ch.12 energy ii: potential energy ch. 12 energy ii: potential energy
TRANSCRIPT
Ch.12 Energy II:Potential energy
12-1 Conservative forces( 保守力 )
Potential energy? It is defined only for a certain class of forces called conservative forces.
Do spring force, gravitational force, and frictional force et al. belong to conservative forces?
Kinetic energy Velocity
What are conservative forces?
Ch.12 Energy II:Potential energy
1. The spring force Fig 12-1
x
0
0
d
d
-d
(a)
(b)
(c)
(d)
(e))(
2
1 22ifs xxkW
ox
Relaxed length
Fig 11-13
The total work done by the spin force is zero in the process from (a) to (e) (round trip).
0
0
Ch.12 Energy II:Potential energy
The total work done by the gravity is zero during the round trip.
2. The force of gravity
If the gravitational force is not constant, is there still such behavior of the work?
See See 动画库动画库 // 力学夹力学夹 /2-04/2-04功功的计算举例的计算举例 (2)(2)
Ch.12 Energy II:Potential energy
3. The frictional force
The total work done by frictional force is not zero in a round trip.
See See 动画库动画库 // 力学夹力学夹 /2-04/2-04功功的计算举例的计算举例 (1)(1)
Ch.12 Energy II:Potential energy
Definition of conservative force:
One particle exerted by a force moves around a closed path and returns to its starting point.
If the total work done by the force during the round trip is zero, we call the force ‘a conservative force’ , such as spring force and gravity.
If not, the force is a nonconservative one.
Ch.12 Energy II:Potential energy
Two Mathematical statements:
a
a
b
b1
2
2
1
Fig 12-4
(a)
(b)
If is a conservative force, we have:
F
0WW ba.2ab.1
0sdFsdFa
b
b
a
(12-1)Path1 Path2
Statement 1
sdFsdFsdFb
a
a
b
b
aPath1 Path2Path2
(12-3)
0 sdF
Statement 2
sdFsdFb
a
b
aPath1 Path2
Ch.12 Energy II:Potential energy
To every action, there is an equal and opposite To every action, there is an equal and opposite reaction.reaction.
Newton’s third lawNewton’s third law
Note:
(1) Both the action and reaction forces belong to the system.(2) The total work done by action and reaction forces is independent of the reference frame chosen (even in non-inertial frame).
Prove point (2):
Ch.12 Energy II:Potential energy
1f
2f
O
1r
2r
z
y
x
m1 m2221121 rdfrdfWWW
SIn S frame:
In S’ frame with velocity ofrelative to S frame:
ssv '
'rdf'rdf'W'WW' 221121
)()( 2211 dtvrdfdtvrdf s'ss's
dtv)ff(rdfrdf s's
212211 21 ff
2211 rdfrdf
W
Ch.12 Energy II:Potential energy
12-2 Potential energy1.Definition
When work is done in a system (such as ball and earth) by a conservative force, the configuration of its parts changes, and so the potential energy changes from its initial value to its final value . We define the change in potential energy associated with the conservative force as:
iU fU
sdFWUUΔU if
(12-4)
Ch.12 Energy II:Potential energy
2. The potential energy of gravityFor the ball-Earth system, we take upward direction to be y positive direction
)()()()( 12
2
112 yymgdymgyUyUUy
y
The physically important quantity is ,ΔU not or .)( 2yU )( 1yU
0)0( 1 yU
mgyyU )( (12-9)
If We set
We have
(the reference zero point of U is at O)
y2
y1
y
sdFWUUΔU if
mg
)()()( 1212 yymgyUyU , dependent on )( 1yU
Ch.12 Energy II:Potential energy
3. The potential energy of spring force
00 u
x
kx)dx(FdxU(x)0
0
2
2
1)( kxxU
o x
Relaxed length
Fig 11-13
When the spring is in its relaxed state, and we can declare the potential energy of the system to be zero ( )
(12-8)
The reference zero point of potential is at x=0.
2
1
Ch.12 Energy II:Potential energy
i.The physically important quantity is . Not or .
