Download Lecture 16

Phys 207
Announcements
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Hwk 6 is posted online; submission deadline = April 4
Exam 2 on Friday, April 8th
Today’s Agenda
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Freshman Interim Grades
Review
Work done by variable force in 3-D
Í Newton’s gravitational force
Conservative forces & potential energy
Conservation of “total mechanical energy”
Í Example: pendulum
1
Work by variable force in 3-D:
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Work dWF of a force F acting
through an infinitesimal
displacement ∆r is:
F
∆r
.
dW = F ∆r
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The work of a big displacement through a variable force will
be the integral of a set of infinitesimal displacements:
∫.
WTOT = F ∆r
2
Page 1
Work by variable force in 3-D:
Newton’s Gravitational Force
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Work dWg done on an object by gravity in a displacement dr is
given by:
.
.
^
^ (dR r^ + Rdθ θ)
dWg = Fg dr = (-GMm / R2 r)
dWg = (-GMm / R2) dR
.
.
^
(since r^ θ = 0, r^ r^ = 1)
dR
Rdθ
Fg
^
θ
r^
dr
m
dθ
R
M
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Work by variable force in 3-D:
Newton’s Gravitational Force
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Integrate dWg to find the total work done by gravity in a “big”
displacement:
∫
R2
∫
R2
Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
R1
R1
Fg(R2)
m
R2
Fg(R1)
R1
M
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Page 2
Work by variable force in 3-D:
Newton’s Gravitational Force
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Work done depends only on R1 and R2, not on the path taken.
⎛ 1
1 ⎞⎟
−
Wg = GMm⎜⎜
⎟
⎝ R2 R1 ⎠
m
R2
R1
M
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Lecture 16, Act 1
Work & Energy
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A rock is dropped from a distance RE above the surface of the
earth, and is observed to have kinetic energy K1 when it hits
the ground. An identical rock is dropped from twice the height
(2RE) above the earth’s surface and has kinetic energy K2
when it hits. RE is the radius of the earth.
ÍWhat is K2 / K1?
(a)
2
(b)
3
2
2RE
(c)
4
3
RE
RE
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Page 3
Lecture 16, Act 1
Solution
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Since energy is conserved, ∆K = WG.
⎛ 1
1 ⎞
WG = GMm ⎜⎜
− ⎟⎟
R
R
⎝ 2
1⎠
⎛ 1
1 ⎞
⎟⎟
∆K = c ⎜⎜
−
R
R
⎝ 2
1⎠
Where c = GMm is the
same for both rocks
2RE
RE
RE
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⎛ 1
1 ⎞
⎟⎟
∆K = c ⎜⎜
−
R
R
⎝ 2
1⎠
Lecture 16, Act 1
Solution
⎛ 1
1 ⎞
1 1
⎟⎟ = c ⋅ ⋅
−
R
2R
2 RE
⎝ E
E ⎠
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For the first rock: K1 = c ⎜⎜
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⎟⎟ = c ⋅ ⋅
−
For the second rock: K 2 = c ⎜⎜
3 RE
⎝ R E 3R E ⎠
⎛ 1
So:
1 ⎞
2
1
2
K2 3 4
= =
K1 1 3
2
2RE
RE
RE
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Page 4
Newton’s Gravitational Force
Near the Earth’s Surface:
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Suppose R1 = RE and R2 = RE + ∆y
⎛ R − R1 ⎞
⎛ (R + ∆y ) − (RE ) ⎞
⎛
⎞
⎟ = −GMm ⎜ E
⎟ ≅ −m⎜ GM ⎟ ∆y
Wg = −GMm ⎜⎜ 2
2
⎟
⎜ (R + ∆y )(R ) ⎟
⎜ R ⎟
⎝ R1R2 ⎠
⎝
⎝ E ⎠
⎠
E
E
⎛ GM ⎞
but we have learned that ⎜ 2 ⎟ = g
⎝ RE ⎠
So:
Wg = -mg∆y
m
RE+ ∆y
RE
M
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Conservative Forces:
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We have seen that the work done by gravity does not
depend on the path taken.
m
⎛ 1
1 ⎞⎟
−
Wg = GMm⎜⎜
⎟
⎝ R2 R1 ⎠
R2
R1
M
m
h
Wg = -mgh
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Page 5
Conservative Forces:
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In general, if the work done does not depend on the path
taken (only depends the initial and final distances between
objects), the force involved is said to be conservative.
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⎛ 1
1 ⎞⎟
−
Gravity is a conservative force: Wg = GMm⎜⎜
⎟
⎝ R2 R1 ⎠
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Gravity near the Earth’s surface: W g = − mg ∆ y
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A spring produces a conservative force: Ws = −
1
k (x22 − x12 )
2
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Conservative Forces:
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We have seen that the work done by a conservative force
does not depend on the path taken.
W2
W1 = W2
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Therefore the work done in a
closed path is 0.
WNET = W1 - W2
= W1 - W1 = 0
W1
W2
W1
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Page 6
Lecture 16, Act 2
Conservative Forces
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The pictures below show force vectors at different points in
space for two forces. Which one is conservative ?
(a) 1
(b) 2
y
(c) both
y
x
x
(1)
(2)
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Lecture 16, Act 2
Solution
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Consider the work done by force when moving along different
paths in each case:
W A = WB
W A > WB
(1)
(2)
Page 7
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Lecture 16, Act 2
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In fact, you could make money on type (2) if it ever existed:
ÍWork done by this force in a “round trip” is > 0!
ÍFree kinetic energy!!
