Physics 111: Lecture 11, Pg 1
Physics 111: Lecture 11
Today’s Agenda
Review Work done by variable force in 3-D
Newton’s gravitational force Conservative forces & potential energy Conservation of “total mechanical energy”
Example: pendulum Non-conservative forces
friction General work/energy theorem Example problem
Physics 111: Lecture 11, Pg 2
Work by variable force in 3-D:
Work dWF of a force F acting
through an infinitesimaldisplacement r is:
dW = F.r The work of a big displacement through a variable force will
be the integral of a set of infinitesimal displacements:
WTOT = F.r
F
r
ò
Physics 111: Lecture 11, Pg 3
Work by variable force in 3-D:Newton’s Gravitational Force
Work dWg done on an object by gravity in a displacement dr isgiven by:
dWg = Fg.dr = (-GMm / R2 r).(dR r + Rd)
dWg = (-GMm / R2) dR (since r. = 0, r.r = 1)
^ ^^
r^
^drRd
dR
R
Fg m
M
d
^ ^ ^ ^
Physics 111: Lecture 11, Pg 4
Work by variable force in 3-D:Newton’s Gravitational Force
Integrate dWg to find the total work done by gravity in a “big”displacement: Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
Fg(R1)
R1
R2
Fg(R2)
R1
R2
R1
R2
m
M
Physics 111: Lecture 11, Pg 5
Work by variable force in 3-D:Newton’s Gravitational Force
Work done depends only on R1 and R2, not on the path taken.
R1
R2
m
M
12
g R
1
R
1GMmW
Physics 111: Lecture 11, Pg 6
Lecture 11, Act 1Work & Energy
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.
What is K2 / K1?
(a)
(b)
(c)
32
43
2
RE
RE
2RE
Physics 111: Lecture 11, Pg 7
Lecture 11, Act 1Solution
Since energy is conserved, DK = WG.
RE
RE
2RE
12G R
1R1
GMm=W
12 R1
R1
c=ΔK
Where c = GMm is the same for both rocks
Physics 111: Lecture 11, Pg 8
Lecture 11, Act 1 Solution
For the first rock:
RE
RE
EEE1 R
121
c2R
1R1
c=K
12 R1
R1
c=ΔK
2RE
EEE2 R
132
c3R
1R1
c=K
For the second rock:
So:34
2132
=KK
1
2
Physics 111: Lecture 11, Pg 9
Newton’s Gravitational ForceNear the Earth’s Surface:
Suppose R1 = RE and R2 = RE + y
but we have learned that
So: Wg = -mgy
y
R
GMm
RyR
RyRGMm
RR
RRGMmW
2
EEE
EE
21
12g
GM
Rg
E2
RE+ y
M
m
RE
Physics 111: Lecture 11, Pg 10
Conservative Forces:
We have seen that the work done by gravity does not depend on the path taken.
R1
R2
M
m
hm
Wg = -mgh
12
g R
1
R
1GMmW
Physics 111: Lecture 11, Pg 11
Conservative Forces:
In general, if the work done does not depend on the path taken, the force involved is said to be conservative.
Gravity is a conservative force:
Gravity near the Earth’s surface:
A spring produces a conservative force: 2
1
2
2sxxk
2
1W
ymgWg
12
g R
1
R
1GMmW
Physics 111: Lecture 11, Pg 12
Conservative Forces: We have seen that the work done by a conservative force
does not depend on the path taken.
W1
The work done can be “reclaimed.”
W2
W1
W2
W1 = W2
WNET = W1 - W2 = 0
Therefore the work done in a closed path is 0.
Physics 111: Lecture 11, Pg 13
Lecture 11, Act 2Conservative Forces
The pictures below show force vectors at different points in space for two forces. Which one is conservative ?
(a) 1 (b) 2 (c) both
(1) (2)
y
x
y
x
Physics 111: Lecture 11, Pg 14
Lecture 11, Act 2Solution
Consider the work done by force when moving along different paths in each case:
(1) (2)
WA = WB WA > WB
Physics 111: Lecture 11, Pg 15
Lecture 11, Act 2 In fact, you could make money on type (2) if it ever existed:
Work done by this force in a “round trip” is > 0!Free kinetic energy!!
