Download Physics 231 Ch 9 Day 1 2013 1 10 11 Ch. 9 Multiparticle Systems

Physics 231
2013 1
Ch 9 Day 1
10
Mon. 11/4
Tues. 11/5
Wed., 11/6
Lab
Fri., 11/8
9.1-.2, (.8) Momentum and Energy in Multi-particle Systems
11
Mon., 11/11 9.4-.5 (.9) The “Point Particle” approximation
Tues. 11/12
9.3 Rotational Energy Quiz 8
L8 Energy Quantization Review Exam 2 (Ch 5-8)
Exam 2 (Ch 5-8)
RE 9.a
HW8: Ch 8 Pr’s 21, 23, 27(a-c)
RE 9.b bring laptop, smartphone, pad,…
Practice Exam 2 (due beginning of class)
RE 9.c
EP8, HW9: Ch 9 Pr’s 34, 40, 43
Load Vpython
Center-of-mass motion
o Air Track, big & small cart bound together by spring, air blower
Lab3 binary w cm.py
09_Krel.py
07_twopucks.py
Wheel
Baton
Weeble wobble
Metal man with barbell
Ch. 9 Multiparticle Systems
Last chapter, we zoomed way in and saw in very small systems that K+U levels
were quantized. Now we’re going to zoom back out again and consider more
carefully how we handle multiparticle systems – systems in which there’s
obviously some net, large-scale motion, but also some inner workings.
Consider a system of many particles, perhaps it is a dust cloud in interstellar space
and each speck of dust is one of our “particles.” On the one hand, if we focus in,
we see that each speck is naturally at a different location, has a different mass,
and is moving with a different velocity. Yet, if we zoom out, we see a single
cloud that behaves somewhat cohesively. If we watch the cloud for a while, as it
moves through space, it makes sense to speak of the whole as having some
velocity and going from some position to another – in short, we can think of it as
a “particle” of its own. For that matter, if we zoom way in, each of our dust
“particles” are themselves made of several, much tinier particles. How do we
reasonably do this – treat a composite of several particles, each more-or-less
doing their own thing, as a single object and yet properly account for the internal
workings? That’s what this chapter is about. We start by considering the motion
and momentum of a multi-particle system, and then we look at the work and
energy.
9.1 Momentum and Center of Mass.
The motion of a multparticle system
The chapter opens with an every-day yet rich example: a person jumping. The
flexing of the system is quite obvious, and somehow that is responsible for you
flying off the ground.
The momentum principle for multiparticle systems.
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Back in Chapter 3, we developed the momentum principle for multiparticle


dptotal
systems. It quite resembled that for the single particle:
Fnet,ext
dt









where ptotal p1 p 2 p3 p 4 ... m1v1 m2 v 2 m3 v3 m4 v 4 ... . Looking at
the right hand side, if you recall, all of the internal forces, between the particles of
the system, cancel thanks to the reciprocity principle (Newton’s 3rd Law:


F1 2
F2 1 ), which leaves only the external forces. This will be our starting
point. So, by analogy to the momentum principle for a single particle,

dp1 

Fnet 1 , ptotal plays the role of the momentum of the single, composite
dt
system. So then, what’s a representative velocity for the whole system? A


velocity such that p system m systemv system . It seems a no-brainer that
m system

v system
M total

p system
msystem
m1
m2 m3 m4 ... so then




m1v1 m2 v2 m3 v3 m4 v4 ...
m1 m2 m3 m4 ...

m1v1

m2 v 2

m3 v3
M total

m4 v 4
...
Weighted Average: Mathematically, this is the velocity averaged over mass.
Demo: Lab_3 binary wo cm.py You may remember writing code to simulate
two stars gravitationally interacting. They went looping through space weaving
around each other. Yet, if we call the two stars together our “system”, since there
is not net external force on the system, the average velocity of its members must
be unchanging – just a constant. What representative point is moving like that?
7.1.1 Center of Mass
We call the representative point in a complex system the “Center of Mass.”

 dr
You’ll see why. So, where is it? It’s the point that satisfies v
.
dt








m1 r1 m2 r2 m3 r3 m4 r4 ... m1 r1 m2 r2 m3 r3 m4 r4 ...

