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PHY 6200
Theoretical Mechanics
Dynamics of a System of Particles
Prof. Claude A Pruneau
Notes compiled by L. Tarini
Physics and Astronomy Department
Wayne State University
Introduction
• So far dealt with point-like objects, or at
least assume we could…
• Let’s begin to consider objects with
finite size or extent.
• First discuss a system of “n” particles.
• Then discuss continuous limit for rigid
bodies.
CM
1
rCM
r1
2
r2
3
r3
o
Center of Mass
Discrete System of “n” masses:
1
R
M
r
 m r

Continuous System:
1 r
R   rdm
M
Total mass:
M   m

or
M   dm
Problem: Calculate the CM position of a solid (uniform
density) half-sphere.
y a y
2
2
dy
Solution:
a
z
Density:

M
2 3
a
3
By virtue of symmetry
x

x 


z 


1
M
1
M
a
 xdm  0
a
a
 zdm  0
a
a
Along “y” :
y
1
ydm

M 0
1

M

a
a
1
2
2
y

dv

y

(a

y
)dy
0

M0
a
  y 2 a 2
a
4
y
2
2

ydy(a

y
)



M 0
M  2 0 4
a 4
3a
y

4M
8
Answer:
3
rcm  (0, a, 0)
8


0

a
Definitions
F(e)
External force acting on particle  .
Internal force - particle  acting on 
r
f   f Sum of internal forces acting on  .
f
 
r (e) r
F  F  f ,
Total force on particle 
Newton’s 3rd Law
Newton’s 2nd Law
f
r
  f
r&& r (e) r
p  m r  F  f
r (e)
d2
r
(m r )  F   f
2
dt

Sum over :
r
r (e)
d2
r
m r   F   f
2   
dt 

 ,
1 42
43
 
r
r
r&
r (e)
&
MR   F   ( f  f )

   1 4 2 43
1 2r 3
0
Conclusion:
r& r
&
MR  F
F
Conclusion (1)
The CM moves as if it were a single particle, of mass
equal to the total mass of the system, acted on by the
total external force and independent of the nature of
the internal forces provided:
r
f   f
r
r&
d
r& d
P   m r   m r  (MR)  MR
dt 
dt

r&& r
P  MR  F
Conclusions (2)
Linear momentum of the system is the same as if a single
particle of mass M were located at the position of the
center of mass and moving in the manner the CM moves.
Conclusions (3)
The total linear momentum for a system free of external
forces is constant and equal to the linear momentum of the
CM.
A
Example:
A
B
A chain of uniform mass density , length b, and mass M
hangs from the ceiling. End b is released at t=0.
x
B
CM
Solution:
Density:
CM

M
b
Equation of motion:
t=0
p  Mg  T
 b  x
x&
Right side in motion: p   

 2 
p

  x&2  &
&  x) 
x(b
2
Free fall
x  gt  2gx
&
x& g
t2
xg
2
t>0
p

2
(gb  3gx)  Mg  T
T 
Mg  3x 
 1


2 b
b2
Uo   g
4
1
U   g(b 2  2bx  x 2 )
4

K

4
K  U  U0
1
1
2
2
(b  x) x&   g(b  2bx  x )   gb 2
4
4
4
2
(b  x) x&2
Solve for
x
2
g(2bx  x )
x 
bx
2
g(2bx  x )
&
x& g 
2
2(b  x)
2
2
Mg 1
T
(2b 2  2bx  3x 2 )
4b (b  x)
Energy
b2
U   g
4
1
U   g(b 2  2bx  x 2 )
4
K

(b  x) x&2
4
K  U  U0

1
1
2
2
(b  x) x&   g(b  2bx  x )   gb 2
4
4
4
2
2
g(2bx

x
)
2
x 
bx
g(2bx  x 2 )
&
x& g 
2(b  x)2
Mg 1
T
(2b 2  2bx  3x 2 )
4b (b  x)
Angular momentum of a system of n particles
r r
r  R  r
r
r
L  r  p
r
r
r
r
r
L   L   (r  p )   r  m r&




r
r
r& r&
  (r  R)  m (r  R)