Notes:
01 xx UUΔU
1xU 0xU
iii.Potential energy belongs to the system (Such as ball-Earth) and not of any of the individual objects within the system.
ii.We are free to choose the reference point at any convenient location for the potential energy.
Ch.12 Energy II:Potential energy
iv. The inverse of Eq(12-4) allows us to calculate the force from the potential energy (12-7)
dx
xdUxFx
)()(
x
x FdxUUΔU00
Eq(12-7) gives us another way of looking at the potential energy:
“The potential energy is a function of position whose negative derivative gives the force”
Eq(12-4)
mgyyU )( yFmgdy
ydU
)(
2
2
1)( kxxU xFkxkx
dx
d
dx
dU )
2
1( 2
Ch.12 Energy II:Potential energy
An elevator cab of mass m=920 Kg moves from street level to the top of the World Trade Center in New York, a height of h=412 m above ground. What is the change in the gravitational potential energy of the cab-Earth system?
Sample problem 12-1
.107.34128.9920 6 JmghymgU
Ch.12 Energy II:Potential energy
12-3 Conservative of mechanical energy
dxxFWUUΔU if )(
KΔU K
mvmvW ifnetF
22
2
1
2
1
From the definition of potential energy, we have:
K
0 KU (12-14)
0)()( ifif KKUU
ffii UKUK (12-15)
Mechanical energy
When can Eq. (12-15) be satisfied?
is a conservative forceF
Ch.12 Energy II:Potential energy
Eq(12-15) is the mathematical statement of the law of conservation of mechanical energy:
“In a system in which only conservative forces do work, the total mechanical energy of the system remains constant.”
Such as the systems of: Ball-Earth system; Block-spring system on frictionless table.
Ch.12 Energy II:Potential energy
How to write the formula of conservation of mechanical energy for:
Ball-Earth system / m+M system (M>>m) ?
No other forces exerted in the system.
.2
1
2
1)( 22 constMVmvRrU
m and v are the mass and speed for Ball, respectively.M and V are the mass and speed for Earth, respectively.
.2
1 2 constmvmgh or
Which one is correct or both correct?(1)If of M is zero, M is an inertial frame. Take M as our reference frame. Two eqs. are equivalent.
a
(2) If of M is not zero, CM of the system is an inertial frame. Since M>>m, the position of CM is very near M. In CM frame, V~0, R~0. So two eqs. are equivalent.
a
Ch.12 Energy II:Potential energy
Sample problem 12-5
Using conservation of mechanical
energy, analyze the Atwood’s
machine (sample problem 5-5) to
find v and a of the blocks after they
have moved a distance y from rest.
Solution: We take the two blocks
plus the Earth as our system. For
simplicity, we assume that both
o
y
1m 2m
Ch.12 Energy II:Potential energy
blocks start from rest at the same level, which we define as y=0, the reference point for gravitational potential energy.
Thus
Solving for the speed v, we obtain
0)2
1()
2
1( 2
221
21 gymvmgymvmUK ff
0,0 ii UK
gymm
mmv
21
122
gmm
mm
dt
dvay
21
12
,0 ii UK
Ch.12 Energy II:Potential energy
When the climber goes down, she must transfer potential energy to other kinds of energy, such as thermal energy.
Ch.12 Energy II:Potential energy
12-4 Energy conservation in rotational motion
We still restrict our analysis to the case in which the rotational axis remains in
the same direction in space as
the object moves. Fig 12-7 shows an arbitrary body of mass M .