rCM
m1 m2 m3 m4 ...
M total
for if you take its derivative, you get back our relation for the system’s velocity.
Demo: Lab_3 binary w cm.py
o Note that the center of mass is just a mathematically defined point – it’s
not a fixed part of either object.
Q:In this simulation, the blue star’s mass is 2e30kg and it starts out at the origin;

1.5e11,0,0 m .
the yellow star’s mass is 1e30kg and it starts out at ry
Initially, where’ the center of mass?
CW
Here are two masses located on the x axis:
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What is the x component of the location of the center of mass of the system
consisting of both masses?


m1r1 m2 r2

10 kg 0m xˆ 2kg 3m xˆ
rcm
0.5mxˆ
m1 m2
10 kg 2kg
Three uniform-density spheres are positioned as follows:
• A 5 kg sphere is centered at ‹ 11, 25, −6 › m.
• A 10 kg sphere is centered at ‹ 7, −19, 11 › m.
• A 12 kg sphere is centered at ‹ −9, 14, −15 › m.
What is the location of the center of mass of this three-sphere system?
cm
=
m
o
Center of mass of large objects
o Now, in reality, two stars are themselves collections of ‘particles’ – atoms,
electrons, etc. Each has its own center of mass, quite obviously at the
geometric center. Let’s call the two stars “Yellow” and “Blue”, then
noting which particles in our system are a part of which star, we have


M b rb M y ry

rCM
Mb M y






m1b r1b m2b r2b m3b r3b ... m1r r1 y m2 y r2 y m3 y r3 y ...
m1b m2b m3b ... m1 y m2 y m3 y ...
o So, the relation scales: you can sum over fundamental particles, or over
more convenient “objects.”
8.X.2. Now for a composite system. Say we make a T of two meter sticks. Taking the
origin to be at the base of the T, where’s the system’s center of mass?
First, where’s the center of mass of the vertical stick?
Where’s the center of mass of the horizontal stick?


m1 r1 m2 r2

Now apply rcm
where both objects have the same mass.
m1 m 2
Center of Mass = “Center of Gravity”=“Particle” location. If you’re still
feeling a bit like: “nice definition, but what does it mean / what good is it”,
consider this. We have defined it such that when a net force is applied to an
object, independent of where on the object it is applied, the center of mass moves
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just like a point object of the whole object’s mass.




dptotal
dvcm
d 2 rcm
Fnet
M total
mtotal
dt
dt
dt 2
o Example: Tossed Baton. Toss a baton – if you trace the trajectory of any
point, other than the center of mass, it follows a complicated path through
space, but the center of mass follows the smooth arc you’d predict for a
tossed point mass.
Red = center of mass trajectory, Blue = baton end trajectory
o Example: Balancing a meter stick. Gravity pulls equally on each
morsel of an object. If you push up on the object’s center of mass, or in
line with it, then there’s an equal amount of mass on the left and an equal
amount on the right, so the object falls neither way, but balances. For this
reason the center of mass is also known as the “center of gravity.”
 Invasion of the weeble-wobbles!
Who doesn’t fall down, no matter how much they may
wobble? Weebles wobble but they don’t fall down. Say you
tip a weebl wobble forward, what does it do?
o Wobble backward.
Where must its center of mass be relative to its contact
point?
o Behind.
 Demo: metal man with bent barbell – wobbles like a weeble
wobble.
Motion of Center of Mass
o Demo/Example: Standing Broad Jump. Here comes the calisthenics
portion of the lecture. Crouch down, and then jump. What significant
external forces are there on you?
 Gravitational force pulling you down and the force of the floor
(normal and friction) pushing you up and forward (the forces
associated with air pressure cancel and we’ll neglect those
Ffl
associated with drag.)
Ffl
 While different parts of your body follow different, complicated
trajectories (say, your arms whindmill while your legs pump in and
Mtotg
M
g
tot
Mtotg
out), your center of mass follows the same smooth ark that a ball
would follow. Its trajectory follows from