r& r r
r r&
r r& r
  m {r  r  r  R  R  r&
  R  R}
1 44 2 4 43

r
 m r '  0

d

r  r&
r 
  m r   R  R  dt   m r 

1 4 4 4 4 44 2 4 4 4 4
4 43
0
r r& r r&
L   m {r  r  R  R}

r r&
r r
thus L  1
M
4R2 43R   r  p

r r
r r
L  R  P   r  p

The total angular momentum about an origin is the sum
of the angular momentum of the CM about that origin and
the angular momentum of the system about the position
of the CM.
d r r
r& r
r r&
L  (r  p)  r  p  r  p
123
dt
0
r
r (e)
r
L  r  (F   f )

r&
v r (e)
L   L   (r  F ) 


r
r
 (r  f )
 ,
 
1 r4 r2 4r 3r
But
Thus
 ( r  f  r  f )
r r
r  r  r
r
r
f   f
r
r
 (r  f ) 
 ,
 
 
r
r r
 (r  r )  f
 
r
r
  (r  f )
 
Assume f is along
Finally
r
 r
r
r
 r  f  0
r (e)
r
L   (r  F )

  N(e)  N (e)

If the net resultant external torques about a given axis
vanish, then the total angular momentum of the system
about that axis remains constant in time.
Energy of the System
Consider the work done on a system in moving all
particles from a certain configuration (1) to another (2)
when all positions are specified.
2
r
1
r
W12   F  dr    d( m v2 )  T2  T1

1
2
1
T   T   m v2

 2
r& r&
r  r  R
r&
&   R)
&
r  r  v2  (r&
  R)  (r&
&  R&2
 r&
  r&
  2(r&  R)

1
&  R&2
T   m r&
  r&
  2(r&  R)
 2

r&
1
1 r&2
  m r&
  MR
  r&
  2 Rg m r&
2
 2

=0
1
1
1
2
2
T   m v   m v  MV 2
2
 2
 2
The total energy of the system is equal to the sum of the
kinetic energy of a particle of mass M moving with the
velocity of the center of mass and the kinetic energy of
motion of the individual particles relative to the center of
mass.
2
W12   

If
r (e) r
f  dr 
1
2

  
, 
r
r
f  dr
1
F(e) and f are conservative forces,
r
F  U
r
r
f  U
(e)
2
  F
(e)

r
r
r
 dr     (U )  dr   U
2

1
2

  
, 
r
f dr 
1

1
2
1
2
r
r
r
r
  ( f  dr  f  dr )
  1
2


  1
2


  1
r
r
r
f  dr  dr


r
r
f  dr
r
r
r
with dr  dr  dr
U  U(x , x )
dU  
i
U
x ,i
dx ,i  
i
U
x ,i
dx ,i
 U  dr  U  dr
U   f
U  U
U  U   f  f
dU
r
r
  f  dr
W12   U
2
1

  U
 
U  U   U

W12  U
 
2
1
 U1  U2
2
1
T2  T1  U1  U2
T1  U1  T2  U 2
E1  E2
The total energy for a conservative system is constant.
The term
 U
 
corresponds to internal potential energy.
Example: Projectile M explodes while in flight into 3 fragments.
Assume:
m1  M 2 ; m2  M 6 ; m3  M 3;3  0
Ereleased  5Tinitial
3) Find v1 and v2
m1 
M
2
r
v1   1v
M
m2 
6
r
v2   2 v
M
m3 
3
v3  0
M 
E
M
M
 1   2
2
6
1
1M 2 2 1M 2 2
M 2 
1 
 2
2
2 2
2 6
1
1
2

2
6
 6  3 1   2   2  3 1  6
1
1
1M 2 2 1M
5( M  2 )  M  2 
1 
(3 1  6)2  2
2
2
2 2
2 6
 12  3 1  0
 1  0;
1  3
 2  6;  2  3
r
r
 v1  3v
r
v2  3v
v3  0
Example: A rope of uniform linear density and mass
wrapped one complete turn around a hollow cylinder of
mass M and radius R. Cylinder rotates freely about its axis.
Rope’s end is at . When P is at , system is slightly
displaced. Find angular velocity.
Solution:

M
M

L 2 R
 x
 x  Rsin  
 r
 ( dx)g  x  Rsin( x R) 
R
W
 g  x  Rsin( x R dx
0
W  gR (
2
2
2
 cos  1)
T
1
1
m(R&)2  M (R&)2
2
2
W  T,

M
2 R
 1
mgR   2

cos


1
 (m  M )R 2&2


2  2
 2
2
mg(

 2 cos  2)
2
 
2 R(m  M )