1. 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.

2. “Physics makes me feel dumb
“Physics makes me feel dumb. Why make such simple things as a ball swinging on a string so difficult?!”

3. 3rd Exam is next Wednesday at 7pm:
Covers lectures (not the stuff we are doing today) Sign up for the conflict if you need to before the end of this week. As before, Dr. Jain if you have a double conflict
Next Tuesdays class will be a review (Fall 2010 exam) .
Don’t forget about office hours !!

4. “There is another theory which states that this has already happened.”
“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

5. Why 42 ?? Drill a hole through the earth and jump in – what happens?
Just for fun – you don’t need to know this.

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!
k = mg/RE

7. 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).

8. 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.

9. Today’s Concept: Simple Harmonic Motion: Motion of a Pendulum
Physics 211 Lecture 22
Today’s Concept:
Simple Harmonic Motion: Motion of a Pendulum

10. Torsion Pendulum
“If it wasn't a toy or found on a playground, I have trouble conceptually understanding it. Please explain a torsion pendulum..”
wire   I
Suppose we have a disk suspended by a vertical wire attached to the center of the disk, parallel to its symmetry axis, as shown. Suppose the other end of the wire is attached to a fixed support so that if the disk is turned the wire is twisted. In the same way that a spring doesn’t like to be displaced from its equilibrium length, the wire doesn’t like to be twisted away from its equilibrium orientation and it exerts a torque on the disk which is proportional to the angle that it is twisted by [show tau = -kappa*theta]. The direction of the torque is always in the direction that would return the disk to its equilibrium orientation. If the disk is turned through some initial angle and released, its orientation will oscillate back and forth with simple harmonic motion.
We can analyze this motion by applying Newton's second law for rotations [torque = I alpha]. Since the angular acceleration is just the second derivative of the angular displacement, this reduces to the differential equation for simple harmonic motion if omega squared is equal to the torsion constant divided by the moment of inertia of the disk about its symmetry axis.
We see that this is just the rotational equivalent of a mass oscillating on a spring, where torque has taken the place of force, the angular displacement has taken the place of linear displacement and the torsion constant has taken the place of the spring constant, and the moment of inertia has taken the place of the mass. The most general solution for the angular displacement of the pendulum as a function of time therefore also has the same sinusoidal form as it did for the mass on the spring.
[we may not want to include this part, though the picture is interesting]
This kind of a pendulum is often used in clocks since the period of oscillation can be very finely tuned. The picture shows such a clock. The pendulum is the disk hanging in the horizontal plane beneath the face of the clock. The speed of the clock can be tuned by changing the moment of inertia of this pendulum, done by adjusting the distance of the two smaller masses from the central axis.
“Can you explain the difference between angular frequency and angular velocity again?”

11. 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!”

12. CheckPoint
If we adjust the disks so that they are closer to the rotation axis, the clock runs
A) Faster B) Slower
I decreases, so the angular f increases. period shorter, the clock faster..

13. Pendulum For small q q RCM Mg RCM XCM q XCM arc-length = RCM q 13
Another important and very useful class of harmonic oscillation is that which is caused by gravity acting on a pendulum. We have already seen that if we take any object and hang it from a frictionless pivot, gravity will exert a torque on the object which is proportional to the horizontal displacement of the center of mass away from the pivot, in the direction that tends to reduce this horizontal displacement [show diagram without R and theta]. Our experience also tells us that if we pull a pendulum to one side and release it, it will swing back and forth in a very regular way [show this]. We will now prove that this, too, is simple harmonic motion as long as the angle that the pendulum makes with the vertical axis is not too big.
We start with Newton's second law for rotations, and we evaluate both sides for motion around the pivot. On the left hand side, the torque is the weight of the pendulum times the horizontal displacement of the center of mass, and the minus sign tells us that that the torque opposes the displacement. As long as the angle between the vertical direction and the line connecting the pivot to the center of mass isn't too big, the horizontal displacement can be approximated by the arc-length as shown, allowing us to write the displacement in terms of the angle and the distance from the pivot to the center of mass. The right hand side of the equation is the moment of inertia times the second derivative of the angle, just as it was for the torsion pendulum on the previous slide.
With the proper definition of omega, we again arrive at the familiar differential equation for simple harmonic motion. This time the variable undergoing simple harmonic motion is the angle between the vertical direction and the line connecting the pivot to the center of mass. [show oscillation]
On the next slide we will examine a very simple example of a pendulum, namely that of a mass at the end of a string. 13

