Your Comments
So I actually watched the video and did the prelecture questions this time....i kinda get it,
but lets hope you're right and i understand this section better tomorrow!
The derivations were a bit intimidating at first, but after a moment of re-analyzing the
equations, the process and formulas made (for the most part) sense.
Pretty simple, just plug and chug for the most part
How to get a date: Girl, you must be sin(pi/2), because you are the 1. Works every time.
3 things: 1) We're nearing the end of the semester. 2) I love physics. 3) I haven't had a
comment up even once. Negate one of the three statements(except the first two), pretty
please?
I actually understood everything very well, because I did this so early when I could
actually take the time to understand it. The Stanley Cup playoffs should be on all the
time...it motivated me to get this done before they started on Wednesday. Go
Blackhawks! :D
Can we discuss the physics behind Star Trek? Please?
I feel like the guy from office space now..... I'm totally hypnotized into not caring about
physics, life, college, etc. Let's go destroy a copy machine!
While watching these videos, the constant pendulum motions hypnotized me and I
passed out. When I finally came to, I was in the Loomis basement strapped to a table
with an axe blade on a thin rod oscillating over my head. A hooded figure told me to
determine the angle of the oscillation before I was beheaded. When I asked if he
accounted for air resistance and friction, he let me go.
Mechanics Lecture 8, Slide 1
“Physics makes me feel dumb. Why make such simple things as a
ball swinging on a string so difficult?!”
Mechanics Lecture 8, Slide 2
3rd Exam is next Wednesday at 7pm:
Covers lectures 14-20 (not the stuff we are doing today)
Sign up for the conflict if you need to before the end of
this week. As before, email Dr. Jain if you have a double
conflict (rdjain1@illinois.edu).
Next Tuesdays class will be a review (Fall 2010 exam) .
Don’t forget about office hours !!
Mechanics Lecture 17, Slide 3
“There is a theory which states that if ever anybody
discovers exactly what the Universe is for and why it is here, it
will instantly disappear and be replaced by something even more
bizarre and inexplicable.”
“There is another theory which states that
this has already happened.”
Douglas Adams
Mechanics Lecture 8, Slide 4
Why 42 ??
Drill a hole through the earth and jump in – what happens?
Just for fun – you don’t need to know this.
Drill a hole through the earth and jump in – what happens?
You will oscillate like a mass on a spring with a period of 84 minutes.
It takes 42 minutes to come out the other side!
k = mg/RE
Mechanics Lecture 8, Slide 6
Drill a hole through the earth and jump in – what happens?
You will oscillate like a mass on a spring with a period of 84 minutes.
It takes 42 minutes to come out the other side!
The hole doesn’t even have to go through the middle – you get the same
answer anyway (as long as there is no friction).
Mechanics Lecture 8, Slide 7
This is also the same period of an object orbiting the earth right at
ground level.
Just for fun – you don’t need to know this.
Mechanics Lecture 8, Slide 8
Physics 211
Lecture 22
Today’s Concept:
Simple Harmonic Motion: Motion of a Pendulum
Mechanics Lecture 8, Slide 9
Torsion Pendulum
“If it wasn't a toy or found on a playground, I have
trouble conceptually understanding it. Please explain a
torsion pendulum..”
 = I

d 2
I 2
dt
wire


d 2
2
=



2
dt
=
“Can you explain the difference
between angular frequency and
angular velocity again?”

I
I
 (t ) =  max cos(t   )
Mechanics Lecture 8, Slide 10
CheckPoint
A torsion pendulum is used as the timing element in
a clock as shown. The speed of the clock is adjusted
by changing the distance of two small disks from the
rotation axis of the pendulum. If we adjust the disks
so that they are closer to the rotation axis, the clock
runs:
A) Faster
B) Slower
Small disks
“My grandma has a torsion pendulum clock!”
Mechanics Lecture 8, Slide 11
CheckPoint
If we adjust the disks so that they are closer to
the rotation axis, the clock runs
A) Faster
B) Slower

=
I
I decreases, so the angular f increases. period shorter, the clock
faster..
Mechanics Lecture 8, Slide 12
Pendulum
 = I
For
small 

RCM
 MgX CM
d 2
 MgRCM  I 2
dt
2
MgRCM
d
=

2
dt
I
d 2
2
=



2
dt
MgRCM
=
I
XCM
Mg

RCM
XCM
arc-length
= RCM 
Mechanics Lecture 8, Slide 13
The Simple Pendulum
pivot
RCM
The simple case


L
CM
MgRCM
=
I
MgL
=
=
2
ML
g
L
“what is the difference between a simple pedulem and a physical
pedulem. mathematically?”
“Can we discuss the differences between a simple pendulum and a
physical pendulum. How would the two compare if they both had
equal lengths (like in the Pre-lecture Question)?”
Mechanics Lecture 8, Slide 14
CheckPoint
A simple pendulum is used as the timing element
in a clock as shown. An adjustment screw is used
to make the pendulum shorter (longer) by moving
the weight up (down) along the shaft that
connects it to the pivot.
If the clock is running too fast, the weight needs
to be moved A) Up B) Down
Adjustment screw
Mechanics Lecture 8, Slide 15
CheckPoint
If the clock is running too fast, the weight needs to
be moved A) Up B) Down
g
=
L
Because angular frequency is inversely proportional to
length, the pendulum will swing slower if the rod is moved
down because the length will be increased.
Mechanics Lecture 8, Slide 16
The Stick Pendulum
pivot

RCM
CM
MgRCM
=
I
L
2
=
1
2
ML
3
g
2
L
3
M
L
2
L
3
Same period
Demo
Mechanics Lecture 8, Slide 17
CheckPoint
Case 1
Case 2
m
m
m
In Case 1 a stick of mass m and length L is
pivoted at one end and used as a pendulum. In
Case 2 a point particle of mass m is attached to
the center of the same stick. In which case is the
period of the pendulum the longest?
A) Case 1
B) Case 2
C) Same
C is not the right answer.
Lets work through it
Mechanics Lecture 8, Slide 18
Case 1
Case 2
1
L
2
L
m
=
In Case 1 a stick of mass m and length L
is pivoted at one end and used as a
pendulum. In Case 2 a point particle of
mass m is attached to a string of length
L/2?
g
2
L
3
=
g
1
L
2
In which case is the period of the
pendulum longest?
A) Case 1 B) Case 2
C) Same
Mechanics Lecture 8, Slide 19
T2
m
T1
m
m
Suppose you start with 2 different pendula,
one having period T1 and the other having
period T2.
T1 > T2
Now suppose you make a new pendulum by
hanging the first two from the same pivot and
gluing them together.
What is the period of the new pendulum?
A) T1 B) T2
C) In between
Mechanics Lecture 8, Slide 20
Case 1
Case 2
m
m
m
In Case 1 a stick of mass m and length L is
pivoted at one end and used as a pendulum. In
Case 2 a point particle of mass m is attached to
the center of the same stick. In which case is the
period of the pendulum the longest?
A) Case 1
B) Case 2
C) Same
Now lets work through it in detail
Mechanics Lecture 8, Slide 21
Case 2
Case 1
m
m
m
MgRCM
Lets compare  =
I
L
mg
2
for each case.
L
2mg
2
Mechanics Lecture 8, Slide 22
Case 2
Case 1
m
m
m
MgRCM
Lets compare  =
I
1 2
mL
3
for each case.
1 2
4
2
mL  mL = mL2 (A)
3
3
1
1
5
2
2
mL  mL = mL2 (B)
3
2
6
2
1 2
7
L
mL  m  =
mL2 (C)
3
12
2
Mechanics Lecture 8, Slide 23
MgRCM
So we can work out  =
I
Case 2
Case 1
=
m
g
2
L
3
g
m =
7
L
12
m
In which case is the period longest (i.e.  smallest)?
A) Case 1
B) Case 2
C) They are the same
L
2
L
3
Same period
Mechanics Lecture 8, Slide 24
The Small Angle Approximation

RCM
XCM
arc-length
= RCM 
“Can we go over the "small angle" ordeal? Why
can't we use the sin(theta)in our equation?”
% difference between  and sin
d 2
2
=


sin 
2
dt
Angle (degrees)
1 3 1 5 1 7
1 3
1 5
1
sin  =         ... =  
 
 7  ...
3!
5!
7!
6
120
5040
Mechanics Lecture 8, Slide 25
Clicker Question
A pendulum is made by hanging a thin hoola-hoop of radius R on
a small nail. What is the moment of inertia of the hoop about an
axis through the nail?
(Recall that ICM = mR2 for a hoop)
pivot (nail)
A) I = mR 2
B)
I = 2mR 2
C)
I = 4mR 2
R
D
Mechanics Lecture 8, Slide 26
Clicker Question
A pendulum is made by hanging a thin hoola-hoop of radius R on
a small nail. What is the angular frequency of oscillation of the
hoop for small displacements?
g
A)  =
2R
B)
2g
=
R
C)
=
pivot (nail)
D
2g
R
Mechanics Lecture 8, Slide 27
The angular frequency of oscillation of the hoop for small
mgRCM
displacements will be given by  =
I
Use parallel axis theorem: I = ICM  mR2
= mR2  mR2 = 2mR2
=
mgR
=
2
2mR
g
=
2R
pivot (nail)
g
D
R
g
So  =
D
X
CM
m
Mechanics Lecture 8, Slide 28