Download v Relate force to potential energy

Physics 207 – Lecture 16
Lecture 16
• Chapter 11 (Work)
v Employ conservative and non-conservative forces
v Relate force to potential energy
v Use the concept of power (i.e., energy per time)
• Chapter 12
v Define rotational inertia
v Define rotational kinetic energy
Assignment:
l HW7 due Tuesday, Nov. 1st
l For Monday: Read Chapter 12, (skip angular
momentum and explicit integration for center of mass,
rotational inertia, etc.)
Exam 2
7:15 PM Thursday, Nov. 3th
Physics 207: Lecture 16, Pg 1
Definition of Work, The basics
Ingredients: Force ( F ), displacement ( ∆ r )
Work, W, of a constant force F
acts through a displacement ∆ r :
W = F ·∆
∆r
F
θ
∆r
(Work is a scalar)
m
ce
a
l
p
dis
en
Work tells you something about what happened on the path!
Did something do work on you?
Did you do work on something?
Physics 207: Lecture 16, Pg 2
Page 1
t
Physics 207 – Lecture 16
Units:
Force x Distance = Work
Newton x Meter = Joule
[M][L] / [T]2 [L]
[M][L]2 / [T]2
mks
cgs
Dyne-cm or erg
= 10-7 J
N-m or Joule
Other
BTU = 1054 J
calorie= 4.184 J
foot-lb = 1.356 J
eV
= 1.6x10-19 J
Physics 207: Lecture 16, Pg 3
Net Work: 1-D Example
(constant force)
l
A force F = 10 N pushes a box across a frictionless floor
for a distance ∆x = 5 m.
F
Finish
Start
θ = 0°
∆x
l Net Work is F ∆x = 10 x 5 N m = 50 J
l1Nm
1 Joule (energy)
l Work reflects positive energy transfer
Physics 207: Lecture 16, Pg 4
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Physics 207 – Lecture 16
Net Work: 1-D 2nd Example
(constant force)
l
A force F = 10 N is opposite the motion of a box across a
frictionless floor for a distance ∆x = 5 m.
Finish
Start
F
θ = 180°
∆x
l
Net Work is F ∆x = -10 x 5 N m = -50 J
l
Work again reflects negative energy transfer
Physics 207: Lecture 16, Pg 5
Work: “2-D” Example
(constant force)
l
An angled force, F = 10 N, pushes a box across a
frictionless floor for a distance ∆x = 5 m and ∆y = 0 m
F
Finish
Start
θ = -45°
Fx
∆x
l (Net)
Work is Fx ∆x = F cos(-45°) ∆x = 50 x 0.71 Nm = 35 J
Physics 207: Lecture 16, Pg 6
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Physics 207 – Lecture 16
Exercise
Work in the presence of friction and non-contact forces
A box is pulled up a rough (µ > 0) incline by a rope-pulleyweight arrangement as shown below.
v How many forces (including non-contact ones) are
doing work on the box ?
v Of these which are positive and which are negative?
v State the system (here, just the box)
v Use a Free Body Diagram
v Compare force and path
v
A. 2
l
B. 3
C. 4
D. 5
Physics 207: Lecture 16, Pg 7
Work and Varying Forces (1D)
l
Area = Fx ∆x
F is increasing
Here W = F ·∆
∆r
becomes dW = F dx
Consider a varying force F(x)
Fx
x
∆x
xf
W
= ∫ F ( x ) dx
xi
Physics 207: Lecture 16, Pg 8
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Physics 207 – Lecture 16
Example: Work Kinetic-Energy Theorem with variable force
•
How much will the spring compress (i.e. ∆x = xf - xi) to bring
the box to a stop (i.e., v = 0 ) if the object is moving initially at a
constant velocity (vo) on frictionless surface as shown below ?
