Transcript Physics 111 - New Jersey Institute of Technology

Chapter 9, System of Particles
Center of Mass
Linear Momentum and Conservation
Impulse
Rocket
Center of Mass for a System of Particles
The center of mass of a body or a
system of bodies moves as though
all of the mass were concentrated
there and all external forces were
applied there.
Where is the center of mass of this meter stick?
(a) 30cm
(b) 50cm
(c) 90cm
The center of mass of a symmetric, homogeneous body
must lie on its geometric center.
How to locate the
center of mass of
an irregular planar
object?
com
Suspend the object from two
different points. Drop a plumb
line and mark on the object. The
center of mass coincides with the
intersection of two lines.
2 bodies, 1 dimension
com
xcom
x2=d
x1=0
For example, if x1=0, x2=d, and
m2=2m1, we find that
com
xcom
xcom
m1 0  2m1d 2m1d 2


 d
m1  2m1
3m1
3
Example 1
X1=0
X2=d
Earth
Sun
xcom 
m1 x1  m2 x2 m1 0  m2 d m2 d


m1  m2
m1  m2
m1
5.981024 kg 1.5 108 km

 450km
30
1.9910 kg
Center of Mass for a System of Particles
2 particles, 1 dimension
n particles, 3 dimensions
m1 x1  m1 x2  ...  mn xn
1
xcom 

m1  m2  ...mn
M
n particles, 3 dimensions, vector equation
n
m x
i 1
i i
Example 2
xcom 
m1 x1  m2 x2  m3 x3 (1kg  2m)  (3kg  2m)  (4kg  4m) 12kg  m


 1.5m
m1  m2  m3
1kg  3kg  4kg
8kg
ycom 
m1 x1  m2 x2  m3 x3 (1kg  2m)  (3kg 1m)  (4kg  2m)  2kg  m


 0.25m
m1  m2  m3
1kg  3kg  4kg
8kg
Center of Mass for a Solid Body
dm is the differential mass element
Uniform density
Newton’s 2nd Law for a System of Particles
particle
system of particles


Fnet , x  Macom, x

Fnet , y


Fnet  ma


Fnet  Macom



 Macom, y Fnet , z  Macom, z
The center of mass moves like an imaginary particle of
mass M under the influence of the net external force on
the system.
A firework rocket explodes
Linear Momentum and Impulse
Particle


p  m (kg m/s)


 dp
Fnet  ma 
dt
System


P  Mcom (kg m/s)



dP
Fnet  Macom 
dt




F (t )dt  dP
F (t )dt  dp


 tf 




pf 
tf
Pf

J   F (t )dt   dp  p J   F (t )dt   dP  P
t
p
i
i
ti
Pi
Impulse–Linear Momentum Theorem:
Impulse = Force  Duration of the force = Change in Momentum




tf 
J   F (t )dt  Pf  Pi  P
ti
Collision of two particle-like bodies

F 
tf
ti
Fdt
t
J  F t  P
Conservation of Linear Momentum

Fnet

dP

dt

Fnet  0

P  constant
 
Pi  Pf
If no net external force acts on a system of particles, the total
linear momentum P of the system cannot change.

Fnet , x

Fnet , y

Fnet , z
 
dPx Fnet , x  0

dt
 
dPy Fnet , y  0

dt 

dPz Fnet , z  0

dt


Pxi  Pxf


Pyi  Pyf


Pzi  Pzf
If the component of the net external
force on a closed system is zero along an
axis, then the component of the linear
momentum along that axis cannot
change.
 Closed system (no mass enters or leaves)
 Isolated system (no external net force)

v1i

p1i

p2 i 
v2i
m1
m2
Before collision:

v1 f

p1 f

p2 f 
v2 f
m1
m2
After collision:
Sum of initial momentums = Sum of final momentums




p1i  p2i  p1 f  p2 f




m1v1i  m2v2i  m1v1 f  m2v2 f


Closed system (no mass enters or leaves)
Isolated system (no external net force)
Collisions in closed, isolated systems
• Elastic
• both colliding objects does not change their shapes
• total linear momentum is conserved
• total kinetic energy is conserved
• Inelastic
• the shape of one or both of object changes during the collision
• total linear momentum is conserved
• total kinetic energy is NOT conserved
• Completely inelastic
• the colliding objects stick together after the collision
• total linear momentum is conserved
• total kinetic energy is NOT conserved
In a closed, isolated system containing a collision, the
linear momentum of each colliding body may change
but the total momentum P of the system cannot
change, whether the collision is elastic or inelastic.
Completely Inelastic Collisions in 1D
m1v1i  m2v2i  (m1  m2 )v f
m1v1i  m2v2i
vf 
m1  m2
m1
if v2i  0 , then v f 
v1i
m1  m2
Velocity of Center of Mass
Sample Problem 9-8
M  5.4 kg
m  9.5 g
h  6.3 cm
v?
1
2
mv >
? (m  M ) gh
2
m+M
Step 1 total linear momentum is conserved
m
mv  (m  M )V  V 
v
mM
Step 2 total mechanical energy is conserved
1
2
(m  M )V 2  (m  M ) gh  V  2 gh
m
mM
2 gh 
v v
mM
m
2 gh  630 m/s
Elastic Collisions in 1D
if m1  m2 , then 1 f  2i and2 f  1i
if 2i  0 then1f 
m1  m2
2m1
1i and2f 
1i
m1  m2
m1  m2
if m1  m2 (a massiveprojectile), 1f  1i
if m1  m2 (a massive target),
Equ.s (9-67), (9-68)
and 2f  21i
2m1
1f  1i and2f 
1i
m2
Sample Problem 9-10
m1  30g
m2  75g
h1  8.0cm
v1 f  ?
Step 1
Step 2
1
2
m1v12i  m1gh1  v1i  2gh1  1.252 m/s
m1  m2
v1 f 
v1i  0.537 m/s
m1  m2
Rocket Propulsion
A Rocket System (rocket + its ejected combustion products)
is a closed, isolated system.
The rocket is accelerated as a result of the thrust it receives
from the ejected gases.
Rocket Equations
First rocket equation:
T  Rvrel  Ma
where R is the fuel consumption rate,
vrel is the velocity of ejected fuel with respect to the rocket
M is the instantaneous mass of the rocket
a is the acceleration of the rocket
T is referred to as the thrust of the rocket engine
Second rocket equation:
Mi
v f  vi  vrel ln
Mf
where vi and vf are initial and final velocities of the rocket
Mi and Mf are the initial and final masses of the rocket
Summary #1
Center of mass:
xcom
1

M
Solid object
xcom
1

M
Solid object with
uniform density
xcom 
System of n particles
n
m x
i 1
i i
 xdm
1
xdV

V
ycom
1

M
ycom
1

M
ycom 
n
m y
i 1
i
i
 ydm
1
ydV

V
Newton’s second law:


Fnet  Macom
Linear momentum:


P  Mcom
Impulse-momentum theorem:
n
zcom
1

M
m z
zcom
1

M
 zdm
zcom 
i 1
1
zdV

V

 tf 


Pf
J   F (t )dt  dP  P
ti
Pi
i i
Summary #2
Conservation of momentum:
Elastic collisions:
Inelastic collisions:
Completely inelastic
collisions:
Rocket equations:
 
Pi  Pf (closed, isolated system)

P is conserved,and Ekinetic is conserved.

P is conserved,and Ekinetic is NOT conserved.

P is conserved,and Ekinetic is NOT conserved.
T  Rvrel  Ma
Mi
v f  vi  vrel ln
Mf