Transcript 9 Systems of Particles - Florida State University

9. Systems of Particles
Center of Mass
Center of Mass


 mi ai   Fi
i
i
Consider a system of interacting
particles. For each particle, the
forces that act on it can be split
into internal and external ones.
Fi  Fi ,int  Fi ,ext
Internal
force
External
force
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Center of Mass
 Fi   Fi ,int   Fi ,ext
i
i
i
  Fi ,ext  Fnet ext
i
Internal
force
The internal forces
sum to zero. Why?
The net force on the
system of particles
is the sum of the
external forces only.
External force
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Center of Mass
From Fnet ext   mi ai
i
d 2r d 2
  mi 2  2
dt
dt
i
m r
i
i
This suggests the definition of
center of mass
Internal
force
m r  M r
i i
cm
i
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Center of Mass
where
m
i
 M is the total mass of the system.
i
from Fnet ext
2
d
 2
dt
2
d rcm
i mi r  M dt 2
we arrive at the 2nd law for the
motion of the center of mass.
Fnet ext  Macm
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Finding the Center of Mass
Example – A Space Station
What is the center of mass of
a space station, made of
three modules of mass m, m
and 2m?
mx1  mx3 m( x1  x1 )
xcm 

0
4m
4m
my1  my3 2my1 1
ycm 

 y1
4m
4m 2
3

L  0.43L
4
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Example – A Skier’s Motion
The different parts of the
skier’s body follow complex
trajectories.
But, the center of mass
follows a simple parabolic
motion. In general, the
center of mass moves as if it
were a simple particle.
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Continuous Mass Distributions
Consider splitting an object into lots of small pieces
of mass Δmi. By definition, the center of mass is given
rcm 
 m r
i i
i
M
If we make the pieces smaller and
smaller we get
rcm  lim
mi 0
 m r
i i
i
M
r dm


M
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Example – An Aircraft Wing
Consider an aircraft wing shaped like an isosceles
triangle. Where is its center of mass? Consider the mass
of a strip of width dx
dm (w/L)x dx 2x dx
 1
 2
M
L
2 wL
xcm
1
1

x
dm

M
M
 2Mx 
x
 0  L2 dx 
L
2 L 2
2 x3 L 2 L3 2
 2  x dx  2
 2 L
L 0
L 3 0 3L 3
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Example – Scale
(1)
What does the scale read as the block slides down
the wedge? We need to compute
the normal force, Fn
This looks very complicated!
But, we know that the motion
of the center of mass is often
simple. So, let’s try to solve
the problem that way.
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Example – Scale
(2)
Consider motion in the vertical
direction:
F
i ,ext,y
 Macm,y
i
and taking up to be +ve,
we have
Fn  m1g  m2 g  Macm,y
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Example – Scale
(3)
But note that
Macm,y  m1a1 y  m2 a2 y
 m1a1 y  0
so we get…
Fn  (m1  m2 ) g  m1a1y
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Example – Scale
(4)
The magnitude of the acceleration
down the slope is
a  g sin 
therefore, the vertical
acceleration of block 1 is
a1y  g sin 
2
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Example – Scale
(5)
So, finally, we deduce that the
scale reads:
Fn  (m1  m2  m1 sin  ) g
2
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Conservation of Momentum
Conservation of Momentum
Start with Newton’s 2nd Law and assume
that the external forces sum to zero
dvi
i mi dt  i Fi,ext  0
Integrate both sides
p
i
i
 constant
where
p  mv
is the momentum
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Conservation of Momentum
If the net external force on a system of particles is
zero, the total momentum of the system is constant, that
is, conserved.
p
i
 constant
i
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Kinetic Energy of a System
Kinetic Energy of a System

2
1
Consider the total kinetic energy K 
2 mi vi
i
of a system of particles. This can
be rewritten as K = Kcm + Kint, where the first term
is the kinetic energy of the center of mass and
the second is the kinetic energy relative to
the center of mass.
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Collisions
Momentum in Collisions
Start with the 2nd Law
Then integrate over
a short period
from ti to tf
dp
F  ma 
dt

tf
ti
Fdt  
tf
ti
dp
dt
dt
tf
  dp  p f  pi  p
ti
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Momentum in Collisions
The integral of a force over a short period
from ti to tf is called the impulse
tf
I   Fdt
ti
 p
Average force
I
Fav 
t
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Example – Impulse
(1)
A car drives into a wall at 25 m/s (about 56 mph).
What’s the average force of the seatbelt on a
crash dummy of mass 80 kg?



I
p
Fav 

t t
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Example – Impulse
(2)
Impulse p  mv f  mvi  0  mvi
 2000 N  s iˆ
Estimate collision time, and use average speed of 12.5 m/s
x
1m
t 

 0.08s
vav 12.5m/s
Force
Fav  25,000 N iˆ
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Totally Inelastic Collisions
Perfectly Inelastic Collisions: particles stick together
Before
m1 m2
m1
m2
After
Since the net external force is zero, momentum is
conserved
m1v1i  m2v2i  (m1  m2 )v f
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Totally Inelastic Collisions
However, although momentum is conserved, it turns
out that kinetic energy
is not!
Before m1
m2
m1 m2
After
Where does the energy go?
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Example
The Ballistic Pendulum
This is a device to measure the speeds of fast-moving
objects, like bullets. The bullet gets lodged in the
wooden block and causes the block to move upwards
a height h. Let’s break the problem into two parts:
1. An inelastic collision
between bullet and block
2. The rise of the block
v
m
h
M
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Example
The Ballistic Pendulum
Inelastic collision: Since the net external force is zero,
momentum is conserved:
mv  (m  M )V
V is the velocity of the bullet plus block. But, note,
kinetic energy is not:
Ki  12 mv 2
K f  (m  M )V
1
2
 m

 mM

 Ki

2
v
m
h
M
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Example
The Ballistic Pendulum
The rise: In moving up by an amount h, momentum is
no longer conserved. Why? But mechanical energy is.
The final energy = initial energy
1
(m  M )V 2  (m  M ) gh
2
therefore,
 mM 
v
 2 gh
 m 
v
m
h
M
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Elastic Collisions
Elastic Collisions
Momentum before
m1v1i  m2v2i
m1
m2
m1
m2
Momentum after
m1v1 f  m2v2 f
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Elastic Collisions
Momentum after
m1
=
m2
Momentum before
m1
m2
m1v1 f  m2v2 f  m1v1i  m2v2i
We have 2 unknowns → therefore, need 2 equations
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Perfectly Elastic Collisions
Perfectly Elastic Collisions:
Kinetic energy after = kinetic energy before
m1
1
2
m1
m2
mv  m v
2
1 1f
1
2
2
2 2f
m2
 mv  m v
1
2
2
1 1i
1
2
2
2 2i
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Perfectly Elastic Collisions
m1
m1
m2
m2
m1v1 f  m2v2 f  m1v1i  m2v2i
We have 2 unknowns and we now have 2 equations
1
2
mv  m v
2
1 1f
1
2
2
2 2f
 mv  m v
1
2
2
1 1i
1
2
2
2 2i
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Perfectly Elastic Collisions
m1
m2
m1
m2
From the two equations, we can derive the following:
v1 f  v2 f  (v1i  v2i )
This shows that the final relative velocity between the
particles is the reverse of the initial relative velocity,
but has the same magnitude.
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Summary

The center of mass of a system of particles is
defined by rcm  (1/ M ) mi ri
i



The center of mass satisfies Fnet = Macm,
where M is the total mass and acm is the
acceleration.
If the net external force is zero, momentum is
conserved.
Kinetic energy is conserved only in perfectly
elastic collisions.
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