Post on 28-Mar-2018
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CHAPTER 4
DYNAMICS OF A SYSTEM OF PARTICLES
• We consider a system consisting of n particles
• One can treat individual particles, as before; i.e.,one can draw FBD for each particle, define a coordinate system and obtain an expression of the absolute acceleration for the particle. One can then use Newton’s second law and proceed to get n second-order coupled ODEs.
• Focus here is on overall motion of the system-also a precursor to rigid body dynamics.
2
4.1 Equations of Motion:
Consider a system with:
• n particles
• masses -
• positions -
There are two types
of forces acting:
• External forces -
• Internal forces -
r i
Fi ;
m i
f ij
Z
X
Y
O
r1
F1
rC
ri
F2
Fi
mi
m2m1
mC
f12 f21
f2i
fi2
i
3
- force on the particle due to its
interaction with the particle
• Newton’s 3rd law
Also
• Newton’s 2nd law for particle:
f ij i th
jth
i th
internal forcesareequaland opposi( )
0 when , . .
te
0
ij ji
ij ii
f f
f i j i e f
1
, 1, 2,3,. . . .,n
i i i ij
j
m r F f i n
4
Now, for 3-dimensional motions, the position
of each particle (in Cartesian coordinates) is:
Thus, each equation in Newton’s second law
has 3 scalar second-order ordinary diff.
equations. 3n scalar second-order o.d.e.’s
for the system
In order to solve for the motion, one needs to
know:
• external forces on each of the particles
• nature of internal forces
F i
f ij
, 1,2,3, . . . ,i i i ir x i y j z k i n
5
e.g., Newton’s law of gravitation:
We also need:
• initial conditions:
The general solutions to these nonlinear
ODEs are unknown; they are difficult to
solve except for in some very simple cases
and small n.
2
3
( )
, ( ) /
j ii j
ij
j ij i
ij i j i j j i
r rm mf G
r rr r
or f Gm m r r r r
(0), (0), 1,2, . . . . ,i ir r i n
6
Suppose we would like to get overall motion of
the system, not those of individual particles.
Adding the n equations:
Now, (net interaction force is zero)
• - total mass
• - defines center of mass;
note that it is a function of
time since the particles move.
1 1 1 1
n n n n
i i i ij
i i i j
m r F f
1 1
0n n
ij
i j
f
1
n
i
i
m m
1
( ) ( )n
C i i
i
mr t m r t
7
• - total external force
• Equation of motion for
the center of mass
Internal forces do not affect the motion of
the center of mass.
1 1
Thus,addition of Eqns.n n
i i i C
i i
F m r mr
1
Letn
i
i
F F
1
n
i C
i
F F mr
8
4.2 Work and Kinetic Energy
• The motion of individual particle is defined by
• The motion of center of mass C is defined by
where the total mass is
1
, 1,2,3,. . . .,n
i i i ij
j
m r F f i n
1
n
i C
i
F F mr
1
n
i
i
m m
9
Consider a motion of
the system. The initial
state is A, and the final
state is B. Let AC and BC
denote the positions of
the CM.
• Now, for the CM
• work-energy statement for the CM
2( / 2)C C
C
C
C C
C
B BB
C C C C AA A
F mr
F dr mr dr mv
Y
Z
X
OrC
ri
mi
m
Ai
Bi
AC
BC
10
Note that is only the work done by
external forces, and it is related to the change
in translational kinetic energy associated with
the CM
• Let work done on the particle by all
the forces acting on it in moving from Wi i th
A to Bi i
C
C
B
C
A
F dr
1
( )i
i
B n
i i ij i
jA
W F f dr
11
Now:
where - position of particle relative to
the CM of the system
• Total work done=sum of the work done on all
particles:
i thi C ir r
i
1
1 1
1 1 1 1 1
( ) ( )
, ( ) ( )
i
i
i C
i C
n
i
i
Bn n
i ij C i
i jA
B Bn n n n n
i ij C i ij C
i j i i jA A
W W
W F f dr d
Now F f dr F f dr
12
work done by summation of the
total ext. forces work done on all the
through the displ. particles through their
of the CM displacements relative
to the CM
• For each particle, the work done is:
1 1
( )C i
C i
B Bn n
C i ij i
i jA A
W F dr F f d
1 1( ) ( )
2 2
i i
i i
B B
i i i i i C i C i
A A
W m r r m r r
13
- the sum of increase/change in
KE of the system.
