Transcript potential energy - s3.amazonaws.com

This Week
7/12 Lecture – Chapter 8
 7/13 Recitation – Bungee

 Problems:

8.4, 8.23, 8.44
7/14 Lab – Kinematics in 1-D
 Homework
#4 Due @ 5pm
7/15 Lecture – Chapter 9
 7/16 Recitation – ???

 Problems:
7/12/04
???
1
Chapter 8
Potential Energy and
Conservation of Energy
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2
Review:
Kinetic Energy:
Work (const force):
Work (variable force):
Work-KE theorem
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K  mv
1
2
2
 
W  F  r  Fr cos
 
W   F  dl
K  W
3
Potential Energy

What happens to the energy I use when I
do work on an object?
know about kinetic energy…
 In certain circumstances, it is stored as
potential energy
 Already
U  W
 In
the case of non-conservative forces, the
energy can be lost (usually as heat)
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Non-Conservative Forces
Non-conservative if the force does not
reverse the energy transfer when the path
is reversed
 Path dependent!
 Examples:

 Friction
 Air
resistance
 John
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Kerry ????
5
Conservative Forces
Conservative force does reverse the
energy transfer when the path is reversed
 Path independent, work depends only
upon position


Examples:
 Gravity
 Springs
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Gravitational Potential Energy
Recall the work done lifting an object against
gravity:


Wgrav  Fgrav  d  mgh
Lifting:


Dropping: Wgrav  Fgrav  d  mgh
Ugrav = -W = +mgh (if lifting)
Ugrav = -W = -mgh (if dropping)
h
d
Note: We get the same energy
out as we put in (conservative)
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Gravitational Potential Energy
An easier way to get the sign right…
U  mgy
U 2  U1  mgy2  mgy1
So the gravitational potential energy can be
written as:
U  mgy
But isn’t the choice of y = 0 arbitrary?
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Potential Energy from Spring
Energy associated with distortion of spring from
equilibrium length
F
x=0
x=xf
xf
Wspring   F ( x)dx  
0
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xf
0
1
2
(kx)dx   kx f
2
1
2
U spring  Wspring  kx f
2
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Important Points About Potential Energy
1. U depends on the position/arrangement
of objects
U between two arrangements does
not depend on path
2. Only U has physical meaning -- the
numerical value of U itself is arbitrary
This means you decide for yourself where
U = 0 (like you decide where x = 0)
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Important Points about Potential Energy
3. U is only defined for "conservative" forces, which do
not dissipate energy
No U for frictional forces
4. Can rewrite conservation of energy:
Old: Kinitial + Wtot = Kfinal
New: Kfinal – Kinitial = Wtot = Wc + Wnc
But: Wc = -ΔU = Uinitial – Ufinal
 Kinitial + Uinitial + Wnc = Kfinal + Ufinal
Define E = K + U  Einitial + Wnc = Efinal
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Conservation of Mechanical Energy
If there are no non-conservative forces:
Kinitial + Uinitial + Wnc = Kfinal + Ufinal
This gives us conservation of mechanical energy:
Kinitial + Uinitial = Kfinal + Ufinal
Einitial = Efinal
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Example: Ball Thrown Upwards
(neglect air resistance)
E = K + U = ½mv2 + mgy = constant
Set U = 0 at floor
vmax
ymax
y = ymax  E = U = mgymax
y = 0  E = K = ½mvmax2
Energy
E
U
K
vmax = 2gymax
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y
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Example: Pig on a Curved Track
No friction
Starts from rest
How fast is it going at
the bottom?
h
Kinitial + Uinitial = Kfinal + Ufinal
Kfinal = Uinitial - Ufinal = - U
Kfinal = mgh
½mvf2 = mgh
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Only endpoints
matter, curve of
track doesn't!
vf = 2gh
14
Pendulum: Qualitative View
Turning Points
energy

E
K
U

K can never be negative
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Motion is bounded
15
Pendulum Problem
L

