Introduction
Here it is really important for the student to draw diagrams of initial and final states
and mark the positions of the center of mass in the two states, then determine the
distance the point particle (at the center of mass) moves. And it is crucial that they see
the importance for calculating work done by individual forces of marking on the
diagrams the distances each force acts through.
A significant innovation is to introduce the point-particle system much earlier than is
done in Ch. 8, in the context of momentum. Don't wait for the energy context.
Describe the momentum principle for multiparticle systems as being the principle
governing the motion of the "crushed" system having the mass of the real system but
with all forces applied at the point.
Discussion: Momentum Principle review
Since we have been mostly working with the energy principle for several weeks,
review the momentum principle by treating projectile motion in a vacuum, or at
low speed:
d
px , py , pz = 0, -mg, 0
dt
This implies px and vx , pz and vz are constant (and equal to initial values).
Therefore x = xi + vxi Dt and z = zi + vzi Dt .
Evidently py = pyi - mgDt , and vy = vyi - gDt and substituting that into the definition
of average velocity, we get
æ vyi + vyf ö
y f - yi = ç
Dt
è 2 ÷ø
(
)
æ vyi + vyi - gDt ö
æ 2vyi - gDt ö
y f - yi = ç
÷ Dt = ç
÷ø Dt
2
2
è
è
ø
y f = yi + vyi Dt -
1
2
g ( Dt )
2
This is something some students have seen in a previous physics course.
Emphasize that there are many simplifying assumptions that go into this result:
neglect air resistance, nearly constant gravitational force (because near surface of
Earth), speed small compared to c. The results are NOT fundamental in the way that
the momentum principle is fundamental.
Chapter 9
1
Discussion: Center of mass review
Remind students that the rate of change of total momentum is equal to the net force,
exactly as for a point particle (comment that proof is in book, depends on
cancellation of forces internal to the system).
Definition of center of mass position
velocity of center of mass
total momentum
P_total = Mv_cm
State without proof that the total momentum (v << c) can be expressed as the total
mass times the velocity of the center of mass point. Comment that there is a
formula for calculating the center of mass of an arbitrarily complicated object, but
we’ll only actually deal with situations where the center of mass is obvious (center of
a sphere, etc.). Show just for orientation two equal spheres a distance apart, ask for
center of mass; then ask where the center of mass would be (qualitatively) if one of
the spheres has much more mass than the other.
Demo: Motion of multiparticle systems
Earth-Moon_system.mov, Fireworks.mov, Pencil.mov
What if the projectile isn’t a point particle but some blobby floppy thing? Throw a
floppy thing and observe that in some sense the floppy object follows the same kind of
trajectory. Show movies of fireworks, pencil, Earth/Moon and baseball bat. It is the
“center of mass” that follows the point-like trajectory, with other internal motions
relative to that center of mass. Why is this?
Have a, b, c groups describe movies as I display their laptop screen? (Group A: Moon,
Group B: pencil, Group C: Fireworks). Little time at end, we discussed what Fnet was
for each system and how it predicted v_cm.
Ponderable: Two pucks
Clickers Q91a, b
09_twopucks.py
TwoPucksDemo.rm
Like example in book, but with more detail.
Have even tables do point particle system. Odd tables to real system.
Two identical pucks pulled on ice, with same force for same amount of time. Center
of mass of 1 and 2 will move the same distance, d, and L string will unwind from puck
1.
Chapter 9
2
L+d
1
x
2
x
x
F
F
x
F
Use GOAL protocol
Point particle system:
F
d
L
dC
L
M
DEpoint particle system = DK translation = W + Q
DK translation = Fnet external dCM
Real system:
DEreal system = DK total = W + Q
(
DK translation + DK rotation = F Lpoint of application + dCM
)
So combining the two analyses:
DK rotation = FLpoint of application
The key point is that it matters whether you calculate the work done by an individual
force acting through the individual displacement of its point of application, or the
work done (on the point particle system) by the net force through the displacement of
the center of mass.
Having gotten at the issues, now show the movie. In the version of this physical demo
built at NCSU (and shown in the video), massive pulleys are mounted on PASCO
carts and pulled by strings that pass over pulleys in such a way that, neglecting
rotational inertia, the forces applied to the rotating and nonrotating “pucks” are the
same. When things work right, the centers of the two systems move together, as
predicted by the multiparticle version of the momentum principle. Tapes attached to
the string let you see that the force acting on the rotating puck moves through a
distance about 3 times that of the force acting on the nonrotation puck, so that if the
nonrotating puck gets 10 J of (translational) kinetic energy, the rotating puck gets 30 J
of kinetic energy, 10 J translational and 20 J rotational.
