Angular Momentum
(of a particle)
The angular momentum of a particle, about the reference
point O, is defined as the vector product of the position,
relative to the reference point, and momentum of the
particle


l
 

p
l  r p

r
O
Torque
Torque , about the reference point O, due to a force F
exerted on a particle, is defined as the vector product of
the position relative to the reference point and force

F

r
O


  
  rF
Newton's second law V
(angular momentum of a particle)





dr   dp  dp  
d  
dl
p  r
 r
 r  Fnet   net
 r  p  
dt
dt
dt
dt dt
(In an inertial reference frame) the net torque,
exerted on a particle, is equal to the rate of change
of its angular momentum

dl 
  net
dt
Example. Kepler’s second law

L
dA

r

dr
The gravitational torque (about the sun) exerted
by the sun on the planets is a zero vector.
 
dA 1 r  dr
1 

L


r  mv  L  const .
dt
2 dt
2m
2m
Newton's second law VI
(angular momentum of a system)




dL d
dl




  li   i    net ,i   int, i   ext ,i   ext ,i   ext
dt i
dt
i
i
i dt
i
i
(In an inertial reference frame) the net external torque,
exerted on a system of particles, is equal to the rate of
change of its (total) angular momentum

dL  
ext
dt
Example. What is the final angular velocity?
initial

i
Initial total angular momentum (magnitude)
L a  2  a  mv a = 2a 2ma
Final total angular momentum
a
final
?
Lb  2  b  mv b = 2b2 mb
b
From conservation of angular momentum
(zero external torque):
b 
a2
b
2
a
Puzzle: Total kinetic energy
K tot
2

mb 2b2
ma 22a
a
2
 2
 2
 ma  2  12a  0
2
2
b

Who performed the work?
Rigid Body
A system in which the relative position of all particles is time
independent is called a rigid body.

vi
The motion can be considered as a
superposition of the translational
motion of a point and the rotational
motion around the point.

rAi
A

i
 
 
vi  v A    rAi
Angular Momentum
and Angular Velocity


 
l 

L
r’
In general, each component of the total
angular momentum depends on all the
components of the angular velocity.



 
 


2


r

m


r


m
r


r
 i i
L   ri  mi vi   i
i
i
i ri
i
i
i
L x     mi zi x i  ;
 i



L y     mi zi yi  ;
 i


Lz   mi ri2  zi   zi    mi ri2  zi2
i
i
    m r'
i
2
i i 


effect of symmetry
Only for object with appropriate
symmetry the direction of
angular momentum is consistent
with the direction of angular
velocity of the object


L x     m i z i x i   0
 i



L y     m i z i y i   0
 i


2
L z    mi r 'i 
i

I
2
 mi r 'i
i
is called the moment of inertia
(rotational inertia) of the body
about the axis of rotation.
Newton’s Law VII
(for rotational motion of a rigid body)


d dL 


 ext
I  I
dt
dt
For symmetrical rigid bodies, the angular acceleration
is proportional to the net external torque.
 
I    ext
Fixed and Instantaneous Axis of Rotation
(Newton’s second law VIII)







F

F
torque
I
d
dL
II   ext ,  ext , 
dt
dt
The angular acceleration, of an
object rotating about a fixed axis
or instantaneous, is proportional to
the component, along the axis of
rotation, of the net external torque.
Moment of Inertia
(rotational inertia)
A
system of particles:
IA 
2
m
r
'
 i i
i
r’
ri’
dm
m
i
continuous body
A
I A   r ' dm
2
body
Example. Moment of inertia of a uniform thin rod
about an end
L
Iy   x2
y
dx
0
3 L
M
M x
dx 

L
L 3
0
1
 ML2
3
x
L
about the center
L/2
Icm
3 L/2
M x
M
  x
dx  
L 3
L
L / 2
2
L / 2
1
ML2

12
Example. Moment of inertia of a uniform circle
r

dr
d
R 2
M
2
2
r

rddr 
I A   r dm   
2
R
circle
0 0
M R 3  2 
1
M R4
2

r
d

dr

MR
2








2
R 2 0  0 
R 2 4
Parallel - axis theorem
C
  2
I A   r dm   D  r ' dm 
A
2
body
body

D

r
 
  D dm   r ' dm  2  D  r ' dm 
2

r'
body
dm
2
body
 
 MD  IC  2D  0
body
2
If the moment of inertia of a rigid body about an axis through the center of
mass is IC, then the moment of inertia, about a parallel axis separated by
distance D from the axis that passes through the center of mass, is given by
IA = MD2 + Ic
Center of a force
lift
buoyancy
lift
weight
If a certain body exerts a force on
several particles of a given system,
the center of the force is defined
by position such that for any point
of reference


 
rcf   f i   ri  f i
i

i

Example. Center of gravity
 

 

   ri  mi g    miri   g 
i

i

 

 Mrcm  g  rcm  Mg

ri

Wi
The center of gravity in a uniform gravitational field is at the
center of mass.
Note: Not applicable to a nonuniform
gravitational field
Equilibrium of a rigid body
F1
O
A
F3
F2
A rigid object is in equilibrium, if and only if the following
conditions are satisfied:
(a) the net external force is a zero vector,
(b) the net external torque is a zero vector.
Rotational kinetic energy
The total kinetic energy of a system rotating about the
point of reference is called the rotational (kinetic) energy
K,o =  Ki,o
rotational energy and angular velocity


The rotational kinetic energy is
related to the magnitude of
angular velocity and the moment
of inertia of the body
K , o
K ,o   K o,i
i
1
 I, o2
2
1 2
1
1
1
2
2 2
2
  mi vi   mi r 'i     mi r 'i   I
2
2 i

i 2
i 2
Total Kinetic Energy of a Rigid Body
K tot
1
  2

1
2
  mi vi   mi v A    ri  
i 2
i 2
1
1
1
  2
 

2
  mi vA   mi   ri    mi  2 v A    ri  
i 2
i 2
i 2
 2 1

1
 
2 2
   mi v A    mi r 'i   v A      mi ri 
2 i
2 i




i
If the center of mass is at point A
K tot
1
21
2

 KMv

K
I

T cm 
, cm
, cm 0
2
2
work and power in rotational motion

dr

F
d

d


The differential work in a
rotational motion depends on
the torque about the point of
rotation


dW    d


    
dW  F  dr  F  d  r  
  
 
 d  r  F     d
The power delivered to a rigid
body depends on the applied
torque and the angular velocity
of the body
 
P  
Transformation of torque

A , i
A

Fi

rA

B, i

rB
B




  

A ,i  rA ,i  Fi   AB rB,i   Fi 



 

 
 AB Fi  rB,i  Fi  AB Fi  B,i

F
conclusion (total force)




A  AB Ftot  B
If the total force applied to a body is zero,
the torque of this force about any point
has the same value.
d
-F
torque transmission





F