W = 15 J
W = 0
W = -5 J
W = 0WNET = 10 J = DK
Physics 111: Lecture 11, Pg 16
Potential Energy
For any conservative force F we can define a potential energy function U in the following way:
The work done by a conservative force is equal and opposite to the change in the potential energy function.
This can be written as:
W = F.dr = -U ò
U = U2 - U1 = -W = - F.dròr1
r2
r1
r2 U2
U1
Physics 111: Lecture 11, Pg 17
Gravitational Potential Energy
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
The change in potential energy of this object is therefore:
U = -Wg = mg y
ym
Wg = -mg y j
Physics 111: Lecture 11, Pg 18
Gravitational Potential Energy
So we see that the change in U near the Earth’s surface is: U = -Wg = mg y = mg(y2 -y1).
So U = mg y + U0 where U0 is an arbitrary constant.
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.
y1
m
Wg = -mg y j y2
Physics 111: Lecture 11, Pg 19
Potential Energy Recap:
For any conservative force we can define a potential energy function U such that:
The potential energy function U is always defined onlyup to an additive constant.
You can choose the location where U = 0 to be anywhere convenient.
U = U2 - U1 = -W = - F.drS1
S2
Physics 111: Lecture 11, Pg 20
Conservative Forces & Potential Energies (stuff you should know):
Force F
WorkW(1-2)
Change in P.E
U = U2 - U1
P.E. functionU
Fg = -mg j
Fg = r
Fs = -kx
^
^
12R
1
R
1GMm
2
1
2
2xxk
2
1
-mg(y2-y1) mg(y2-y1)
12R
1
R
1GMm
2
1
2
2xxk
2
1
2R
GMm
mgy + C
GMm
RC
1
22kx C
Physics 111: Lecture 11, Pg 21
Lecture 11, Act 3Potential Energy
All springs and masses are identical. (Gravity acts down). Which 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)
Physics 111: Lecture 11, Pg 22
Lecture 11, Act 3Solution
The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg)
(1) (2)
d 2d
0
Physics 111: Lecture 11, Pg 23
Lecture 11, Act 3Solution
(1) (2)
d 2d
0
The potential energy stored in (1) is 212
kd kd2 2
The potential energy stored in (2) is 12
k 2d 2kd2 2
The spring P.E. is twice as big in (2) !
Physics 111: Lecture 11, Pg 24
Conservation of Energy If only conservative forces are present, the total kinetic
plus potential energy of a system is conserved.
E = K + U is constant!!!
Both K and U can change, but E = K + U remains constant.
E = K + U
E = K + U = W + U = W + (-W) = 0
using K = Wusing U = -W
Physics 111: Lecture 11, Pg 25
Example: The simple pendulum
Suppose we release a mass m from rest a distance h1 above its lowest possible point. What is the maximum speed of the mass and where
does this happen? To what height h2 does it rise on the other side?
v
h1 h2
m
Physics 111: Lecture 11, Pg 26
Example: The simple pendulum
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
v
h1 h2
y
y = 0
Physics 111: Lecture 11, Pg 27
Example: The simple pendulum
E = 1/2mv2 + mgy. Initially, y = h1 and v = 0, so E = mgh1. Since E = mgh1 initially, E = mgh1 always since energy is
conserved.
y
y = 0
Physics 111: Lecture 11, Pg 28
Example: The simple pendulum
1/2mv2 will be maximum at the bottom of the swing. So at y = 0 1/2mv2 = mgh1 v2 = 2gh1
v
h1
y
y = h1
v gh 2 1
y = 0
Physics 111: Lecture 11, Pg 29
Example: The simple pendulum
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
Physics 111: Lecture 11, Pg 30
Example: The simple pendulum
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
Bowling
Physics 111: Lecture 11, Pg 31
Example: The simple pendulum
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.
v
h1 h2
y
y = 0
Physics 111: Lecture 11, Pg 32
Example: The simple pendulum
E = 1/2mv2 + mgy. Initially, y = 0 and v = 0, so E = 0. Since E = 0 initially, E = 0 always since energy is conserved.
y
y = 0
Physics 111: Lecture 11, Pg 33
Example: The simple pendulum
1/2mv2 will be maximum at the bottom of the swing. So at y = -h1 1/2mv2 = mgh1 v2 = 2gh1
v
h1
y
y = 0
y = -h1
v gh 2 1
Same as before!