T TB A
1 1
1
1 1
1
/ 2
/ 2
, 0 0
/ 2 / 2
i
C
C
i
i
i
i
C
C
i
Bn n
B
i C C C i iAi i A
Bn
i i i
i A
n n
i i i i
i i
Bn
B
C C i i iAi A
W W mr r r m
m
Now m m
W mr r m
14
work-energy principle for
the system of particles
T = K.E. at any instant
Recalling the work-energy principle for the CM:
Work done by all forces (external as well as
internal) in relative motion KE for relative
motion
A B B AW T T
2
1
/ 2 / 2n
C i i i
i
mv m
1 1 1
( ) / 2
ii
i i
BBn n n
i ij i i i i
i j iA A
F f d m
15
Important: In general, internal forces do
work in any motion of the system. Sometimes,
net work (that on the whole system) may be
zero even though there is work done on
individual particles.
Ex: Consider the force in a spring
connecting two moving bodies - there is
net work done by the spring force -
evaluated by potential function .
f ij
sp
16
Ex 1: Consider two particles connected by
a massless rigid (inextensible) rod, and
acted upon by a force F.
FBDs for individual
particles are:
F
1
2r1
r2
rC
C
O
m1
m2
m1
m2
F
Cf12
f21Note: f12=- f21
17
• Work done in relative motion by internal
forces:
• constraint
• Differentiate:
Now:
12 1 21 2 12 1 2( )dW f d f d f d d
22
12 2 1 2 1( ) ( )r l
2
12 2 1 2 1( ) ( ) ( ) 0d r d d
12 12 1 2 1 2
12 1 2
( ) /
Thus, ( ) 0
f f
f d d
18
Ex 2:
Consider the system
shown here. A slider
moves on a rough
guide, and a pendulum
is attached to it at A.
• connected by a massless rigid link.
• Coulomb friction between and the
horizontal guide. Force P acts on the block A.
m m1 2,
m1
m
1
m
2
x
l
xyA
P
rough guide
19
The FBDs are:
The positions of the two particles can now
be defined: 1
2
( ) ( )
( ) { ( ) sin } cos
r t x t i
r t x t l i l j
m1g
PTf
N
yx
T
m2g
m2
20
The equations of motion for the individual particles are:
• Try to write the equation of motion for the
CM of the system.
1 1 1
2
2 2
2
2
: ( sin ) ( cos )
where (sgn( ))
: {( cos sin )
( sin cos ) } sin
( cos )
m m x i P f T i N m g T j
f N x
m m x l l i
l l j T i
T m g j
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4.3 Conservation of Mechanical Energy
• Suppose that the External forces are conservative, that is, are conservative.
for the CM of the system
Total energy conserved for motion of the CM
• Suppose that Internal forces also conservative:
E = T + V
Total energy conserved for the whole system.
1
n
i
i
F F
C C CE T V
22
Ex. 3 (4.2): Consider the system shown.
• connected by a massless spring.
• A constant force F applied to at t = 0.
• No friction between the floor and the blocks
Find:
IC (t = 0), spring
unstretched
1 2andm m
m1
1 1 2( ) ; when massesareequal :x t m m m
1 2 1 2 0;x x x x
m1 m2
k
xC
x1
F Cx2
23
Motion of the CM:
Motion of the block m1:
FBD:
1 2
1 2
2
( ) / ( ) / 2
Newton'sSecond law :
( 2 )
/ 2 ; Init.Conds. are : (0) 0
( / 2 ) ; ( / 2 ) / 2
C i i C
x C
C C C
C C
r m r m x x x
F F mx m m m m
x F m x x
x F m t x F m t
F
xk(x2- x1)m1
24
Newton’s law for block m1:
2 1 1
2 1
1 1
1 1
1 1
1
( ) ; ICs.: (0) 0
Also,note that 2
2 ( ) (1)
Also: / 2 (2)
(1) (
Har
2)
m
( ) 2 ( ) / 2
ICs: [ (0) (0)] [ (0) (0)] 0
: ( ) {1 cos 2 / }/Sol
o
n 4
(
x C C
C
C
C
C C
C C
C
F F k x x mx x x
x x x
F k x x mx
mx F
m x x k x x F
x x x x
x x F k m k
nicoscillation)
25
Aside (steps involved in the solution):
1Theeqn.is : 2 / 2 where ( )
Thesolution is ( ) ( ) ( )
: 2 / 2 / 4
: ( ) cos sin , where 2 /
(0) 0 / 4 0 / 4
(0) 0 0 0
: ( ) {Sol
Harmonicoscill
1 s }
(
n co / 4
C
h p
p p p
h h n n n
n
n
my ky F y x x
y t y t y t
y ky F y F k
y y t A t B t k m
y A F k A F k
y B B
y t F t k
ation)
26
Or
Energy considerations: (verification)
Recall that
2
1( ) ( / 4 ) {1 cos 2 / }/ 4x t F m t F k mt k
2 2 2
2
CWork done on CM (W )=changein K.