Chapter 9
3
VPython Demo: Relative velocity
09_kRel.py (first) with Clickers Q9.1a , Q9.1b, Q9.2a
09_RotateVibrateTranslate.py (second) with Q92b-h
Talk briefly about v_cm and v_relative. v_relative is what you see from your POV if
you are “riding along” with the center of mass. Relate to the fireworks video.
Clickers 9.3a, 9.3b, 9.3a
Discussion: Translational kinetic energy
So the momentum principle holds for the motion of the center of mass of arbitrarily
complicated objects. This is why our analysis of a point particle projectile is also fine
for the center of mass of a floppy object. What we’ve just seen in the motion of the
ball and the floppy object makes these results plausible.
It can be useful to think in terms of Superman crushing the actual system of interest
down to a point particle located at the center of mass, having the mass of the actual
system, and subjected to the same net force as the actual system. The motion of this
“point particle system” will be the same as the motion of the center of mass of the
2
actual system. The kinetic energy of the point particle 12 M total vCM
is called
“translational kinetic energy.”
So you know from the familiar momentum principle applied to the point particle
system something about the way a complicated multiparticle system will move. What
about the motion relative to the center of mass? The rotation, vibration and general
floppiness of the system?
The last property of multiparticle systems that we need is that if you add up the kinetic
energy of every atom, which we naturally call the total kinetic energy of the system,
it turns out that this is numerically equal to the kinetic energy of the point particle
2
system 12 M total vCM
plus the kinetic energy relative to the center of mass, involving
speeds relative to the center of mass.
Ktot = Ktrans + Krel
Simple example: a nonrotating but “translating” wheel has kinetic energy equal to the
translational kinetic energy. If it rotates without translating it has “rotational” kinetic
energy, associated with speeds relative to the center. If the wheel rotates while it
translates the two terms simply add to give the total kinetic energy, which is equal to
the total kinetic energy of all the atoms. The fact that kinetic energy splits up this way
sounds plausible but the proof is rather difficult; the proof is given in an appendix to
Ch 8.
Chapter 9
4
Ponderable: Activity - 2-particle system
WID 1156844
twoparts
Consider a 2-particle system like this:
1) mass 2 kg, velocity < 3, 5, 0 > m/s
2) mass 3 kg, velocity < -4, 2, 0 > m/s
Have students use whiteboards to sketch
situation and then calculate the total momentum:
< 6, 10, 0 > kg m/s + < -12, 6, 0 > kg m/s = < -6, 16, 0 > kg m/s.
This is equal to total mass times cm velocity, so students calculate velocity of the
center of mass:
Ptot » Mvcm
-6,16, 0 kg×m
Ptot
s
vcm =
=
= -1.2, 3.2, 0 ms , which looks plausible when drawn.
M
5 kg
They calculate Ktot = .5(2)(32+52)+.5(3)((-4)2+22) = 64 J
They calculate Ktrans = Ptot2/2m = (62+ 162)/(2*5) = 29.2 J or ½ mtot vcm2
(They’ll probably need some guidance here.)
They calculate Krel = 34.8 J. Ktot = Ktrans + Krel
(Calculating Krel directly of course requires getting the velocities relative to the center
of mass and then finding the corresponding kinetic energies. This is a check on getting
Krel by subtraction. Unfortunately it requires saying something about relative
velocities, which is not a topic we’ve covered.)
Ponderable: Skater vs. wall
Clickers Q9.4f-g
(Like example in book, but with more detail)
Skater pushes against wall with force F, moving cm distance d.
P > 0 although external work done is zero!
Use GOAL protocol and divide up by even/odd tables again, but have even tables do
the real system this time.
Chapter 9
5
System is skater (real system gives info about internals)
E f - Ei = Wexternal + Q = 0
( mc
2
) (
d
)
+ K f + Einternal, f - mc 2 + K i + Einternal, i = 0
K f = Einternal, i - Einternal, f = -DEinternal
Since K f > 0, Einternal, i - Einternal, f > 0, so Einternal, i > Einternal, f
System is point particle (gives info about cm motion)
E f - Ei = Wexternal + Q = Wexternal
( mc
2
) (
)
d
+ K cm, f - mc 2 + K cm i = Fdcm
K cm, f = Fdcm
So K cm, f = Fdcm = -DEinternal
d
Ponderable: Activity - Two disks colliding
WID 1157010 (Problem 9.P.40)
dragon
d-b
b
x
x
F
“extra distance” gives “extra work”
F
d
initial
final
a) Since we are interested just in the speed v of the stuck-together disks, which is also
the speed of their center of mass, we should consider the point-particle system. The
center of mass of the two-disk system moves a distance b, so we have this:
Chapter 9
6
DK trans = Fnet, x DxCM = Fb (point-particle system)
1
( 2M ) v 2 - 0 = Fb
2
Fb
v=
M
b) Consider the real system of the two disks. You do an amount of work Fd, and the
main energy change is the translational kinetic energy and the thermal energy:
DK trans + DEthermal = Fd (real system)
Fb + DEthermal = Fd
DEthermal = F ( d - b )
This is positive, corresponding to the fact that the temperature of the two disks rises.
