classical mechanics lecture 22
TRANSCRIPT
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Classical Mechanics Lecture 22
Today’s Concept: Simple Harmonic Mo7on: Mo#on of a Pendulum
Mechanics Lecture 8, Slide 1
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so the omega can stand for both the oscilla7on frequency or angular velocity right? its really confusing some7mes
What does phi represent again? Also, what is the difference between angular velocity and angular frequency? They look and sound to be the same.
Who's the genius who decided omega should have two meanings? Did they run out of Greek leMers? Why don't they fly over there and get some more? It would probably help boost their economy at this point.
I am finding it difficult to understand how the moment of iner7a and the radius are both being used in the equa7on. Isn't the moment of iner7a dependent on the radius?
Is the period propor7onal to Rcm then?
talking about harmonic mo7ons , Lets all dance "GANGNOM style " , its a perfect prac7cal example !
Why my roommate keeps on ramming Grape Fantas into my mini fridge is way more confusing than anything that we've covered this year
Can we do a potluck on Friday?
Your Comments
Mechanics Lecture 8, Slide 2
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Conceptual Problems workbook
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Don’t Panic!
http://www.youtube.com/watch?v=7rOMGIbY-9s
http://www.youtube.com/watch?v=Iexp-OwOs0M
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Drill a hole through the earth and jump in – what happens?
Just for fun – you don’t need to know this.
I want to know why the answer to life is 42!
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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!
Drill a hole through the earth and jump in – what happens?
k = mg/RE
Mechanics Lecture 8, Slide 5
I want to know why the answer to life is 42!
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The hole doesn’t even have to go through the middle – you get the same answer anyway as long as there is no fric7on.
Mechanics Lecture 8, Slide 6
I want to know why the answer to life is 42!
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!
Drill a hole through the earth and jump in – what happens?
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This is also the same period of an object orbi7ng the earth right at ground level.
Mechanics Lecture 8, Slide 7Just for fun – you don’t need to know this.
I want to know why the answer to life is 42!
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Panic!
“Is there such a thing as Rota7onal Harmonic Mo7on? There beMer not be...”
Yes there is.
Are you ready?
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I
wire
θ
τ
Torsion Pendulum
Q: In the prelecture the equa7on for restoring torque is given as τ=-‐κθ in clockwise direc7on..so if the restoring torque is in counter clockwise direc7ons then would τ be posi7ve?
Mechanics Lecture 8, Slide 8
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A torsion pendulum is used as the 7ming element in a clock as shown. The speed of the clock is adjusted by changing the distance of two small disks from the rota7on axis of the pendulum. If we adjust the disks so that they are closer to the rota7on axis, the clock runs:
A) Faster B) Slower
Small disks
CheckPoint
Mechanics Lecture 8, Slide 9
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If we adjust the disks so that they are closer to the rota7on axis, the clock runs
A) Faster B) Slower
B) T = 2pi * sqrt(I/MgRcm). If Rcm decreases, T will increase, making the clock run slower.
A) The moment of iner7a decreases, so the angular frequency increases, which makes the period shorter and thus the clock faster.
Mechanics Lecture 8, Slide 10
CheckPoint
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Grading
Unit 14 and 15 Ac7vity Guides will not be graded
Please turn in:! Unit 14 WriMen Homework on Monday! The Mini-‐labbook on your SHM or Karate Project, April 25
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Triumf LecturesThis weekend, Sat. Nov. 29
! 10 am Earthquakes! 11 am Earthquake engineering
snacks
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Pendulum
Forsmall θ
θ
XCM
RCM
Mg
θ
arc-‐length = RCM θ
XCM
RCM
Mechanics Lecture 8, Slide 11
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CM
pivot
θ
RCM
The Simple Pendulum
The general case
θL
The simple case
Mechanics Lecture 8, Slide 12
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If the clock is running too fast, the weight needs to be moved A) Up B) Down
If the clock is running too fast then we want to reduce it's period, T, and to do that we need to increase omega, the frequency it moves with and to do that we need the posi7on of the center of mass to be further from the pivot, which is achieved by moving the weight down.
Mechanics Lecture 8, Slide 14
CheckPoint
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M
pivot
θRCM
The Stick Pendulum
CM
Mechanics Lecture 8, Slide 15
Same period
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In Case 1 a s7ck of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point par7cle of mass m is aMached to the center of the same s7ck. In which case is the period of the pendulum the longest?
A) Case 1 B) Case 2 C) Same
Case 1 Case 2
C is not the right answer.
Lets work through it
CheckPoint
m
m
m
Mechanics Lecture 8, Slide 16
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T = 2⇡
s23 Lg
T = 2⇡
s12 Lg
In Case 1 a s7ck of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point par7cle of mass m is aMached to a string of length L/2?
In which case is the period of the pendulum longest?
A) Case 1 B) Case 2 C) Same
Case 1 Case 2
m
Mechanics Lecture 8, Slide 17
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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
m
Suppose you start with 2 different pendula, one having period T1 and the other having period T2.
T1
T2
T1 > T2
Mechanics Lecture 8, Slide 18
mm
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In Case 1 a s7ck of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point par7cle of mass m is aMached to the center of the same s7ck. 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
Case 1 Case 2
m
m
m
Mechanics Lecture 8, Slide 19
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Case 2m
m
mCase 1
Lets compare for each case.
Mechanics Lecture 8, Slide 20
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(A)
(B)
(C)
Case 2m
m
mCase 1
Lets compare for each case.
Mechanics Lecture 8, Slide 21
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In which case is the period longest?
A) Case 1
B) Case 2
C) They are the same
m
Case 1
So we can work out
Case 2
m
m
Mechanics Lecture 8, Slide 22
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Angle (degrees)
% differen
ce between θ and
sinθ
- Exact expression
The Small Angle Approximation
θ
arc-‐length = RCM θ
XCM
RCM
Mechanics Lecture 8, Slide 23
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A pendulum is made by hanging a thin hoola-‐hoop of diameter D on a small nail. What is the angular frequency of oscilla7on of the hoop for small displacements? (ICM = mR2 for a hoop)
A)
B)
C)
D
pivot (nail)
Clicker Question
Mechanics Lecture 8, Slide 24
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The angular frequency of oscilla7on of the hoop for small
displacements will be given by
R
XCM
Use parallel axis theorem: I = ICM + mR2
m
= mR2 + mR2 = 2mR2
pivot (nail)
Mechanics Lecture 8, Slide 25
So