( / 2 )
K.E.of CM (2 ) / 2 ( / 4 )
Work doneon CM ( / 4 )
(for a constant forc
E.of C
)
M
e
C C
C C
C C
x v F m t
T m v F m t
W Fx F F m t
27
Now, consider for the whole system:
2 2 2
1 2
2 2 2 2
2 2
1 2 1
2 2
2 2 2
[(2 ) {( ) ( ) }] / 2
or ( / 4 ) ( / 8 )sin ( 2 / )
( ) / 2 2 ( )
(Work done byinternal forces)
or ( / 8 ){1 cos( 2 / )}
( /
TotalKE:
PotentialEnerg
4 ) ( / 4 )
:
co
y
{1
C C C
C
T m x m x x x x
T F m t F k k mt
V k x x k x x
V F k k mt
T V E F m t F k
s( 2 / )}k mt
28
W = work done by the external force2 2 2
1 ( / 4 ) ( / 4 ){1 cos( 2 / )}
(0 0)
Work done byall forces (externaland internal)
(final totalenergy) (initial totalenergy)
Fx F m t F k k mt
W T V
29
4.4 Linear Impulse and Momentum
Let, - lin. impulse of external forces
Considering Newton’s laws for motion of CM:
Let - total linear momentum of the
system at a given instantThen
2
1
ˆ ( )
t
t
F F d
2 2
1 1
2 1ˆ ( ) ( ) ( )
t t
C C C
t t
F F d mr d m v v
1
( ) ( )n
i i C
i
p t m v t mv
2 1 2 1ˆ ( ) ( ) ( )C CF m v v p t p t
30
4.5 Angular Momentum:
The key point to consider here is the point about
which the moment can be taken.
• Moment about a
fixed reference point:
Z
X
Y
O
r1
F1
rC
ri
F2
Fi
mi
m2m1
mC
f12 f21
f2i
fi2
i
(angular momentum
of theith particleabout
point O)
iO i i iH r m r
31
1 1
1 1 1
1
Rateof chan
Totalangular mome
geof angular momen :
Now,using Newton'second law for a part
ntum of the syst
icl
t
em:
e :
um
n n
O iO i i i
i i
n n n
O i i i i i i i i i
i i i
n
i i i ij O
j
H H r m r
H r m r r m r r m r
m r F f H r
1 1
1
( )
or
n n
i i ij
i j
n
O i i O
i
F f
H r F M
32
• Reference point as the center of mass:
Let
1
1 1
1
1
( ) ( )
( )
n
O i C i C i
i
n n
C C C i i i i C
i i
n
i i i
i
n
O C C i i i
i
H m r r
mr r r m m r
m
H mr r m
i C ir r
33
(Ang. momentum with respect to the CM, as
viewed by a nonrotating observer moving
with the CM)
Now, differentiating:
1
Thus,
where
O C C C
n
C i i i
i
H r mr H
H m
1
1
n
O C C i i i
i
C n
i ii
r F
F
H r mr m
34
• (very convenient for rigid bodies)
1
1
for motion of CMNow ( )
about fixed pointO
ab
Reviewing : ( )
out C, the( )CM
n
CO C i i C C
i
C C C
n
CC i i i
i
OO
OO
CM
M r F F H r mr
r F r mr
M H m
M H
M H
35
• About an arbitrary reference point P:
Let P be an arbitrary
point (could be moving).
Let
Then, one can show
that
- angular momentum about P
Z
X
Y
O
rP
rC
ri
Fi
mi
mC fi2
iC
P
and
i P i
C P C
r r
r r
P C C CH m H
36
And
Choosing an arbitrary point for moments
of forces results in an additional term in the
moment equation.
• If P is a fixed point
• If P is the center of mass
• If P is such that and are parallel
throughout the motion
P C P PM mr H
( 0)P P PM H r
( 0)P P CM H
Pr C
P PM H
37
• Computation of Kinetic energy using P as
a reference point:
The kinetic energy is: 1
22
1
22
1
/ 2
Now, ,
[ 2 ] / 2
If : [ ] / 2 (as before)
n
i i i
i
i P i i P i
n
P i i P C
i
n
C i i
i
T m r r
r r r r
T m r m r m
P C T m r m
38
Ex. 4 (4.7):Consider a particle traveling at a speed ‘v’ to
the right. It strikes a stationary dumbbell (two
particles connected by a massless rigid rod).
The masses are:
Assumption:
• Perfectly elastic impact in (e=1).
Find: motion of the particles just after impact.
m m m m1 2 3
m m1 2,
x
y
C
Ovm1
m2
m3
l/2
l/2
45º
39
FBDs:
Observe that during impact:
• Net force on the whole system = 0
linear momentum conserved for the system
• Resultant moment about O (a fixed point)= 0
angular momentum about O conserved for the system
x
y
C
O
m2
m3
l/2
l/2
45ºF̂m1 F̂
40
Set up of the problem:
Motion before impact:
Motion after impact: It is convenient to think
in terms of the motion
of the CM, and rotational
motion about CM. Use
the triad
to define the motion of
the CM and the particles.