xf
Wbox
to vo
xi
xf
m
= ∫ − kx dx
Wbox
spring at an equilibrium position
xi
∆x
V=0
t
= ∫ F ( x ) dx
F
m
spring compressed
1
2
k ∆x 2 = 12 mv02
Physics 207: Lecture 16, Pg 9
Compare work with changes in potential energy
Consider the ball moving up to height h
(from time 1 to time 2)
l How does this relate to the potential energy?
l
Work done by the Earth’s gravity on
the ball)
W = F • ∆x = (-mg) -h = mgh
mg
h
∆U = Uf – Ui = mg 0 - mg h = -mg h
∆U = -W
mg
Physics 207: Lecture 16, Pg 10
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Physics 207 – Lecture 16
Conservative Forces & Potential Energy
l
If a conservative force F we can define a potential energy
function U:
W =
∫ F ·dr = - ∆U
The work done by a conservative force is equal and opposite to
the change in the potential energy function. r
Uf
f
(independent of path!)
ri
Ui
Physics 207: Lecture 16, Pg 11
A Non-Conservative Force, Friction
l Looking down on an air-hockey table with no air
flowing (µ > 0).
l Now compare two paths in which the puck starts
out with the same speed (Ki path 1 = Ki path 2) .
Path 2
Path 1
Physics 207: Lecture 16, Pg 12
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Physics 207 – Lecture 16
A Non-Conservative Force
Path 2
Path 1
Since path2 distance >path1 distance the puck will be
traveling slower at the end of path 2.
Work done by a non-conservative force irreversibly
removes energy out of the “system”.
Here WNC = Efinal - Einitial < 0
and reflects Ethermal
Physics 207: Lecture 16, Pg 13
A child slides down a playground slide at
constant speed. The energy transformation
is
A.
B.
C.
D.
E.
U
K
U
ETh
K
U
K ETh
There is no transformation because
energy is conserved.
Physics 207: Lecture 16, Pg 14
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Physics 207 – Lecture 16
Conservative Forces and Potential Energy
l So we can also describe work and changes in
potential energy (for conservative forces)
∆U = - W
l Recalling (if 1D)
W = Fx ∆x
l Combining these two,
∆U = - Fx ∆x
l Letting small quantities go to infinitesimals,
dU = - Fx dx
l Or,
Fx = -dU / dx
Physics 207: Lecture 16, Pg 15
Equilibrium
l Example
v Spring: Fx = 0 => dU / dx = 0 for x=xeq
The spring is in equilibrium position
l In general: dU / dx = 0
for ANY function
establishes equilibrium
U
U
stable equilibrium
unstable equilibrium
Physics 207: Lecture 16, Pg 16
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Physics 207 – Lecture 16
Work & Power:
l
l
l
l
l
Two cars go up a hill, a Corvette and a ordinary Chevy
Malibu. Both cars have the same mass.
Assuming identical friction, both engines do the same amount
of work to get up the hill.
Are the cars essentially the same ?
NO. The Corvette can get up the hill quicker
It has a more powerful engine.
Physics 207: Lecture 16, Pg 17
Work & Power:
Power is the rate, J/s, at which work is done.
l Average Power is,
W
l
P =
P=
∆t
dW
dt
l
Instantaneous Power is,
l
If force constant, W= F ∆x = F (v0 ∆t + ½ a∆t2)
and
P = W / ∆t = F (v0 + a∆t)
Physics 207: Lecture 16, Pg 18
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Physics 207 – Lecture 16
Work & Power: Example
Average
Power:
P =
W
∆t
1 W = 1 J / 1s
Example:
l A person, mass 80.0 kg, runs up 2 floors (8.0 m) at
constant speed. If they climb it in 5.0 sec, what is the
average power used?
l Pavg = F h / ∆t = mgh / ∆t = 80.0 x 9.80 x 8.0 / 5.0 W
l P = 1250 W
Physics 207: Lecture 16, Pg 19
Exercise
Work & Power
Starting from rest, a car drives up a hill at constant
acceleration and then suddenly stops at the top.
