MA4248 Weeks 6-7.
Topics Work and Energy, Single Particle Constraints,
Multiple Particle Constraints, The Principle of Virtual
Work and d’Alembert’s Principle
Work and Energy: the English words originated, via
Germanic and Greek branches, respectively, from the
Proto Indo-European word Werg about 7 k years ago
my homepage under courses/Ussc2001/Energy1.pdf

The work done on a particle
that
is
displaced
by


 
in a constant force field F equals F  
This work has units of energy.
1
CONSTANT FORCE FIELDS
Let us consider this situation in detail. Let A denote
affine space and choose a point q  A
then constructthe function U : A  R by


U(p)  F r where p  q r , p  A
Then
 U(p  )  U(p) F   hence
F  grad U and if F is the net force
 on a
particle whose trajectory is p( t )  q  r ( t )


then U( p( t ))  m r ( t )  r ( t ) / 2 is constant
since Newton’s second law implies that



d mr (t )  r (t ) / 2  F  r (t )
2
dt
and
CONSTANT FORCE FIELDS
d U ( p( t ))  lim U ( p( t  t ))  U ( p( t ))
dt
t
t 0

U ( p( t )  r ( t ) t ))  U ( p( t ))

lim
t 0
t

[grad U ( p( t ))][ r ( t ) t ]
 lim
t
t 0
 


 grad U(p( t ))  r ( t )  F  r ( t )
3
CONSERVATIVE FORCE FIELDS
The argument on page 20 in Week 1-3 Vufoils show
that this total energy is constant for any conservative
force field. Let us consider the following converse
There exists a force field F and a function U that are
functions on A (time independent) such that


U(p( t ))  mr ( t )  r ( t ) / 2 is constant for
every particle of mass m that moves with net force F.
Then





grad U (p( t ))  r ( t )  F(p( t ))  r ( t )  0
hence F  grad U is conservative.
4
WORK OVER A SMOOTH CURVE
Thomas,
p.
1062.
The
work
done
by
a
force
(field)

F
over a smooth curve parameterized by a smooth

vector valued function r on the interval [a,b] is

If
t b 
t b  


F  dr 
F  r ( t )dt
W
t a

F  grad U
W 
t a
is conservative then
t b
t a

Udt  U(p(b))  U(p(a ))
5
PATH INDEPENDENCE AND COMPONENT TEST

F
Thomas, p. 1072 A vector field
is conservative
if and only if W depends only on the endpoints
p(b) of the curve.

Thomas, p. 1074 F  M i  N j  P k
p (a )
and
is conservative if and only if
P  N ,
y z
M  P ,
z
x
N  M
x
y
6
SINGLE PARTICLE CONSTRAINT
In Tutorial 3, Prob. 5 you computed the trajectory of
a ring sliding down a straight rod by assuming either
1. that the total energy is conserved, or
2. that the force of constraint is orthogonal to the rod
Why do these assumptions yield the same trajectory?
Consider a particle having mass m that is constrained
to move along a curve
 parameterized by a function
c(s)  q  h (s), s  [a , b]
of a variable s, called a generalized coordinate. This
may be the case if the particle consists of a ring that
slides along a rigid wire. We will first assume that the
curve does not move so that it is independent of time 7
SINGLE PARTICLE CONSTRAINT
Therefore, the trajectory of the particle must equal

p( t )  q  h (s( t )), s( t )  [a , b]
where t denotes time and s(t) is a function of time
Newton’s second law implies that
 
2 
d
m 2 h (s( t ))  Fa  Fc
 dt
where F is the applied force that would be there
a
if
 the physical constraint (wire) was removed, and
F is the force of constraint defined by this equation
c
8
SINGLE PARTICLE CONSTRAINT


Define r ( t )  h (s( t ))

The work performed by F
c

r [t1, t 2 ]
over the curve
Wc (t1, t 2 ) 
is
t t 2 
 t  t 2  

Fc  d r 
mr  Fa  d r
t  t1
t  t1




9
SINGLE PARTICLE CONSTRAINT
Then
t t 2
t t
t t 2
1
  t  t 2  
mr  d r 
mr  r dt 
t t
1
d 1 mr  r  E ( t )  E ( t )
kin
2
kin
1
dt
2
t  t1

is the change of kinetic energy over the time interval
10
SINGLE PARTICLE CONSTRAINT

If the applied force is conservative F  grad U
a
and
t t 2
t t
1

t t 2



 Fa  d r 
grad U  r dt
 
t t 2
t t
t t
1

Udt  U( t 2 )  U( t1 )
1
Wc (t1, t 2 )
is the change intotal energy

It equals zero for all time intervals iff F  d r  0
c
11
hence
SINGLE PARTICLE CONSTRAINT
In this case, it is very convenient to use arc length
parameterization of the curve, then
E kin
  1 2
1
 mr  r  ms
2
2
and energy conservation implies that the trajectory is
determined, up to the initial position, by the first order
differential equation
s ( t ) 
2 E  U ( t ) 
m
12
SINGLE PARTICLE CONSTRAINT
We now consider the case where the particle is
constrained to move along a curve that is moving
with time as in Tutorial 5, Problems 4 and 5.
In this case the force of constraint may perform work
on the particle, yet it is reasonable to assume that at
each value of time the force of constraint is orthogonal
to the curve described at that time.
Note: the curve at a specific time does not describe the
actual trajectory of the particle, this very important fact
is illustrated in Fig. 2.04 on page 31 of the textbook.
13
SINGLE PARTICLE CONSTRAINT
Therefore, the trajectory
of the particle must equal

p( t )  q  h (s( t ), t ), s( t )  [a ( t ), b( t )]
since we now have a time varying family of curves.
Newton’s second law implies that
 