1 2 3; 0v v i v v
( , , )t a be e e
x
y
C
O
m2
m3
l/2
l/2
45ºet
ea
41
Expressing velocities
in terms of
x
y
C
O
m2
m3
l/2
l/2
45ºet
ea
( , , )t a be e e
2
2
3 3
3
( / 2)
( / 2)
( / 2)
Similarly, ( / 2)
( / 2)
C a a t t
C CO
C a
C t
a a t t
C C C t
a a t t
v v e v e
v v k r
v k le
v le
v v e v l e
v v k r v l e
v v e v l e
42
linear momentum conserved for the system:
(a vector equation 2 scalar equations)
angular momentum conserved for the system:
1 2 3
1 2 2 (1)a a t t
mvi mv i mv mv
or v i v i v e v e
3 30 [ ( / 2) ]
( / 2) / 2 (2)
O a a a t t
t t
r v le v e v l e
l v l k v l
43
Note: The vectors can be expressed
in terms of the unit vectors i and j as:
In equations (1) and (2),
are unknowns but there are only 3 equations.
Thus one more relation is required.
• coefficient of restitution:
1 , , ,t av v v
andt ae e
cos 45 sin 45 ( ) / 2
cos 45 sin 45 ( ) / 2
t
a
e i j i j
e i j i j
vAx vBx
A B x
44
Aside: central impact: Consider two particles
A and B that collide with each other. The
geometry and definitions of terms are:
FBDs: No y-comp of force
vA1 vB1
A
B
vA2 vB2 line of impact
plane of impact
A Bx
F̂
F̂
45
1 1 2 2
Let and
be velocities; , before,and , after.
The i
the ration
s then defined as
of the relative velocity after impact
to the relative velocity b
coefficient of resti
efore i
:tution
m
A Ax Ay B Bx By
A B A B
v v i v j v v i v j
v v v v
2 2 2
1 1 1
, for
velocities along the line of impact:
( ) ( )
pact
( ) ( )
Bx Ax Bx Ax
Bx Ax Bx Ax
v v v ve
v v v v
46
• For the system at hand, elastic impact: e = 1.
Also,
1 1 2 1 2 1
2 1 2
1
1
1
, =0, 0, ,
( 2 ) / 2, 0. Thus
(( 2 ) / 2 )1
(0 )
/ 2 2 (3)
Solving (1), (2), and (3)
(2 2 / 7) , (2 2 / 7) , / 7
Ax Ay Ay Bx Ax
Bx a t By By
a t
a t
a t
v v v v v v v
v v v v v
v v ve
v
v v v v
v v v v v v
47
Ex. (Problem 3.19)
Consider a cylinder rotating
at a constant rate .
• A thin, flexible and
massless rope goes
around the drum.
• There is no gravity,
and the rope does not
slip relative to the drum ;
At
Find: Tension in the rope as a function of time.
t r0 0 0 0, ( ) , ( )
O
Pm
r
l
48
Setup:
Consider a triad
for the moving reference
frame , with coordinate
system located at
• Let be angular
velocity of the moving
reference frame or triad.
• The fixed reference
frame is with origin O.
e e et n b, ,b g
O .
Y
X
O
P
m
r
O’
R
l
l
49
- triad for moving coordinate system
Let be angular velocity of the moving frame.
e e et n b, ,b g
Y
X
O
P
r
l
O’
R
Schematic:
50
•Use the general formulation to express aP:
Let us now consider the various terms:
( ) ( ) 2 ( )P r ra R
( ) and
/
( / ) , ( / )
Position : ,
( / ) ( / )
Also ( / ) ( / )
b
b b
t
t t
b t n
n n
e l r
l r l r
l r e l r e
deR re R r r e
dt
R r l r e e r l r e
R r l r e r l r e
51
2
2
2
2
Thus ( ) ( / )
Now, ( ) ( )
Als
The
o =( / ) ( ) ( / )
( ) ( / )
2 ( ) 2 ( / )
[ ( / ) / 2 ( / )]
( / )
mp:I
n t
n r n r n
b n t
n
r t
P t
n
R l e r l r e
le le le
l r e le ll r e
l l r e
l l r e
a r l r ll r l l r e
l l r e
is acceleration relative to
52
Now, applying Newton’s Second Law:
Initial conditions:
Integration 2 2r t
( ) , ( )0 0 0 r
2
2 2 2
2 2 2 2
: ( / ) / 2 ( / ) 0
or 0
( ) ( )
n
t
F ma Te ma
e r l r l l r l l r
l l l r
d l l dt r d l l r dt