Your Comments
Dark side of OZ
It is/will be my birthday during Thursday's lecture. Coincidentally, it is also the last lecture
before break. Can you make an effort to do something particularly funny or memorable
or amusing? Happy Birthday
Pendulum happens to be one of my favorite bands, so that's nice.
Happy Thanksgiving!
Can you do a demo with a turkey?
Alright, in all seriousness, this stuff is difficult. I could have spent another hour watching
this prelecture and it would not have clicked. I'm still hung up on x =Asin(wt+phi). How
do you know when to use sin or cos? And what is phase change (phi) ? IE
for some odd reason this and the stuff exam three is about to be on is easier for me to
understand than the first two exam materials
I'm terribly sorry for spending so little time on the prelecture — all of my classes decided
to pile on the work right before break.
I wonder if i played these lectures on a loop during my sleep, could I literally learn in my
sleep?
Is the 80% deadline for homework next tuesday despite the break, or is it the week
tuesday after since there will be office hours.
I want to congratulate my friend Jon Henricks on getting engaged!!
Apparently pendula is the plural of pendulum. I didn't know that.
Mats, where are you going for break? Punta Allen
Mechanics Lecture 8, Slide 1
3rd Exam is on Wednesday Nov 28th
Covers lectures 14-20 (not the stuff we are doing now)
Sign up for the conflict exam if you need to
If you need a double conflict, email me as usual.
I am having a really really hard time taking in all of these formulas.
Not just on harmonic motion, but in general. I wish there was
something like the equation sheet, but that goes into a bit more
detail so that I could try to start memorizing them and thus be able
to use them more effectively…
Mechanics Lecture 17, Slide 2
“I AM SO CONFUSED.. YOU MUST HELP ME.... OR
IM GONNA GO NUTS!!!!!!!!!!”
Lots of office hours & tutoring available, especially
during exam weeks.
Mechanics Lecture 8, Slide 3
What is the answer to the question of life, the universe and everything?
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
“How does torsion produce torque if the lever
arm is zero?”
 = I
d 
2

I
dt
2
wire

d 

2
dt
2
=  
 =
2

I
I
 ( t ) =  max cos(  t   )
d
Angular
velocity
dt
=    m ax sin( t   )
Angular frequency
WTF, the omegas
aren't the same? ...
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
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
less moment of inertia means shorter period and faster rotation
Mechanics Lecture 8, Slide 12
Pendulum
 = I
 MgX
For
small 
dt
d 
2
2
2
RCM
CM
 MgR CM 
d 

=
I
dt
MgR CM
XCM
2
Mg

I
d 

RCM
2
dt
 =
2
=  
2
MgR CM
XCM
arc-length
= RCM 
I
Mechanics Lecture 8, Slide 13
The Simple Pendulum
pivot
RCM
The simple case


L
CM
 =
MgR CM
I
 =
M gL
ML
2
=
g
L
Simple Harmonic Motion? But this stuff is pretty damn complicated.
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
This increases the moment of inertia which
increases the period.
Mechanics Lecture 8, Slide 16
The Stick Pendulum
pivot
L

RCM
 =
2
MgR CM
I
CM
1
 =
ML
2
g
2
L
3
3
M
2
L
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
“…Are the answers to the checkpoint
questions supposed to be obvious?”
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
 =
g
1
L
In which case is the period of the
pendulum longest?
2
A) Case 1 B) Case 2
C) Same
L
3
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.
Using your intuition, 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
Lets compare  =
MgR CM
for each case.
I
mg
L
2
2 mg
L
2
Mechanics Lecture 8, Slide 22
Case 2
Case 1
m
m
m
Lets compare  =
MgR CM
for each case.
I
1
mL  mL =
2
2
3
1
3
1
mL
2
3
4
mL
2
3
mL 
2
1
mL =
2
2
5
mL
2
(A)
(B)
6
2
7
L
2
2
mL  m   =
mL (C)
3
12
2
1
Mechanics Lecture 8, Slide 23
So we can work out  =
MgR CM
I
Case 2
Case 1
 =
g
2
m
L
3
m
 =
g
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
2
L
L
3
Same period
Mechanics Lecture 8, Slide 24
The Small Angle Approximation
d 
“Can we go over the "small angle" ordeal? Why
can't we use the sin(theta)in our equation?”
2
=   sin 

% difference between  and sin
dt
2
2
RCM
XCM
arc-length
= RCM 
Angle (degrees)
sin  =  
1
3!
 
3
1
5!
 
5
1
7!
  ... =
7
1
6
 
3
1
120
 
5
1
  ...
7
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
B)
I = 2mR
2
C)
I = 4mR
2
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?
A)
=
g
pivot (nail)
2R
B)
=
C)
=
2g
R
D
2g
R
Mechanics Lecture 8, Slide 27
The angular frequency of oscillation of the hoop for small
displacements will be given by  =
mgR CM
I
Use parallel axis theorem: I = ICM  mR2
= mR2  mR2 = 2mR2
 =
m gR
2mR
2
g
=
=
2R
pivot (nail)
g
D
R
So  =
g
D
X
CM
m
Mechanics Lecture 8, Slide 28