Fig12-7
0
y
x
cmr
nr
'nr
c
p
Ch.12 Energy II:Potential energy
1. Relative to o, the kinetic energy of is ,
and total kinetic energy of the body is
(12-16)
From Fig12-7,we see that Then:
nm2
2
1nnvm
2
1 2
1nn
N
n
vmK
'ncmn rrr
'ncmn vvv
)2(2
1
)()(2
1
2
1
'2'2
1
''2
ncmncmn
N
n
ncmncmnnn
vvvvm
vvvvmvmK
(12-17)
Ch.12 Energy II:Potential energy
In Eq(12-17):
(1) ( )
(2)
(3)
22
2
1
2
1cmcmn Mvvm Mmn
22'2'
2
1)(
2
1
2
1 cmnnnn Irmvm
0)2(2
1 ''
nncmncmn vmvvvm
)'(
cmnn vvv
Ch.12 Energy II:Potential energy
Thus (12-18)
Eq(12-18) indicate that the total kinetic energy of the moving object consists of two terms, the pure translational of Cm, and the pure rotation about Cm. The two terms are quite independent.2.Rolling without slipping For this case,
22
2
1
2
1 cmcm IMvK
Rvcm 22222
2
1
2
1)(
2
1
2
1 cmcm
cmcm IRMR
vIMvK (12-19)
Ch.12 Energy II:Potential energy
3.When an object rolls without slipping, there may be a frictional force exerted at the instantaneous point of contact between the object and the surface on which it rolls. However, this frictional force does no work on the object.
Ch.12 Energy II:Potential energy
Sample problem 12-8
Using energy conservation
find the final speed of the
rolling cylinder in Fig 9-32
when it reaches the bottom
of the plane.
h
mg
fN
c
Fig 9-32
Ch.12 Energy II:Potential energy
Solution: For our system we take the cylinder and
the Earth. The friction does no work and so it cannot change the mechanical energy. ,
Setting
with
mghUKE iii 0
0)(2
1
2
1 22 R
vIMvUKE cmcmcmfff
ghvcm 3
4
2
2
1MRI cm
mghR
vIMv mcmcm 22 )(
2
1
2
1fi EE
Ch.12 Energy II:Potential energy
12-5 One-dimensional conservation system: the complete solution1. How to read the curve of potential energy? If only conservative forces do work in the
system, we have , (E is constant) (12-20)UKE
EmvxU x 2
2
1)(
(12-21))]([2
xUEm
vx )(xUE
Ch.12 Energy II:Potential energy
dx
dUF
6x 5x
(a)
4x3x 2x1x
0E
3E
4E
F
1E
(b)
2E
0x
Fig12-8
U(x)
)]([2
xUEm
vx
EU
K=E-U
x
x
Ch.12 Energy II:Potential energy
Sample problem 12-10
The potential energy function for the force between
two atoms in a diatomic molecule can be expressed approximately as follows
where a and b are positive constant and x is the distance between atoms.(a)Find the equilibrium separation between the atoms, (b) the force between the atoms, and (c) the minimum energy necessary to break the molecule apart.
612)(
x
b
x
axU
Ch.12 Energy II:Potential energy
Solution: (a) Fig12-9 shows u(x) as a function of x.
Equilibrium occurs at the coordinate
, where is a
minimum.
that is
mx)( mxU
0)( mxxdx
dU
0612713
mm x
b
x
a 6
1
)2(b
axm
x
x0
0
U
(a)
(b)
mx
xF
Fig (12-9)
Ch.12 Energy II:Potential energy
(b)
(c) The minimum energy needed to break up the
molecule into separate atoms is called the
dissociation energy , .
713
612)(
x
b
x
a
dx
dUxFx
0)( ExUE mddE
a
bEd 4
2
612)(mm
mdx
b
x
axUE ))
2(( 6
1
b
axm
Ch.12 Energy II:Potential energy
2. General solution for x(t)From Eq(12-21) ,with we have (12-22))]()[/2( xUEm
dxdt
dt
dxvx
)]()[/2(0 xUEm
dxt
x
x
Suppose , , we have (12-23)After carring out the integration of Eq(12-23), we would obtain t=t(x), or x=x(t).
00)( xx t xx t )(
)]([2
xUEm
vx
Ch.12 Energy II:Potential energy
In one dimension, the magnitude of the gravitational force of attraction between two particles of mass m1 and mass m2 is given by
where G is a constant and x is the distance between the particles. (a) What is the potential energy function U(x)? Assume that U(x)0, as x ∞. (b) How much work is required to increase the separation of the particles form x=x1 to x=x1+d?
221
x
mmGFx
Exercise: