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Physics 207 – Lecture 16
Lecture 16
Goals:
• Chapter 11 (Work)
Employ conservative and non-conservative forces
Relate force to potential energy
Use the concept of power (i.e., energy per time)
• Chapter 12
Define rotational inertia (“mass”)
Define rotational kinetic energy
Define center of mass
Assignment:
HW7 due Tuesday, Nov. 2nd
For Wednesday: Read Chapter 12, Sections 1-3, 5 & 6
Exam 2
7:15 PM Thursday, Nov. 4th
Physics 207: Lecture 16, Pg 3
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)
“Scalar or Dot Product”
r r
F ⋅ dr
m
ce
a
l
p
dis
en
If we know the angle the force makes with the
path, the dot product gives us F cos θ and ∆r
∆W
=
r r
v r
F ⋅ ∆r = F ∆r cos θ
Physics 207: Lecture 16, Pg 4
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t
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.
How many forces (including non-contact ones) are
doing work on the box ?
Of these which are positive and which are negative?
State the system (here, just the box)
Use a Free Body Diagram
Compare force and path
v
A. 2
B. 3
C. 4
D. 5
Physics 207: Lecture 16, Pg 5
Work with varying forces or curved paths
If the path is curved or if the force varies with
position then at each point
dW
=
r r
F ⋅ dr
and
W
=
r
rf
r
r
∫r F ⋅ d r
ri
Physics 207: Lecture 16, Pg 7
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Physics 207 – Lecture 16
Work and Varying Forces (1D)
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
Example: Spring-Mass energy transfer
•
How much will the spring compress (u = x – xequilibrium) 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 ?
uf
Wbox
to vo
ui
uf
m
Wbox
spring at an equilibrium position
∆u
= ∫ − ku du
ui
= - 12 ku 2 |uf
ui
1
2
Wbox = - 2 k ∆u = ∆K
Wbox
V=0
t
= ∫ F (u ) du
F
m
spring compressed
- 12 k ∆u 2 = 12 m0 2 − 12 mv 02
Physics 207: Lecture 16, Pg 10
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Physics 207 – Lecture 16
Relate work to potential energy
Consider the ball moving up to height h
(from time 1 to time 2)
How does this relate to the potential energy?
Work done by the Earth’s gravity on the ball)
W = F • ∆x = -mg h
mg
∆U = Uf – Ui = mg h - mg 0 = mg h
h
∆U = -W
This is a general result for all
conservative forces
(also path independent)
mg
Physics 207: Lecture 16, Pg 11
Conservative Forces & Potential Energy
For any conservative force F we can define a potential energy
function U in the following way:
W =
∫ F ·dr = - ∆U
The work done by a conservative force is equal and opposite to
the change in the potential energy function. r
U
f
ri
f
Ui
Physics 207: Lecture 16, Pg 12
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Physics 207 – Lecture 16
Conservative Forces and Potential Energy
So we can also describe work and changes in
potential energy (for conservative forces)
∆U = - W
but in 1D
W = Fx ∆x
Combining these two,
∆U = - Fx ∆x
Letting small quantities go to infinitesimals,
dU = - Fx dx
Or,
Fx = -dU / dx
Physics 207: Lecture 16, Pg 13
A Non-Conservative Force, Friction
Looking down on an air-hockey table with no air
flowing (µ > 0).
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 15
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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 W NC = Efinal - Einitial < 0 and reflects Ethermal
Physics 207: Lecture 16, Pg 16
Work & Power:
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 22
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Physics 207 – Lecture 16
Work & Power:
Power is the rate at which work is done.
Average
Power:
Instantaneous
Power:
P = W∆t
P = dW
dt
Units (SI) are
Watts (W):
1 W = 1 J / 1s
Example:
A person of mass 200. kg walks up 10.0 m. If he/she climbs
in 10.00 sec what is the average power used (g = 10 m/s2)
Pavg = F h / t = mgh / t = 200. x 10.0 x 10.0 / 10.00
Watts
P = 2000 W
Physics 207: Lecture 16, Pg 23
Work & Power:
Average Power:
Instantaneous Power:
Constant force
P = Paverage = W∆t
P = dW
dt
r r
r r
dW d ( F ⋅ x ) F ⋅ dx r r
P=
=
=
= F ⋅v
dt
dt
dt
Physics 207: Lecture 16, Pg 24
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Physics 207 – Lecture 16
Work & Power:
Example 2 :
Engine of a jet develops a trust of 15,000 N when plane is
flying at 300 m/s. What is the horsepower of the engine ?
r r
P = F ⋅v
P=Fv
P = (15,000 N) (300 m/s) = 4.5 x 106 W
= (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp !
Physics 207: Lecture 16, Pg 25
Exercise
Work & Power
Starting from rest, a car drives up a hill at constant
acceleration and then suddenly stops at the top.
The instantaneous power delivered by the engine during this
drive looks like which of the following,
Power
mg∆h
P = dW
=
dt
∆t
A. Top
Power
C. Bottom
Z3
time
Power
B. Middle
time
time
r r
P = F ⋅v
Physics 207: Lecture 16, Pg 26
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Physics 207 – Lecture 16
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
Rotation is common in the world around us.
Many ideas developed for translational motion are transferable.
Physics 207: Lecture 16, Pg 30
Rotational Dynamics: A child’s toy, a physics
playground or a student’s nightmare
A merry-go-round is spinning and we run and jump on
it. What does it do?
We are standing on the rim and our “friends” spin it
faster. What happens to us?