l The instantaneous power delivered by the engine during this
drive looks like which of the following,
B. Middle
time
Power
A. Top
Power
l
Z3
C. Bottom
Power
time
time
Physics 207: Lecture 16, Pg 20
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Physics 207 – Lecture 16
Exercise
Work & Power
l
P = dW / dt and W = F d = (µ mg cos θ − mg sin θ) d
and d = ½ a t2 (constant accelation)
So W = F ½ a t2
P=Fat=Fv
(A)
Power
l
l
(B)
Power
time
Z3
l
(C)
Power
time
time
Physics 207: Lecture 16, Pg 21
Chap. 12: Rotational Dynamics
Up until now rotation has been only in terms of circular motion
with ac = v2 / R and | aT | = d| v | / dt
l Rotation is common in the world around us.
l Many ideas developed for translational motion are transferable.
l
Physics 207: Lecture 16, Pg 22
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Physics 207 – Lecture 16
Rotational motion has consequences
Katrina
How does one describe rotation (magnitude and direction)?
Physics 207: Lecture 16, Pg 23
Rotational Variables
l
l
Rotation about a fixed axis:
v Consider a disk rotating about
an axis through its center:
θ
ω
Recall :
ω=
dθ 2π
=
(rad/s) = vTangential /R
dt
T
(Analogous to the linear case v =
dx )
dt
Physics 207: Lecture 16, Pg 24
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Physics 207 – Lecture 16
System of Particles (Distributed Mass):
Until now, we have considered the behavior of very simple
systems (one or two masses).
l But real objects have distributed mass !
l For example, consider a simple rotating disk and 2 equal
mass m plugs at distances r and 2r.
l
ω
l
1
2
Compare the velocities and kinetic energies at these two
points.
Physics 207: Lecture 16, Pg 25
For these two particles
1 K= ½ m v2
ω
2 K= ½ m (2v)2 = 4 ½ m v2
l Twice the radius, four times the kinetic energy
K = 12 m v 2 + 42 m v 2 = 12 m (ω r1 ) 2 + 12 m (ω r2 ) 2
K = 12 [ mr1 + mr2 ]ω 2
2
2
Physics 207: Lecture 16, Pg 26
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Physics 207 – Lecture 16
For these two particles
1 K= ½ m v2
ω
2 K= ½ m (2v)2 = 4 ½ m v2
K = 12 [ mr1 + mr2 ]ω 2
2
2
K = 12 [ ∑ mr 2 ]ω 2
All
mass
points
Physics 207: Lecture 16, Pg 27
For these two particles
ω
K = 12 [ ∑ mr 2 ]ω 2
All
mass
points
K Rot = 12 Iω 2 (where I is the moment of inertia)
Physics 207: Lecture 16, Pg 28
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Physics 207 – Lecture 16
Rotation & Kinetic Energy...
l
The kinetic energy of a rotating system looks similar
to that of a point particle:
Point Particle
Rotating System
K = 12 I ω 2
K = 12 mv 2
v is “linear” velocity
m is the mass.
ω is angular velocity
I is the moment of inertia
about the rotation axis.
I = ∑ m i ri
2
i
Physics 207: Lecture 16, Pg 29
Calculating Moment of Inertia
N
I ≡ ∑ m i ri
2
where r is the distance from
the mass to the axis of rotation.
i =1
Example: Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
m
m
m
m
L
Physics 207: Lecture 16, Pg 30
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Physics 207 – Lecture 16
Calculating Moment of Inertia...
For a single object, I depends on the rotation axis!
l Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2
l
I1 = 2mL2
m
m
m
m
I2 = mL2
I = 2mL2
L
Physics 207: Lecture 16, Pg 31
Lecture 16
Assignment:
l HW7 due Tuesday Nov. 1st
Physics 207: Lecture 16, Pg 32
Page 16