     
L L L r p r p

19. May 15, 2023
Example: calculating angular momentum for particles
Two objects are moving as shown in the figure .
What is their total angular momentum about point
O?
Direction of L is out of screen.
m2
m1
net 1 1 1 2 2 2
1 1 2 2
2
sin sin
2.8 3.1 3.6 1.5 6.5 2.2
31.25 21.45 9.8 kg m /s
L rmv r mv
rmv r mv
q q
 
 
     
  
net 1 2 1 1 2 2
     
L L L r p r p

20. May 15, 2023
 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 ?
Angular Momentum for a Car
A) 5.0  102
B) 5.0  106
C) 2.5  104
D) 2.5  106
E) 5.0  103 )
sin(
pr
r
p
p
r
L q


 

P
A
B

21. May 15, 2023
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
net
d d
m m
dt dt
    
v p
F F a
L
t


 

m

p v
net t
   
I F p

22. May 15, 2023
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
 and are to be measured about the same origin
 The origin must not be accelerating (must be an
inertial frame)
net
d
dt
  
p
F F net
d
dt
  
L
τ τ
τ L

23. May 15, 2023
Demonstration
 Start from
 Expand using derivative chain rule
)
(
)
( v
r
dt
d
m
p
r
dt
d
dt
L
d 








dt
L
d
net





 

dt
p
d
F
Fnet






 
a
r
v
v
m
dt
v
d
r
v
dt
r
d
m
v
r
dt
d
m
dt
L
d 

























 )
(
  net
net
F
r
a
m
r
a
r
m
a
r
v
v
m
dt
L
d























 )
(

24. May 15, 2023
What about SYSTEMS of Rigid Bodies?
• i = net torque on particle “i”
• internal torque pairs are
included in sum
i

 L
Lsys

 • individual angular momenta Li
• all about same origin

 


i
i
sys
dt
L
d
dt
L
d




i
BUT… internal torques in the sum cancel in Newton 3rd law
pairs. Only External Torques contribute to Lsys
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.
dt
L
d i
i
:
body
single
a
for
law
Rotational
nd




2
Total angular momentum
of a system of bodies:
net external torque on the system
net
, 





 
i
ext
i
sys
dt
L
d

25. May 15, 2023
a

a

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. gR
m
ext 1



26. May 15, 2023
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.
As sphere falls, pulley rotates, block slides
• Constraints:
for pulley
Equal 's and 's for block and sphere
/
v a
v ωR α d dt
a αR dv / dt

 
 
• Ignore internal forces, consider external forces only
• Net external torque on system:
• Angular momentum of system:
(not constant)
ω
MR
vR
m
vR
m
Iω
vR
m
vR
m
Lsys
2
2
1
2
1 





gR
m
τ
MR)a
R
m
R
(m
α
MR
aR
m
aR
m
dt
dL
net
sys
1
2
1
2
2
1 







1 about center of wheel
net m gR
 
2
1
1
m
m
M
g
m
a



 same result followed from earlier
method using 3 FBD’s & 2nd law
I
a

a


27. May 15, 2023
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
constant or
tot i f
 
L L L
0
tot
ext
d
dt
  
L
τ

28. May 15, 2023
Angular Momentum Conservation
 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,
 For an isolated system that is deformable
1 2 3 constant
tot n
      
L L L L L
constant

 f
f
i
i I
I 

constant or
tot i f
 
L L L

29. May 15, 2023
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.

30. May 15, 2023
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?
i f

L L
0.2
2 4m/s
0.1
i
f i
f
r
v v
r
  
( )
m
   
L r p r v
i
i
i
i
i v
mr
v
mr
L 
 
90
sin f
f
f
f
f v
mr
v
mr
L 
 
90
sin
N
80
1
.
0
4
5
.
0
2
2




f
f
c
r
v
m
ma
T

31. May 15, 2023
constant
0 

 L
τ axis
about z -
net



 

final
f
f
initial
i
i ω
I
ω
I
L

Moment of inertia
changes
Isolated
System

32. May 15, 2023
Controlling spin () by changing I (moment of inertia)
In the air, net = 0
L is constant
f
f
i
i I
I
L 
 

Change I by curling up or stretching out
- spin rate  must adjust
Moment of inertia changes

33. May 15, 2023
Example: A merry-go-round problem
A 40-kg child running at 4.0
m/s jumps tangentially onto a
stationary circular merry-go-
round 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.

34. May 15, 2023
 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.
The Merry-Go-Round

35. May 15, 2023
Solution: A merry-go-round problem
I = 20 kg.m2
VT = 4.0 m/s
mc = 40 kg
r = 2.0 m
0 = 0
r
v
m
r
v
m
I
I
L T
c
T
c
i
i
i 


 0


f
c
f
f
f ω
r
m
I
ω
I
L )
( 2



r
v
m
ω
r
m
I T
c
f
c 
 )
( 2
rad/s
78
.
1
2
40
10
2
4
40
2
2








r
m
I
r
v
m
ω
c
T
c
f
tot i i f f
I I
 
 
L ω ω