Download 6.3 The momentum of a system of particles

Lecture 6
6.1
System of particles
6.2
Collision and the Newton’s second law for a system of particles
6.3
The momentum of a system of particles
6.4
Appendix – Centers of gravity of uniform bodies
6.1
System of particles
m1
(1) Two particles in 1-D
0
x1
d
m2
x2
x
M  m1  m2
Total mass:
m1 x1  m2 x 2 m1 x1  m2 x 2

m1  m2
M
Center of mass:
x CM 
If m1  m2  m
then x CM 
m( x1  x 2 ) x1  x 2

2m
2
(2) N particles in 1-D
N
x CM
m x  m2 x 2    mN x N
 1 1

m1  m2  m3    mN
m x
i
i
i 1
M
(3) N particles in 3-D
r1  ( x1 , y1 , z1 )
m1
r2  ( x 2 , y2 , z2 )

r1

1

 xCM  M

1

 y CM 
M

z  1
 CM M

m3

r3
N
m x
i 1
N
i
i
 mi y i
i 1
N
m z
i 1
i i

rCM 
1
M
N
m r
i i
i 1
O

r2
m2
mi

ri
M
Block
x
x CM 
yCM 
1
M
 x m
1
M
 y m
i
x CM 
1
M
 xdm
y CM 
1
M
 ydm
i
i
i
i
rCM 
In general,
When mi  0
i
1
r dm
M
Example
Suppose the masses in the following figure are separated by 0.500
0.500 m
m, and that m1 = 0.260 kg and m2 = 0.170 kg. What is the distance
m2
m1
from m1 to the center of mass of the system?
Answer:
Method 1:
We know that the center of mass lies on the line joining m1 and
between m1 and O. We can write
0.500 m
x1m1  (0.500  x1 )m2
That
is,
0.500 –x1
x1
m2. Now, let O be the center of mass and x1 be the distance
m2
m1
x1 (0.260)  (0.500  x1 )(0.170) ,
and
we
C.M.
obtain
x1  0.198 m.
This is the location of the center of mass.
Method 2:
We take an arbitrary point O as the origin which
has a distance a from m1. Hence, we can write
a  x1 
m1a  m2 (a  0.500)
m1  m2
,
which
gives
x1  0.198 . The position of the C.M. is 0.198 m
from m1.
x1
a
0.500 –x1
0.500 m
O
m2
m1
C.M.
Example***
Find the center of mass of a ring.
Answer***:
xCM 
1
M
 xdm ,
Linear
mass density 
M

2R
dm  ds  
 M   dm    ds    2R  M
x  R cos
y  R sin 
0    2 ,
1 2
R cos  Rd
M 0
R 2 2

 cos d
M 0
R2

 sin  02  0
M
ds  Rd
Check
 ds  R d  2R  R  2
xCM 
1
y dm
M
1 2

R sin   Rd
M 0
R 2 2

 sin d
M 0
R2

 cos 02  0
M
y
 cos d  sin 
ds
d
x
y CM 

rCM  0
 sin  d   cos
Center of mass
Example***
Find the center of mass of a disk.
Answer***:
A disk can be considered as a system of rings. Hence the center of mass of a disk is at the
center of the disk.
y
Or, one can use direct integration.
x CM
1

M
dxdy
M

Density 
R 2
 xdm ,
r
dm  dxdy
x
 x  r cos 

 y  r sin 
1
M
xCM 


R
0
M
1
M
0
2
R
r 2 dr  cos d  0
0

2
  r cos rdrd

y CM 
dxdy  rdrd
0
R
2
  r sin rdrd
0
M
R
0
0
2
r 2 dr  sin d  0
0
Example
You are given two rigid bodies, one block and one disk. Find the center of mass of this
system.
y
m2
m1
d
Origin
Answer:
x CM
a
a
m2 (d  R  )
m x  m2 x 2
2
 1 1

m1  m2
m1  m2
y CM 
m1 y1  m2 y 2
0
m1  m2
x
R
Example
y
Find the center of mass of a special disk. Note that a
part of it is empty.
2R
A
R
x
B
Answer:
The disk of radius 2R consists of two parts A and B, namely, the larger disk A, and the smaller
disk B (empty). The center of mass of the large disk is at xCA  2R, and yCA  0 . The center
of mass of the smaller disk is at xCB  R, and yCB  0 . Note that the mass of the smaller disk
y
is defined as negative.
xCM
m x  mB xCB m A (2R)  mB ( R) 2m A  mB
 A CA


R
m A  mB
m A  mB
m A  mB
M
Origin
2R
A
R
x
Define the mass density  = mass per unit area
xCM
 [2(4R 2 )  R 2 ]

R
 [4R 2  R 2 ]
 xCM 
B
7
R
3
That is, the center of mass is R/3, from the center of the disk, as shown in figure.
xCM
Collision and the Newton’s second law for a system of particles
6.2
Before collision
After collision
m1
m2
m1
m2
v1  0
v2
v1 '
v2 '
If you are at the center of mass frame, you will see that the velocity of center of mass
v CM 
drCM
m v  m2 v 2
d m r  m2 r2
 ( 11
) 1 1
dt
dt m1  m2
m1  m2
Does
not change! (Why?)
For a system of particles
MaCM   Fext
Internal forces

Fext : Vector sum of all external forces.