Lcosθ
If I release the pendulum at rest from
an angle , how fast is it going at the
bottom?
L-Lcosθ
At bottom: y = 0
(set U = 0 there)
At angle : y = L – Lcos 
Kinitial + Uinitial = Kfinal + Ufinal
mg(L – Lcos ) = ½mv2
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v = 2gL(1 – cos )
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Defenestration
n. [Lat., de-,”out of”; fenstra, “window”.]
An act of throwing something or someone
out of a window.
Traditional political custom: Prague
1419
A popular uprising led by the priest
Jan Zelivsky included the throwing
of the city councilors from the windows
of the New Town Hall.
1618
The governors of Bohemia attempted to crush
Czech Protestantism. They were thrown from the
windows of the council room in Hradcany. This
event helped precipitate the Thirty Years War.
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Potential Energy Example
If the royal councilors were
given the heave-ho at 5 m/s
who is going fastest when
they hit the ground?
a
b
c
Ei=Ef
Ei = Ki+Ui = ½mvi2 + mgh
Ef = Kf + Uf = ½mvf2
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½mvi2 + mgh = ½mvf2
vf2 = vi2 + 2gh
They all hit the ground with the same velocity!
18
Graphical Representation of Energy
For a closed system: E = K + U = constant
Can plot U(x) to see how system evolves
Since K cannot be negative, motion is bounded by E
U
bounds
E1
trapped in well
E2
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x
19
Force From Potential Energy
-U  Work done by force
 U  Fx
U
F 
x
In the limit of small x, we get:
dU
F 
dx
Why the minus sign?
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Force From Potential Energy
dU
F 
dx
Force pushes in direction to
decrease potential energy,
i.e., "downhill"
U
F
F
F=0
x
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dU/dx < 0
F>0
dU/dx > 0
F<0
21
How High to “Loop the Loop”?
Ei = mgh
= Ef = mg(2R) + ½mv2
Centripetal: v2/R = g
(minimum v, barely loops)
mgh = 2mgR + ½mRg
v
h
R
h= 2R + 0.5R = 2.5R
(minimum)
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Pig Sliding Up a Frictional Plane
mk = 0.3
v0 = 4 m/s
 = 30
How high does he get?
Set U = 0 at bottom
Kinitial + Uinitial + Wfriction = Kfinal + Ufinal
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Example (continued)
d
Kinitial + Wfriction = Ufinal
h
 = 30
Kinitial = ½mv02
Wfriction = -|Ff|d
= -mkNd
= -mk (mgcos) d
Ufinal = mgh
= mgdsin
½mv02 – mk mgd cos = mgd sin
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Example (continued)
1
2
v0  m k gd cos   gd sin 
2
2
v0
d ( m k g cos   g sin  ) 
2
2
v0
d
2( m k g cos   g sin  )
d
h
 = 30
1
(4 m / s) 2
d
2 (0.3)(9.8 m / s 2 ) cos 30  (9.8 m / s 2 ) sin 30
 1.07 m
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Constraint Forces
The normal force and tension, in many
situations, exert forces which are
perpendicular to the motion
Motion on a fixed surface (e.g. track)
Tension from a rope with one end fixed
(e.g. pendulum)
If F•dx is always zero, no work can be done!
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Constraint Forces
No motion in direction of constraint
No work
No potential energy
Example - pole vaulting
Originally invented by
Dutch farmers
1
Emech = mvi2
2
1 2
mvi = mgymax
2
vi2
ymax =
2g
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Energy in the Pole Vault
More detail: pole acts as a spring
Energy
Pole
mgh
Kinetic
Time
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Estimating the Record
Fast sprinter travels ~10 m/s
Vaulter running:
Erun=½mv2
At top of motion:
Etop=mgh
½mv2=mgh
Record: 6.14 m Sergey
Bubka, 1994.
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h=v2/2g
About 5 m.
Remember-center of mass
is about 1m up at start.
29
Another Cultural
Moment…
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Cow-a-pult
400 kg
Say that a spring of
constant k = 104 N/m is
stretched 2 m to launch
the cow. What is the
max range of the cow if it is released 5m above the
ground?
Uspring,i+Ugrav,i+Ki = Uspring,f+Ugrav,f+Kf
U spring,i  12 kx2
U grav, f  mgh
K f  12 mv2
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Cow-a-pult (continued)
1
2
kx2  mgh  12 mv2
kx2  2mgh
v
m
(10 4 N / m)( 2 m) 2  2(400 kg)(9.8 m / s 2 )(5 m)
v
 30 m / s
400 kg
Maximum range when  = 45:
v0 sin 2 (30 m / s ) 2
R