dptot

Fnet.ext F fl M total g upon launch and just
dt


Ffl
dptot

F
M total g during the flight. Just as would a ball of
net.ext
Ffl
dt
your mass.
Mtotg
Mtotg
Mtotg
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A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick,
at right angles to the stick, with a force F = 7 newtons. The ensuing motion of the meter
stick is quite complicated, but what are the initial magnitude and direction of the rate of
change of the momentum of the stick, dPtot/dt, when you first apply the force?
Magnitude
left
right
up
down
What is the magnitude of the initial acceleration of the center of the stick?
24.1m/s2
Summing up
We’ve learned
o How to relate a system’s velocity and position to the
velocities and positions of its constituents
o That this point, the center of mass, moves in response to
external forces just as a point mass does, according to
the momentum principle.
7.2.4 Application: Pull on two hockey pucks
Consider two pucks, one with a string attached to its center and one with a string
wrapped around its circumference. Both are pulled with the same force, and for
the same period of time.
Q: What does that imply about the changes in momentum of their centers of
mass?
 
o A: the same since p Fnet t
Q: What does that imply about the motion of the center of mass?
o A: The same.
o
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Demo: 07_twodiscs.py
o While the center of mass motion is the same, how do the two pucks
behave differently?
 One has internal motion; it spins.
 Internal motion doesn’t show-up on net momentum.
While the motions of the individual particles of the two
pucks are very different, they have different momenta from
their counterparts, these differences cancel when you sum
pBottom =p
over the whole disc.
ptop + pbottom = 2p
Look at two points at either end of an axis on the red disc,
and the two similar points on either end of an axis on the
blue disc, at any moment, the two on the blue disc have
pTop =p + dp
equal momenta to each other, while one of the points one
the red disc has a little more, the other has an equal amount
less, so summed together, they give the same net.
pTop =p
pBottom =p - dp
ptop + pbottom = 2p
While, for both pucks, the same force was applied for the same
 
time, thus they experienced the same p Fnet t ; what was
different about the application of force? Over different


distances. Of course, E Fnet r
Since the system’s momentum doesn’t speak to its internal
motion, we’ll turn to our other tool for quantifying motion: the
work-energy relation.
Separation of Multiparticle System Energy
As you recall, the work-energy relation is
Wnet.ext
system


Fi.ext dri
Es
Es
i
Where the sum reminds us that we need to add up all the work done by each individual
force acting on the system. Before we get to applying it, let’s focus on the right-handside and consider the forms of energy there are.
On the one hand,
Es
mi c 2 K i
U ij - the total energy of the system is simply the sum of the rest,
i
kinetic, and potential energies of all its members.
On the other hand, it sure would be convenient to be able to separately talk about the
translational motion of the whole system:
2
pcm
2
K cm 12 M systemvcm
2M system
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And, then the energy associated with the ‘internal’ details of the system.
Es
K cm
mi c 2
E int ernal
Ki
U ij
i
mi c 2
Where E int ernal
Ki
U ij
K cm
i
We already know how to describe this down on the sub-atomic level
(electronic states), on the atomic level (vibrational states), on the
molecular level (rotational states), and when all the atoms are jiggling
fairly randomly (thermal energy). But sometimes the internal energy
manifests itself on the macroscopic scale, with macroscopic rotation,
macroscopic vibration. We’ll look at another example and see how to take
these motions / energies into consideration.
Internal Energy
Internal Kinetic Energy
The difference between the total kinetic energy in the system and that of the
whole-sale motion is that associated with internal motions.
K int ernal
K i K cm
i
Section 9.8 demonstrates that

2
1
v
o K int ernal
2 mi v i .rel where i .rel

vi

vcm , i.e., how the particle is
i
moving relative to the center of mass.
Considering only the change in internal kinetic energy, we can take this a step or
two further. Essentially the internal kinetic energy is just the sum of each
particle’s kinetic energy, relative to the center of mass.
2
2
1
K system K cm K int 12 M tot vcm
2 mi vi.rel
all . particles
Let’s see where that comes from before we make use of it.
2
1
o Total kinetic Energy: K total
. We can express the velocity
2 mi vi
all . particles
1
2
=
=
of an individual particle of the object in terms of the velocity of the center
of mass and then the point object’s velocity relative to that.
 