2 
d
m 2 h (s( t ))  Fa  Fc
dt
and the orthogonality condition
 implies that
  
h (s( t ), t )
Fc  h  Fc ( t ) 
s  0
s
There are 4 unknowns and 4 equations
14
SINGLE PARTICLE CONSTRAINT
We now consider a particle that is constrained to move
along a (possibly moving) surface. Then the trajectory
is determined by the principle that constraint force at
time t is orthogonal to the constraint surface at time t
Parameterize the surface
 at time t is by a (possibly time
varying) function h of generalized coordinates
q1 , q 2 so that at time t

p( t )  q  h (q1 ( t ), q 2 ( t ), t )


2 
d
Newton implies m
h
(
s
(
t
))

F

F
a
c
2
dt
15
SINGLE PARTICLE CONSTRAINT
Note that we now have 5 unknowns, the 2 generalized
coordinates and 3 components of the constraint force.
Newton gives us three equations. We need 2 more.
They are provided by the orthogonality principle:
 
0  Fc  r 


 h (q1 ,q 2 , t )
 h (q1 ,q 2 , t )
Fc 
q1  Fc 
q 2
q1
q 2

 h (q1 , q 2 , t )
 Fc 
 0, i  1, 2
q i
16
SINGLE PARTICLE CONSTRAINT
If the applied force is conservative and if the surface is
independent of time then energy is conserved. Energy
conservation is not sufficient to determine the motion
since it provides only 1 additional equation.
If the surface is moving then the forces of constraint
may (and usually do) perform work since

   
W  Fc  d r  Fc    r 


 h (q1 ,q 2 , t )
 Fc 
dt
t

h (q1, q 2 , t )
t

dt 



since W  Fc  r  0
17
SINGLE PARTICLE CONSTRAINT
In the previous discussion of a single particle, we used
generalized coordinates. However, we could have used
rectangular coordinates. If we did then we would have
6 unknown variables – 3 coordinates x, y, z for the
position of the particle and 3 coordinates of the force
of constraint. These 6 variables are determined (by the
solution of differential equations) by the constraint
equations (1 for a surface and 2 for a curve), Newton’s
second law (3 equations), and the principle that the
force of constraint is orthogonal to the constraint set
(2 for a surface, 1 for a curve)
18
MULTIPLE PARTICLE CONSTRAINTS
For a system with N particles, Newton’s 2nd law gives
  net  app  con
mi ri  Fi  Fi  Fi , i  1,...,N
3N equations in 6N variables. We need 3N more!
M=3N-f holonomic constraints, given by equations
 

G  ( r1 , r2 ,..., rN ; t )  0,   1,..., M
give a total of 6N-f equations, we need f more!
These f equations will be provided by the Principle
Of Virtual Work. For 1 particle, this principle says
that the force of constraint is orthogonal to the surface
(M=1) or curve (M=2) that the particle moves on.
19
MULTIPLE PARTICLE CONSTRAINTS
For multiple particle constraints virtual displacements


are any displacements r1 ,..., rN that satisfy
i  N 


 r G    ri  0,   1,..., M

i 1  i


The principle of virtual work says that the total work done by
the forces of constraint over these displacements equals zero
W 
i  N  con 
Fi  ri  0
i 1

20
GENERALIZED COORDINATES
The set of virtual displacements form a vector space
having dimension f = 3N-M, these are the number of
degrees of freedom of the system
q1 ,..., q f
i  1,..., N
If we introduce generalized coordinates
 
ri  ri (q1 ,..., q f ),
then
and we can find a basis for this vector space

 ri 


r
i
f
q  , i
  1 q 

 1,..., N
21
GENERALIZED COORDINATES
Then the principle of work can be expressed by
W 
f
 1


r i
i  N  con
Fi 
q   0
i 1
q 

q  ,   1,..., f iff

i  N  con  r i
Fi 
 0,   1,..., f
i 1
q 
This holds for all choices of

These are the f additional equations that we require.
22
D’ALEMBERT’S PRINCIPLE
Then the principle of work can be expressed by
 W 

f
 1




i  N  app

Fi  mi r i   r i 
i 1


r
i
i  N  app

Fi  mi ri 
q   0
i 1
q 



The work done by the applied forces, plus the work done by


the inertial forces  m r , in a virtual displacement is zero
i i
23
EXAMPLES
Two particles connected
by a light

 rigid rod (pp.29-30)
F1
The constraint
 
| r1  r2 |  a
F2
implies that
   
(r1  r2 )  ( r1  r2 )  0
 
The forces of constraint are proportional to r1  r2


and Newton’s second law implies that F1  F2 hence
  




W  F1  r1  F2  r2  F1  (r1  r2 )  0
24
EXAMPLES
s
Inclined plane p. 34
generalized coordinate
S is the distance down
the inclined plane
g
m

If the block undergoes a virtual displacement s
down the plane then the applied force, gravity, does work
mg sin  s and the inertial force (oriented up the plane)
does work  mss which yields the well-known result
s  g sin 
25