Download The Center of Mass

PHYS 172: Modern Mechanics
Lecture 15 – Multiparticle Systems
Summer 2012
Read 9.1 – 9.2
Quantizing two interacting atoms
Classical harmonic oscillator:
Quantum harmonic oscillator:
U = (1/2)kss2
U = (1/2)kss2
w0 =
E2 = 2 w0 + E0
E1 = w0 + E0
E0 = 12 w 0
ks
m
equidistant spacing
2
E = 12 mv 2 + 12 kx 2 = 12 kAmax
Any value of A is allowed
And any E is possible.
ground state
w0 = k s / m
=
h
= 1.05 ｴ10 34 J s
2p
Energy levels:
EN = N w 0 + 12 w 0
Time to Throw Things
BALL
BATON
We need to understand Center of Mass
The Center of Mass
This is a weighted average
of the positions -- each position
appears in proportion
to its mass
where
The Center of Mass
rcm
=
m1r1 + m2r2 + m3 r3 + …
m1 + m2 + m3 + …
r3
r1
rcm
r2
Motion of the Center of Mass
1) Take one time derivative:
Same as:
(Good!)
Motion of the Center of Mass
1) Take a second time derivative:
This says that the motion of the center of mass looks just like what would
happen if all forces were applied to the total mass, as a point particle
located at the center of mass position!
Motion of the Center of Mass
This says that the motion of the
center of mass looks just like what
would happen if all forces were
applied to the total mass, as a point
particle located at the center of
mass position.
F ne t ,ex t =M tota la cm =
dPtotal
Center of Mass
dt
Center of Mass Motion
Same Tension.
Which puck will move faster?
dPtot
» M acm = Fnet ,surr
dt
The centers of mass experience
the same acceleration!
HOWEVER: Hand #2 has to pull the string farther: W2 > W1.
Where does this energy go?
Rotational energy. The bottom spool is spinning.
Question for Discussion
Question for Discussion
Clicker Question
Through what distance did the force act on the Point Particle system?
Equal masses
A) 0.03 m
B) 0.04 m
C) 0.07 m
D) 0.08 m E) 0.10 m
Clicker Question
Through what distance did the force act on the Real system?
Equal masses
A) 0.03 m
B) 0.04 m
C) 0.07 m
D) 0.08 m E) 0.10 m
Clicker Question
Which is the energy equation for the Point Particle system?
Equal masses
A) Ktrans = F*(0.07 m)
B) Ktrans = F*(0.08 m)
C) Ktrans + Kvib + Uspring = F*(0.07 m)
D) Ktrans + Kvib + Uspring = F*(0.08 m)
Kinetic energy of a multiparticle system
Translational, motion of center of mass
Motion of parts relative to center of mass
Ktot = Ktrans + K rel
Krel = Kvib + Krot
Rotation about center of mass
Vibration
Translational kinetic energy
Ktot = Ktrans + K rel
Krel = Kvib + Krot
Translational kinetic energy:
(motion of center of mass)
Ktrans
2
MvCM
Ptot2
=
=
2
2M
(nonrelativistic case)
Clicker:
A system is initially at rest and consists of a man with a bottle sitting on ice
(ignore friction). The man then throws the bottle away as shown.
The velocity of the center of mass vcm will be:
A) Zero
B) Directed to right
C) Directed to left
dPtot
= Fnet ,ext
dt
http://www.punchstock.com/asset_images/95652058
Translational kinetic energy
Ktot = Ktrans + K rel
Krel = Kvib + Krot
2
Translational kinetic energy:
MvCM
Ktrans =
(motion of center of mass)
2
(nonrelativistic case)
Clicker:
A system is initially at rest and consists of a man with a bottle sitting on ice
(ignore friction). The man then throws a bottle away as shown.
The translational kinetic energy of the system will be:
A) Zero
B) > 0
C) < 0
http://www.punchstock.com/asset_images/95652058
Vibrational kinetic energy
- Net momentum = 0
- Energy is constant (sum of elastic energy
and kinetic energy)
Evib = Kvib + U spring
1
1 2
2
Etot = Mvcm + K vib + ks + 2mc 2
2
2
Rotational kinetic energy
- Net momentum = 0
- Energy is constant
Erot = K rot
Motion around of center of mass
Rotation and vibration
pi
K rel
2
N æ p2
+
p
pi2
rad ,i
=å
= å çç tan,i
2mi
i =1 2mi
i =1 è
K rel
2
N æ p2
æ ptan,
ö
ö
i
rad ,i
= å çç
÷÷ + å çç
÷÷
i =1 è 2mi ø
i =1 è 2mi ø
N
N
CM
K rot º
K vib º
Rotation and vibration and translation
1 2
1
Etot = Mvcm
+ K rot + Kvib + ks 2 + 2mc 2
2
2
ö
÷÷
ø
Gravitational potential energy of a multiparticle system
U g = m1 gy1 + m2 gy2 + m3 gy3 + ...
U g = (m1 y1 + m2 y2 + m3 y3 + ...) g
= Mycm
U g = Mgycm Gravitational
energy near the
Earth’s surface
M
ycm
Example: Rotation and translation
Ktot = Ktrans + Krot
1
2
Mvcm
2
EXAMPLE:
Assume all mass
is in the rim
1
2
Mvrel
2
1
1
2
2
Ktot = Mvcm
+ Mvrel
2
2
Energy principle:
1 2
1 2
1 2
1 2
Mvcm,i + Mvrel ,i + Mgycm,i = Mvcm, f + Mvrel , f + Mgycm, f
2
2
2
2
vcm = vrel
=0
=0
2
Mg Dycm = Mvcm
,f