We are standing on the rim a walk towards the center.
Does anything change?
Physics 207: Lecture 16, Pg 32
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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).
But real objects have distributed mass !
For example, consider a simple rotating disk and 2 equal
mass m plugs at distances r and 2r.
ω
1
2
Compare the velocities and kinetic energies at these two
points.
Physics 207: Lecture 16, Pg 37
System of Particles (Distributed Mass):
1 K= ½ m v 2 = ½ m (ω r)2
ω
The rotation axis matters too!
2 K= ½ m (2v) 2 = ½ m (ω 2r)2
Twice the radius, four times the kinetic energy
K = 12 mv 2 = 12 m(ωr ) 2
Physics 207: Lecture 16, Pg 38
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Physics 207 – Lecture 16
Rotation & Kinetic Energy
Consider the simple rotating system shown below.
(Assume the masses are attached to the rotation axis by
massless rigid rods).
The kinetic energy of this system will be the sum of the
kinetic energy of each piece:
4
K = 12 ∑ mi v i2
i =1
K = ½ m 1v12 + ½ m 2v22 + ½ m 3v32 + ½ m 4v42
m4
ω
r4
m3
r1
m1
r2
r3
m2
Physics 207: Lecture 16, Pg 39
Rotation & Kinetic Energy
Notice that v1 = ω r1 , v2 = ω r2 , v3 = ω r3 , v4 = ω r4
So we can rewrite the summation:
4
4
4
i =1
i =1
i =1
K = 12 ∑ mi v i2 = 12 ∑ miω 2 ri2 = 12 [ ∑ mi ri2 ]ω 2
We recognize the quantity, moment of inertia or I, and
write:
m
4
K = 12 I ω 2
ω
r4
m3
r3
r1
r2
m1
m2
Physics 207: Lecture 16, Pg 40
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Physics 207 – Lecture 16
Calculating Moment of Inertia...
For a single object, I depends on the rotation axis!
Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2
I1 = 2mL2
m
m
m
m
I2 = mL2
I = 2mL2
L
Physics 207: Lecture 16, Pg 41
Calculating Moment of Inertia...
For a discrete collection of point
masses we found:
N
I = ∑ m i ri
2
i =1
For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
dm
An integral is required to find I :
I = ∫ r 2 dm
r
Physics 207: Lecture 16, Pg 42
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Physics 207 – Lecture 16
Moments of Inertia
Some examples of I for solid objects:
dr
r
L
R
Solid disk or cylinder of mass M
and radius R, about
perpendicular axis through its
center.
I = ½ M R2
Physics 207: Lecture 16, Pg 43
Lecture 16, Work & Energy
Strings are wrapped around the circumference of two solid
disks and pulled with identical forces for the same linear
distance.
Disk 1 has a bigger radius, but both are identical material (i.e.
their density ρ = M/V is the same). Both disks rotate freely
around axes though their centers, and start at rest.
Which disk has the biggest angular velocity after the pull?
W = F d = ½ I ω2
(A)
ω2
ω1
Disk 1
(B) Disk 2
(C) Same
start
finish
F
F
d
Physics 207: Lecture 16, Pg 49
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Physics 207 – Lecture 16
Rotation & Kinetic Energy...
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 50
Connection with CM motion
If an object of mass M is moving linearly at velocity VCM
without rotating then its kinetic energy is
2
K T = 12 MVCM
If an object of moment of inertia ICM is rotating in place
about its center of mass at angular velocity ω then its
kinetic energy is
K R = 12 I CMω 2
What if the object is both moving linearly and rotating?
2
K = 12 I CMω 2 + 12 MVCM
Physics 207: Lecture 16, Pg 51
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Physics 207 – Lecture 16
Connection with motion...
So for a solid object which rotates about its center of mass
and whose CM is moving:
2
K TOTAL = K Rotational + 12 MVCM
VCM
ω
Physics 207: Lecture 16, Pg 52
System of Particles: Center of Mass (CM)
If an object is not held then it will rotate about the
center of mass.
Center of mass: Where the system is balanced !
Building a mobile is an exercise in finding
centers of mass.
m1
+
m2
m1
+
m2
mobile
Physics 207: Lecture 16, Pg 53
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Physics 207 – Lecture 16
System of Particles: Center of Mass
How do we describe the “position” of a system made up of many
parts ?
Define the Center of Mass (average position):
For a collection of N individual point-like particles whose
masses and positions we know:
N
r
∑ mi ri
r
RCM ≡ i =1
M
RCM
m2
m1
r1
r2
y
x
(In this case, N = 2)
Physics 207: Lecture 16, Pg 54
Sample calculation:
Consider the following mass distribution:
r
∑ mi ri
r
RCM = i =1
= XCM î + YCM ĵ + ZCM k̂
M
N
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
RCM = (12,6)
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
(12,12)
2m
XCM = 12 meters
YCM = 6 meters
m
(0,0)
m
(24,0)
Physics 207: Lecture 16, Pg 55
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Physics 207 – Lecture 16
System of Particles: Center of Mass
For a continuous solid, convert sums to an integral.
dm
y
r
r
r
∫ r dm ∫ r dm
=
RCM =
M
∫ dm
r
x
where dm is an infinitesimal
mass element.
Physics 207: Lecture 16, Pg 56
Recap
Assignment:
HW7 due Nov. 2nd
For Monday: Read Chapter 12, Sections 7-11
Physics 207: Lecture 16, Pg 57
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