 92 m
2
g
9.8 m / s
2
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Example:
 = 30
A 2 kg piglet on rough plane is compressing a spring
by x = 0.1 meters and is released from rest on this
plane (mk = 0.5).
If v = 4 m/s after traveling 6m, what is k of the spring?
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Example (continued)
 = 30
Kinitial + Uinitial + Wfriction = Kfinal + Ufinal
Ugravity = 0 at start
Uspring = ½kx2 at start
Ugravity = mgd sin at end
Uspring = 0 at end
Wfriction = -|Ff|d
= -mkNd
= -mkmgd cos
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Kinitial = 0
Kfinal = ½mv2
34
Example (continued)
 = 30
Kinitial + Uinitial + Wfriction = Kfinal + Ufinal
0  12 kx2  m k mgd cos   12 mv 2  mgd sin 
1
2
kx2  12 mv2  mgd sin   m k mgd cos 
mv2  2mgd sin   2m k mgd cos
4
k

2
.
7

10
N /m
2
x
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A More Complicated Potential
The bow provides a non-constant
force to the arrow
F
x
What is the kinetic energy of the arrow?
What is the final speed of the arrow?
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Arrows of Outrageous Fortune…
Force vs Distance
Area = Uinitial = 50 J
200.00
Uinitial=Kfinal= ½mv2
180.00
160.00
An arrow has a
mass ~ 0.03 kg…
Force (N)
140.00
120.00
100.00
80.00
60.00
v
40.00
20.00
0.00
0.00
0.10
0.20
0.30
0.40
0.50
2 K final
m
 58 m / s
Distance (m)
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Work Due to Gravity
Near the Earth
Away from the Earth
F  mg
m1 m2
F G 2
x
xf
xf
xi
xi
W =  F dx   -mg dx
xf
xf
m1 m2
W =  F dx    G 2 dx
x
xi
xi
xf
W   mgx x =  mg ( x f -xi )
xf
i
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 1
W =  G m1m2   
 x  xi
 1 1
= G m1m2   
x

x
f
i


38
The End of the World as We
Know It?
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Extinction!
70 Million years ago Dinosaurs
ruled the Earth
They disappeared at the boundary
between the Cretaceous and
Tertiary periods (K-T boundary)
Luis Alvarez
(a Nobel Prize winner in Physics)
suggested an asteroid impact
might be responsible
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The Impact Site
Alvarez calculated the asteroid
would need to be 10 km across
and would leave a crater
150 km in diameter
A huge crater off the Yucatan
peninsula of Mexico has been
identified as a possible impact
site. Research on this crater
has shown it is the result of an
extra-terrestrial impact.
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Assume an asteroid started at rest in the middle of the
inner Oort cloud (~5000 RE-S)
Assume it is acted on primarily by the Sun
Assume mass ~1016 kg (10 km rock)
 1 1
W = G ms ma   
x

 f xi 
= (6.672  10-11 N m 2 /kg 2 )(1.99 1030 kg)(1016 kg)
= 8.9  10 24
2K
v=
=
m
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1
1




11
14
 1.5 10 m 7.5  10 m 
J
2 (8.9  1024 J )
= 42,100 m/s
16
10 kg
1 Ton TNT
=4109J
Asteroid Impact
=2109 MT TNT
Over 80,000 MPH!
42
The Ball Race
1
2
d = h/sinθ
h
θ
Ball 1 falls straight down, ball 2 rolls down a plane
Which reaches the bottom first?
Which is traveling fastest at the bottom?
Ball 1: h=½gt2  t1= 2h/g
Ball 2: a||=g sinθ  t2= 2d/(gsinθ) = 2h/(gsin2θ) = t1/sinθ > t1
Balls 1,2: same Ki=0, Ui=mgh, Uf=0  same Kf  same vf
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