Transcript Energy Dissipation

Fall 2012
PHYS 172H: Modern Mechanics
Lecture 13 – Energy Dissipation
Chapter 7.10– 7.12
Annnouncement:
8:00-9:30 PM
EVENING EXAM II
TUES. OCT 23
Room 112
Energy dissipation
Euniverse = const
Energy dissipation: “loss” of energy into forms less useful to us
Example: air resistance:
h
Free fall of metal ball
1m
0s
Free fall of Paper
t
Terminal speed
Fair
dp
0
dt
mg
v  const
terminal speed
3
Mechanism of air resistance
Simple model:
In average, molecules hit object with larger
relative momentum from below than from
above, the difference is larger for higher
speeds.
Too simple!
Need to take into account macroscopic “wind”-like phenomenon
Fluid dynamics: mathematically complex, beyond the scope of this course
Factors determining air resistance
1. Speed of an object with respect to air?Increases with speed
2. Density of air?
Proportional to density
3. Mass of an object?
- not a factor
3. Area of an object?
Proportional to area
4. Shape of an object?
Depends on shape
Approximate air resistance force:
Fair   12 C  Av 2vˆ,
where 0.3  C  1.0
(for blunt objects, ordinary speeds)
NOTE: this is not a fundamental force law!
Fair  12 C  Av 2
Terminal speed
Fair
v  const
terminal speed
Fair  mg
1
2
C  Av2  mg
mg
vterminal
Person
ρ = 1.26 kg/m3
A = 0.8 m2
C = 0.5
m = 80 kg
Sqrt(19.6*80/0.4*1.26)=
56 m/s = 200 km/hour
2mg

C A
CLICKER: To decrease terminal speed a factor of two, what should be
done with the area of the parachute?
A) Area should be the same (this is wrong answer)
B) Area should be increased 2 times
C) Area should be increased 2 times
D) Area should be increased 4 times
E) Area should be decreased twice
Air resistance example: a car
Fair  12 C  Av 2 ,
where 0.3  C  1.0
http://www.atmosphere.mpg.de/enid/Information_ss/Velocity___air_drag_507.html
Fair  12 C  Av 2 ,
Air resistance example: a car
where 0.3  C  1.0
A=area
Car
C (drag)
Sports
0.27 – 0.31 – 0.38
Performance 0.32 – 0.34 – 0.38
60’s Muscle 0.38 – 0.44 – 0.50
Sedan
0.34 – 0.39 – 0.50
Motorcycle 0.50 – 0.90 – 1.00
Truck
0.60 – 0.90 – 1.00
Trailer
0.60 – 0.77 – 1.20
Examples:
Lamborghini Murcielago: C = 0.32
Toyota Prius 2006:
CLICKER POLL:
C=
A) 0.26 
B) 0.32
C) 0.38
D) 0.44
E) 0.50
Fair  12 C  Av 2 ,
Air resistance example: a car
where 0.3  C  1.0
A=area
Car
C (drag)
Sports
0.27 – 0.31 – 0.38
Performance 0.32 – 0.34 – 0.38
60’s Muscle 0.38 – 0.44 – 0.50
Sedan
0.34 – 0.39 – 0.50
Motorcycle 0.50 – 0.90 – 1.00
Truck
0.60 – 0.90 – 1.00
Trailer
0.60 – 0.77 – 1.20
Box-fish
C = 0.06
Mercedes concept car
C = 0.19
Motion through Earth’s atmosphere
Motion is complex
Spinning ball can curve:
Sliding friction
y
Ftable  FN  f
FN
Forces:
Fspring
Fpull
1. Non-contact: gravity
2. Contact: spring, table
f
mg
x
dpx dp y dpz
,
,
 ks s  f , FN  mg ,0
dt dt dt
=0
=0
Sliding friction force
f max
  FN vˆ
k
(4.8 & 7.10 in the book)
Coefficient of kinetic friction
Sliding friction: energy dissipation
y
Ftable  FN  f
FN
Sliding friction force
Fspring
Fpull
f max
  FN
k
f
Coefficient of kinetic
friction
mg
x
W  Fpull x  fmax x
k
Energy is dissipated – rising temperature
Irreversible – cannot transform it back to motion
Friction and irreversibility
No potential energy can be defined for friction:
internal energy of the objects changes
Is he jumping into water or
Is he kicked out from the water?
Reversibility
Movie taken and modified from: www.theclam.com/Video/JumpingIndex.asp
Friction and spring-mass system
y
dpx
 ks x
dt
+ friction
The energy is dissipated: damped oscillations
Special case: Viscous friction,
x  Ae
x
c

t
2m
cos t 
f  cv
ks  c 



m  2m 
2
Friction and spring-mass system
dpx
  k s x + friction
dt
f  cv
Special case: Viscous friction,
x  Ae
c

t
2m
cos t 
Note: if “free” ωF < c / 2m, ω
Overdamped, no oscillations.
ks  c 



m  2m 
2
becomes imaginary.
Energy goes as x2 or v2, and hence dies out as exp(-ct/m)
The Q value of the system is the number of oscillations which reduce the
energy to 1/e of its original value.
Large Q is when ωF >> c / 2m
Resonance
dpx
 ks  x  D sin Dt    cvx
dt
Driven frequency
Trial solution:
x  A sin Dt   
Can then solve for A and phase:
A
cos  
F2
F2  D2    cD / m 
2
D
2
F2  D2

2
F

   c
2 2
D
F  k s / m
D
/ m
2
Resonance
dpx
 ks  x  D sin Dt    cvx
dt
A
F2
  
2
F
2
D

2
  cD / m 
D
2
F  k s / m
x  A sin Dt   
cos  
F2  D2
  
2
F

2 2
D
  cD / m 
2
At resonance (drive frequency matches the
“free” undamped frequency) the response
is maximal. (Infinite if zero damping) And:
A = D ω2m/cω = D ω m/c
If the damping term c/m is small compared
to ω, the resonant response can be much
larger than the driving amplitude.
Armies march on bridges out of step
Tacoma Narrows Bridge film clip
Resonance in other systems
Radio
Maser
Laser
NMR, MRI
Photosynthesis