M: Total mass of the system

a CM : Acceleration of the center of mass
e.g. For a two-particles system: a CM

m1a1  m2 a 2 d 2 rCM


m1  m2
dt 2
There are three equations for a description of 3-D system, each for different direction
x, y, and z, e.g.
rCM 
m1 r1  m2 r2   mn rn
,
M
M  m1  m2    mn
 MrCM  m1 r1  m2 r2   mn rn
Taking the second derivative with respect to t,
Ma CM  m1a1  m2 a 2         mn a n
 F1  F2         Fn ,
where Fi is the total forces acting on particle i.
Fi  Fi ( ext )  Fi (int)
Fi(int)  The forces exerted by the particle within the system. e.g. action force or reaction
force.
As the action force and the reaction force come in pairs. We have
F
i(int)
0
i
 MaCM   Fext
When a body or a collection of particles is acted on by external forces, the center of mass
moves just as all the masses were concentrated at that point and it is acted on by a resultant
force which equals to the sum of the external forces on the system.
Example***
Given a spherical shell of mass m and its inner radius 2R. A ball of mass m and radius R is
released as shown in figure. The ball will roll back and forth. What will be the displacement
of the shell when the ball is at the bottom of the shell?
Answer***:
From Newton’s second law
F
ext
 2mg  Normal Reaction  2maCM
Along x-direction
 aext , x  0 ,
F
ext , x
a
m
2R
R
0
origin
m
dv
dt
Just before the
ball is released
x
 vCM  constant
Originally, v CM  0 , it remains
v
The ball is at the
bottom of the shell
dx
dt
Frictionless surface
 xCM  constant
x
Before release,
x CM
m  0  m(  R)
R


2m
2
When the ball is at the bottom of the shell,
d
xCM
x CM 
m( d )  m( d )
 d .
2m
Since there are no external forces acting on the system horizontally, hence xCM does not
move, that is xCM = R/2 = d. Or, we can say the shell moves R/2 to the left from its original
position.
6.3
The momentum of a system of particles

For a one particle system
p  mv

p: momentum of a particle, it is a vector.
Newton’s second law for a single particle
dv
 F  m dt


dp
dt
Rate of change of momentum
The momentum of a system of particles
N
N
i 1
i 1
P   pi   mi vi .
N
v CM 
Remember that
 mi vi
i 0
N
m
N
m v
i i

i 0
M
i
i 0
 P  Mv CM

Newton’s second law for a system of particles
F
 Ma CM  M
ext
If
F
ext
 0
P = constant,
dv CM dP

dt
dt
dP
 0 , or P = constant (law of conservation of momentum)
dt
 v CM  constant
Example
The blocks are pulled apart and then released
At any later time after release
from rest.
(a)
m1 v1
Discuss the center of mass of the system
v2
m2
after release.
(b)
What fraction the total kinetic energy of
x
the system will each block occupy at any
later time?
Answer:
Since, there are no external forces acting horizontally, the motions of the blocks are due to
internal force (i.e. the restoring force of the spring), the center of mass does not change.
F
ext , x
 0 ,  Px is conserved.
At t  0 , Px  0
At any later time: Px  m1v1  m2 v2  0

v1
m
 2 ,
v2
m1
f1

K1
K1  K2


1

1
f2 
2
1
2
2
m2 m
( )
m1 m
m1v12
2
1
1
2 m v  2 m2 v 2
1
2
2
1 1
1
m
1 1
m2
1
m v2
1  2 22
m1 v1
m2

m1  m2

K2
m1
1
1
1




2
K1
m2
K1  K2
m1  m2
m1v1
1
1

1
2
K2
m1
m2 v 2
f1  f 2 
m2
m1

1.
m1  m2 m1  m2
6.4
Appendix – Centers of gravity of uniform bodies
Uniform body
Solid tetrahedron, pyramid, cone
Hollow and without base
tetrahedron, pyramid, cone
Arc subtending an angle 2 at center
Semi-circular arc
Circular sector subtending an angle
2 at the center
Semi-circle
Solid hemisphere
Hollow without base hemisphere
Position of center of gravity on the axis of symmetry
h
from base
4
h
from base
3
r sin 
from center

2r
from center

2r sin 
from center
3
4r
from center
3
3r
from plane face
8
r
from plane face
2