 vi vcm vi.rel

 2
2
 

1
1
K total
vi.rel
2vcm vi.rel vi2.rel
2 mi v cm
2 mi v cm
all . particles
2
1
2 mi v cm

all . particles
1
2
2
M tot vcm
1
2
 
1
2 mi v cm vi .rel

vcm

1
2

mi vi.rel
all . particles

all . particles

mi vi2.rel
1
2
all . particles

mi vi2.rel
but the middle term is a sum over the relative momenta of each
particle – this just gives the momentum of the center of mass,
relative to itself – oh, it’s not moving relative to itself =0.
2
2
1
.
K total 12 M tot vcm
2 mi vi.rel
all . particles
o Rotation and Vibration
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

It is convenient to resolve this motion into two components –
radial and tangential.
2
2

1
1
K int
vi2.rad
2 mi v i .rel
2 mi v i . tan
all . particles
2
1
2 mi v i . tan
all . particles

2013 8
all . particles
1
2
all . particles

mi vi2.rad
Vibration: Radial motion would be moving toward or away from
the center of mass. Take for example a diatomic molecule, the two
atoms can jiggle back and forth, i.e. vibrate.
2
1
K int .vib
2 mi vi.rad
all . particles

Rotation: Tangential motion would be moving around the center
of mass. Consider again our diatomic molecule, it can rotate.
2
1
K int .rot
2 mi vi.rad
all . particles


So, K int K vib K rot
Of course many systems do both at the same time, but it’ll be
simplest for us to consider one kind of motion at a time. I don’t
really have much more to say about vibrational, but rotational
motion can be rephrased a bit to make more manageable.
o Rotation. Now, in the case of the puck, or say, that of this wheel.
o Demo: Spin wheel.
o We’ll get into much more detail about the rotational term next time.
Total energy for Translating, Vibrating, Rotating system (such as a
molecule).
o A common place you find systems that do all three is in gas of molecules.
Calling one molecule the “system”, each molecule can zip around
(translate), their atoms can jiggle (vibrate), and the can spin (rotate). The
total energy of such a molecule then could be broken down into terms for
each kind of motion, as well as the potential energy associated with the
atoms bonding.
o Considering just s diatomic molecule:
o Etotal K trans K rot K vib U bond m1c 2 m2 c 2

Where U bond
1
2
k sp s 2 U o
Consider a system consisting of three particles:
m1 = 3 kg, 1 = < 11, -8, 15 > m/s
m2 = 3 kg, 2 = < -12, 11, -5 > m/s
m3 = 5 kg, 3 = < -23, 36, 18 > m/s
(a) What is the total momentum of this system?
kg·m/s
tot =
(b) What is the velocity of the center of mass of this system?
m/s
cm =
(c) What is the total kinetic energy of this system?
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Ktot = 6420 J
(d) What is the translational kinetic energy of this system?
Ktrans = 2910 J
(e) What is the kinetic energy of this system relative to the center of mass?
Krel = 3510J
7.3.1 Gravitational Potential Energy of a multi-particle system. (near the Earth’s
surface)
Ug
m1 gy1 m2 gy2 m3 gy3 ... g m1 y1 m2 y 2 m3 y3 ...
m2 y 2 m3 y3 ...
gM total y cm
M total
o In general, as long as the object is itself small compared to the distance
between it and the center of the object with which it is gravitationally
interacting, you can approximate the strength of the gravitational field as
constant over the whole object.
M total
m1 m1
m1
 U
GM sun
...
GM sun
r1 s r2 s r3 s
rcm s
o If the object is spherically symmetric, then inspite of gravity’s distance
dependence, that dependence cancels out.
o One of the nice things about momentum is that, what you see on the
macroscopic scale equals the sum of what’s going on on the microscopic
scale. Physically, this follows from the reciprocity principle for forces,
and how forces impact momentum. Mathematically, this is a consequence
of its being a vector quantity, allowing both positive and negative
contributions to cancel each other.
gM total
m1 y1
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o The individual particles, are spinning around and around the center of
mass. Furthermore, the object isn’t significantly deforming, meaning that
each particle stays a fixed distance from the center of mass, ri , and that all
particles are moving with the exact same angular speed as each other, .
So, how can we rewrite the sum relative kinetic energies in the case of
rotation?
2
2 2
2
1
1
1
 K int K rot
mi ri 2
2 mi vi .rel
2 mi ri
2
all . particles

all . particles
all . particles
In the case of a spinning disc or wheel, or any object with a
continuous mass distribution, this sum becomes an integral. In the
simple case of discrete masses, as you’ll encounter in pr.10, it’s a
simple sum.
o Work & Energy of Center of Mass
 Let’s consider the puck that spun.
 First let’s just think of the motion of the center of mass.
Remember, the center of mass moves just like if we had a point
particle experiencing the same net force as does our real system.
W cm
Ecm
d