Physics 111: Lecture 11, Pg 34
Example: The simple pendulum
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!
Galileo’sPendulum
Physics 111: Lecture 11, Pg 35
Example: Airtrack & Glider
A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ?
d
M
m
v
v
Physics 111: Lecture 11, Pg 36
Example: Airtrack & Glider
Kinetic+potential energy is conserved since all forces are conservative.
Choose initial configuration to have U=0.
K = -U
d
M
m
v
1
22m M v mgd
Glider
Physics 111: Lecture 11, Pg 37
Problem: Hotwheel
A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1, rises a distance h, and ends up moving horizontally with speed v2. Find v1 and v2.
hd
v1
v2
Physics 111: Lecture 11, Pg 38
Problem: Hotwheel...
K+U energy is conserved, so E = 0 K = - U Moving down a distance d, U = -mgd, K = 1/2mv1
2
Solving for the speed:
hd
v1
v gd1 2
Physics 111: Lecture 11, Pg 39
Problem: Hotwheel...
At the end, we are a distance d - h below our starting point. U = -mg(d - h), K = 1/2mv2
2
Solving for the speed:
hd
v2
v g d h2 2
d - h
Physics 111: Lecture 11, Pg 40
Non-conservative Forces:
If the work done does not depend on the path taken, the force is said to be conservative.
If the work done does depend on the path taken, the force is said to be non-conservative.
An example of a non-conservative force is friction.When pushing a box across the floor, the amount of
work that is done by friction depends on the path taken.» Work done is proportional to the length of the path!
Physics 111: Lecture 11, Pg 41
Non-conservative Forces: Friction
Suppose you are pushing a box across a flat floor. The mass of the box is m and the coefficient of kinetic friction is k.
The work done in pushing it a distance D is given by:
Wf = Ff • D = -kmgD.
D
Ff = -kmg
Physics 111: Lecture 11, Pg 42
Non-conservative Forces: Friction
Since the force is constant in magnitude and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just Wf = -mgL.
Clearly, the work done depends on the path taken.
Wpath 2 > Wpath 1
A
B
path 1
path 2
Physics 111: Lecture 11, Pg 43
Generalized Work/Energy Theorem:
Suppose FNET = FC + FNC (sum of conservative and non-conservative forces).
The total work done is: WNET = WC + WNC
The Work/Kinetic Energy theorem says that: WNET = K. WNET = WC + WNC = K WNC = K - WC
But WC = -U
So WNC = K + U = E
Physics 111: Lecture 11, Pg 44
Generalized Work/Energy Theorem:
The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. E=K+U of system not conserved!
If all the forces are conservative, we know that K+U energy is conserved: K + U = E = 0 which says that WNC = 0, which makes sense.
If some non-conservative force (like friction) does work,K+U energy will not be conserved and WNC = E, which also makes sense.
WNC = K + U = E
Physics 111: Lecture 11, Pg 45
Problem: Block Sliding with Friction
A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is k. How far, x, does the block go along the bottom portion
of the track before stopping?
x
d k
Physics 111: Lecture 11, Pg 46
Problem: Block Sliding with Friction...
Using WNC = K + U As before, U = -mgd WNC = work done by friction = -kmgx. K = 0 since the block starts out and ends up at rest. WNC = U -kmgx = -mgd
x = d / k
x
d k
Physics 111: Lecture 11, Pg 47
Recap of today’s lecture
Work done by variable force in 3-D (Text: 6-1) Newton’s gravitational force (Text: 11-2)
Conservative Forces & Potential energy (Text: 6-4)
Conservation of “Total Mechanical Energy” (Text: 7-1) Examples: pendulum, airtrack, Hotwheel car
Non-conservative forces (Text: 6-4) friction
General work/energy theorem (Text: 7-2)
Example problem
Look at Textbook problems Chapter 7: # 9, 21, 27, 51, 77