WNET = 10 J = ∆K
W=0
W = 15 J
W = -5 J
Note: NO REAL
FORCES OF THIS
TYPE EXIST, SO
FAR AS WE KNOW
W=0
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Potential Energy
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For any conservative force F we can define a
potential energy function U in the following way:
W =
∫ F.dr = -∆U
Í The
work done by a conservative force is equal and
opposite to the change in the potential energy function.
r2
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U2
This can be written as:
r2
∆U = U2 - U1 = -W = -
∫F.dr
r1
r1
U1
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Page 8
Gravitational Potential Energy
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We have seen that the work done by gravity near the Earth’s
surface when an object of mass m is lifted a distance ∆y is
Wg = -mg ∆y
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The change in potential energy of this object is therefore:
∆U = -Wg = mg ∆y
j
m
∆y
Wg = -mg ∆y
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Gravitational Potential Energy
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So we see that the change in U near the Earth’s surface is:
∆U = -Wg = mg ∆y = mg(y2 -y1).
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So U = mg y + U0 where U0 is an arbitrary constant.
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Having an arbitrary constant U0 is equivalent to saying that
we can choose the y location where U = 0 to be anywhere
we want to.
j
m
y2
y1
Wg = -mg ∆y
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Page 9
Potential Energy Recap:
For any conservative force we can define a potential
energy function U such that:
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S2
∆U = U2 - U1 = -W = -
∫F.dr
S1
The potential energy function U is always defined only
up to an additive constant.
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can choose the location where U = 0 to be
anywhere convenient.
Í You
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Conservative Forces & Potential Energies
(stuff you should know):
^
Fg = -mg j
Fg = −
GMm ^
r
R2
Fs = -kx
Change in P.E
∆U = U2 - U1
Work
W(1-2)
Force
F
mg(y2-y1)
-mg(y2-y1)
⎛ 1
1 ⎞⎟
GMm⎜⎜
−
⎟
⎝ R2 R1 ⎠
−
1
k (x22 − x12 )
2
P.E. function
U
mgy + C
⎛ 1
GMm
1 ⎞⎟
− GMm⎜⎜
−
−
+C
⎟
R
⎝ R2 R1 ⎠
1
k (x22 − x12 )
2
1 2
kx + C
2
(R is the center-to-center distance, x is the spring stretch)
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Page 10
Lecture 16, Act 3
Potential Energy
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All springs and masses are identical. (Gravity acts down).
ÍWhich of the systems below has the most potential energy
stored in its spring(s), relative to the relaxed position?
(a) 1
(b) 2
(c) same
(1)
(2)
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Lecture 16, Act 3
Solution
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The displacement of (1) from equilibrium will be half of that of (2)
(each spring exerts half of the force needed to balance mg)
0
d
2d
(1)
(2)
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Page 11
Lecture 16, Act 3
Solution
1 2
2
The potential energy stored in (1) is 2 ⋅ kd = kd
2
The potential energy stored in (2) is
1
2
k (2d) = 2kd2
2
The spring P.E. is
twice as big in (2) !
0
d
2d
(1)
(2)
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Conservation of Energy
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If only conservative forces are present, the total kinetic
plus potential energy of a system is conserved,
i.e. the total “mechanical energy” is conserved.
Í(note: E=Emechanical throughout this discussion)
E=K+U
∆E = ∆K + ∆U
⇒ using ∆K = W
= W + ∆U
= W + (-W) = 0 ⇒ using ∆U = -W
E = K + U is constant!!!
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Both K and U can change, but E = K + U remains constant.
But we’ll see that if non-conservative forces act then
energy can be dissipated into other modes (thermal,sound)
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Page 12
Example: The simple pendulum
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Suppose we release a mass m from rest a distance h1
above its lowest possible point.
Í What is the maximum speed of the mass and where
does this happen?
Í To what height h2 does it rise on the other side?
m
h1
h2
v
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Example: The simple pendulum
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Kinetic+potential energy is conserved since gravity is a
conservative force (E = K + U is constant)
Choose y = 0 at the bottom of the swing,
and U = 0 at y = 0 (arbitrary choice)
E = 1/2mv2 + mgy
y
y=0
h1
h2
v
Page 13
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Example: The simple pendulum
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E = 1/2mv2 + mgy.
Í Initially, y = h1 and v = 0, so E = mgh1.
Í Since E = mgh1 initially, E = mgh1 always since energy is
conserved.
y
y=0
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Example: The simple pendulum
z 1/2mv2
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will be maximum at the bottom of the swing.
1/ mv2 = mgh
So at y = 0
v2 = 2gh1
2
1
v =
2 gh1
y
y = h1
y=0
h1
v
Page 14
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Example: The simple pendulum
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Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum
height on the other side will be at y = h1 = h2 and v = 0.
The ball returns to its original height.
y
y = h1 = h2
y=0
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Example: The simple pendulum
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The ball will oscillate back and forth. The limits on its
height and speed are a consequence of the sharing of
energy between K and U.
E = 1/2mv2 + mgy = K + U = constant.
y
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Page 15
Example: The simple pendulum
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We can also solve this by choosing y = 0 to be at the
original position of the mass, and U = 0 at y = 0.
E = 1/2mv2 + mgy.
y
y=0
h1
h2
v
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Example: The simple pendulum
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E = 1/2mv2 + mgy.
Í Initially, y = 0 and v = 0, so E = 0.
Í Since E = 0 initially, E = 0 always since energy is conserved.
y
y=0
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Page 16
Example: The simple pendulum
z 1/2mv2
z
will be maximum at the bottom of the swing.
1/ mv2 = mgh
So at y = -h1
v2 = 2gh1
2
1
v =
2 gh1
y
Same as before!
y=0
y = -h1
h1
v
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Example: The simple pendulum
z
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Since 1/2mv2 - mgh = 0 it is clear that the maximum height
on the other side will be at y = 0 and v = 0.
The ball returns to its original height.
y
y=0
Same as before!
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Page 17