Fnet.ext drcm
d
0
d


F pull drcm
1
2
2
M cm vcm
.f
0


F pull drcm
1
2
2
M cm vcm
.f
0
2
F pull d 12 M cm vcm
.f
o Work & Energy of whole Puck & String.
1
2
2
M cm vcm
.i
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
Now let’s find the work done on the real system. In this case, the
point at which the external force was applied moved an additional
distance, Lstring. So, an additional amount of work was done.
W system
Es


Fnet.ext drsystem
d L
d
0


F pull drs
0


Fnet.ext drsystem


F pull drs
d L
1
2
2
M cmvcm
.f
1
2
2
M cm vcm
.i
d
d L
1
2
2
M cmvcm
.f
Eint
0
F pull d
L
1
2
2
M cm vcm
.f
Eint
In this case, the change in internal motion is
macroscopically obvious – the disc spins, each particle is
circling the center of mass.
 Comparing the two expressions, we find that
F pull L
E int The change in internal energy is the
difference between the work done on the center of mass and
that done on the whole system.
RW 20a: Ice Skater pushing off wall & work.
In RW 20a, you were asked to consider the work and energy associated
with an ice skater pushing off from a wall. There were some subtleties.
We’ll look at this system from a few different angles. And hopefully
illuminate and explain the subtle points.
Momentum and Center of Mass Motion
o The force of the wall pushes back on her for some time,
accordingly,




dptotal
 
Fnet.ext Fwall F floor mg Fwall
dt


o ptotal M tot vcm

dvcm 
M tot
Fwall
dt
o her center of mass moves just as would a ball of her mass.
Work & Energy – Point mass vs. whole object
o System = Center of Mass.

 A net external force is applied, Fwall and the center of mass

is displaced rcm . As we’ve said, the center of mass’s
Fwall
motion is exactly that of a single particle of the full object’s
mass, so, how much work would get done to this
replacement particle, and what would the resulting change
c.m
in its kinetic energy?
Eint
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W
d

cm
E cm


Fnet.ext drcm
d
0


Fwall drcm
1
2
2
M cm vcm
.f
1
2
2
M cm vcm
.i
0


Fwall drcm
1
2
2
M cm vcm
.f

Fwall
Fwall
This is indeed what the center of mass of the person is
doing; in many cases, you only care about the center of
mass – for example, if you want to know how long it will
take the skater to glide across the ice rink, you don’t really
care what she’s doing with her arms or head – just the
motion of the center of mass.
o System =Whole Skater
 This isn’t the whole story though, and sometimes you do
want to think about what’s going on within the object.
 Let’s consider the whole person. Now the point of
application of the external force is her hand, and her hand
isn’t moving off the wall as long as the force is being
applied, so what does that say about the work done by the
wall on her?
0.
 Then how is it that she goes zipping across the ice if no
external work is done, and what role does the wall play?
 Where does her kinetic energy come from? For the first
question, here’s a hint – which party has to eat three square
meals a day – the wall or her, of course it’s her. She’s the
one who takes in fuel.
E skater
K
Eint ernal 0

K
1
2

M tot v
Eint ernal
2
cm. f
Eint ernal
She transformed internal energy into translational kinetic
energy. The internal energy probably changed in a few
different ways: chemical bonds in her stored fuel were
changed – parts of her deformed (muscles bulged, arms
moved), and if she does this enough, she may notice that
she’s warming up – changes in thermal energy.
 What roll does the wall play? Not to undersell the wall;
certainly, if it weren’t there for her to push against, there’d
be no change in motion – she’d just stand there on the ice
thrusting her arms. The wall provides a “restraining” force
– it prevents her from pushing her hands forward, and in
turn allows her to drive her center of mass backward.
o Similar situations: Other instances of “restraining” forces doing
no work, but influencing motion are the friction of tire on road,
tension in a pendulum’s string, and the track of a rollercoaster.