Clickers Q9.4a-h
Lab: Jumping up
WID 628636
yump
Students apply energy and momentum principle to the real and point particle system
for a person jumping in the air from a crouching position.
3.1
DK tran + DEinternal = F floor × Dr floor + F Earth × Dr CM
0 + DEinternal = 0 + mg(y3 - y1 )cos180 = -862J
Chapter 9
7
DK tran = F net × Dr CM
3.2
DK tran = mg(y3 - y2 )cos180 = -431J
v2 =
3.3
3.3
2(431)
= 2.8 ms
m
DK tran = F net × Dr CM
+431 J = ( Ffloor - mg) (y2 - y1 ) = 2156 N » 540 lb
DK tran = F net × Dr CM
+431 J = ( Ffloor - mg) (y2 - y1 ) = 2156 N » 540 lb
D p = F net Dt
3.4
mv2 = ( Ffloor - mg ) Dt
Dt =
110(2.8)
= 0.29 s
(2156 -110 * 9.8)
Ponderable: Activity - Spring and two masses
WID 1156913 (Problem 9.P.44 – used in M&I labs) boing
Encourage students to use the following process:
 Draw diagrams of initial and final state of the point particle system
o Include all forces acting on the system
o Every force should act on the center of mass
 Write Energy Principle for the point particle system
 Draw diagrams of initial and final state of the real system
o Include all forces acting on the system
 Calculate the work from each force
 Write Energy Principle for the real system
 Use ∆Ktrans from point particle system in real system
You hold up an object that consists of two blocks at rest, each of mass M = 5 kg,
connected by a low-mass spring. Then you suddenly start applying a larger upward
force, of constant magnitude F = 157 N (which is greater than 2Mg). The diagram
shows the situation some time later, when the blocks have moved upward, and the
spring stretch has increased.
The heights of the centers of the two blocks are as follows:
Initial and final positions of block 1: y1i = 0.3 m, y1f = 0.4 m
Chapter 9
8
Initial and final positions of block 2: y2i = 0.7 m, y2f = 1.0 m
It helps to show these heights on a diagram. Note that the initial center of mass of the
two blocks is (y1i + y2i)/2, and the final center of mass of the two blocks is (y1f + y2f)/2.
(a) Consider the point particle system corresponding to the two blocks and the spring.
Calculate the increase in the total translational kinetic energy of the two blocks. It is
important to draw a diagram showing all of the forces that are acting, and through
what distance each force acts.
DEpoint particle system = W + Q
DK translation = Fnet external iDrCM
æ y1f + y2 f y1i + y2i ö
= Fnet, y yCM, f - yCM, i = (F - Fgrav ) ç
2
2 ÷ø
è
(
(
)
(
= 157 N - 2 ( 5 kg ) 9.8
m
s2
)) æçè 0.4 2+ 1.0 - 0.3 +2 0.7 ö÷ø
= (157 N - 98 N ) ( 0.7 m - 0.5 m ) = ( 59 N ) ( 0.2 m )
DK translation = 11.8 J
Many students have difficulty with calculating center of mass because they confuse
which height is which (diagrams are helpful).
For Fgrav , check that students are using 2M (mass of both blocks) instead of M (mass
of only one block).
(b) Consider the real system corresponding to the two blocks and the spring. Calculate
the increase of (Kvib+Us), the vibrational kinetic energy of the two blocks (their kinetic
energy relative to the center of mass) plus the potential energy of the spring. It is
important to draw a diagram showing all of the forces that are acting, and through
what distance each force acts.
DEreal system = W + Q
DK total = å F1 iDr1 + F2 iDr2 + ...
DK translation + DK vibration + DUspring = FDy2 + Fgrav DyCM
Chapter 9
9
(
)
(
)
DK vibration + DU spring = F y2 f - y2i - 2Mg yCM, f - yCM, i - DK translation
= (157 N ) (1.0 m - 0.7 m ) - ( 98 N ) ( 0.2 m ) - 11.8 J
= 47.1 J - 19.6 J - 11.8 J
DK vibration + DU spring = 15.7 J
The biggest confusion is in the fact that in the real system, the forces move through
different displacements
Chapter 9
10