14. The Simple Pendulum The simple case pivot RCM q q L CM
Suppose we replace our simple pendulum by an object having a more complicated shaped. Which properties of this oscillating system will be the same as those for a simple pendulum, and which will change? The answer is that the oscillating behavior is exactly the same as before, the only difference is that the expression for the frequency of the oscillation looks a bit different, depending on the location of the center of mass and the moment of inertia of the object.
Lets see what happens if we consider a stick of mass M and length L pivoted at one end. The distance from the pivot to the center of mass of the stick is half the length of he stick, and we remember from an earlier prelecture the expression for the moment of inertia of a stick about one of its ends. Using this in our general expression for the angular velocity of a pendulum we find that the stick behaves exactly like a simple pendulum having two thirds the length of the stick.
If we did the same calculation using another object for which we knew moment of inertia as a function of its mass and shape we would get the same general result; that the mass would cancel out and that the frequency of its oscillation would depend only on its size and shape.
All of the results we have discussed so far involving a pendulum oscillating due to gravity have depended on the assumption that the angle of oscillation be kept small. Why is this necessary? How small is small? What happens to a pendulum if this condition isn't true? We will discuss these questions a bit more on the next slide.
“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)?” 14

15. 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

16. CheckPoint
If the clock is running too fast, the weight needs to be moved A) Up B) Down
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.

17. The Stick Pendulum M pivot RCM q CM Same period Demo 17
Suppose we replace our simple pendulum by an object having a more complicated shaped. Which properties of this oscillating system will be the same as those for a simple pendulum, and which will change? The answer is that the oscillating behavior is exactly the same as before, the only difference is that the expression for the frequency of the oscillation looks a bit different, depending on the location of the center of mass and the moment of inertia of the object.
Lets see what happens if we consider a stick of mass M and length L pivoted at one end. The distance from the pivot to the center of mass of the stick is half the length of he stick, and we remember from an earlier prelecture the expression for the moment of inertia of a stick about one of its ends. Using this in our general expression for the angular velocity of a pendulum we find that the stick behaves exactly like a simple pendulum having two thirds the length of the stick.
If we did the same calculation using another object for which we knew moment of inertia as a function of its mass and shape we would get the same general result; that the mass would cancel out and that the frequency of its oscillation would depend only on its size and shape.
All of the results we have discussed so far involving a pendulum oscillating due to gravity have depended on the assumption that the angle of oscillation be kept small. Why is this necessary? How small is small? What happens to a pendulum if this condition isn't true? We will discuss these questions a bit more on the next slide.
Demo 17

18. C is not the right answer.
CheckPoint
Case 1
Case 2
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 B) Case C) Same m m
C is not the right answer.
Lets work through it 18

19. In which case is the period of the pendulum longest?
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?
In which case is the period of the pendulum longest?
A) Case 1 B) Case C) Same m 19

20. 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. T2
T1 > T2 T1
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 20

21. Now lets work through it in detail
Case 1
Case 2
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 B) Case C) Same m m
Now lets work through it in detail 21

22. Lets compare for each case.22

23. Lets compare for each case.
(B)
(C) 23

24. In which case is the period longest (i.e. w smallest)? A) Case 1
So we can work out m
Case 1
Case 2 m
In which case is the period longest (i.e. w smallest)?
A) Case 1
B) Case 2
C) They are the same
Same period 24

25. The Small Angle Approximation
“Can we go over the "small angle" ordeal? Why can't we use the sin(theta)in our equation?”
Angle (degrees)
% difference between q and sinq q
arc-length = RCM q
XCM
RCM
The root cause of the requirement that the angular displacement of the pendulum be small was that we needed to find an expression for the horizontal displacement of the center of mass in terms of this angle. The approximation we used was that the length of the arc swept out by the tip of a vector of length R as it is displaced an angle theta from the vertical is very close to the horizontal distance moved by the tip of the vector, as long as the angle is small.
What if we used the exact expression? The exact horizontal displacement of the center of mass for any angle is easily seen to be R-sin-theta. If we had used this exact form, our differential equation would have looked almost the same, except we would have had sin theta rather than theta on the right hand side. The up side of this equation is that it is a perfectly accurate description of our system for any angle, but the down side is that it doesn’t have a simple solution the way our signature equation does.
So, given that we are stuck making the small angle approximation, how small does the angle have to be in odder for this approximation to be valid? We can investigate this by seeing how close theta is to sin theta. You can do this yourself by putting your calculator into radian mode and just comparing how close the sine of an angle is to the value of the angle itself. The plot shows the result of doing this for angles between 0 and 45 degrees, where the vertical axis shows the percentage error we make by using theta rather than sin theta. We see that the error is less than five percent for angles up to about thirty degrees.
What happens to the motion of the pendulum as we make it swing with a bigger and bigger angle so that our small angle approximation gets worse and worse? Nothing very noticeable. It still swings back and forth regularly with a fixed period, but a careful measurement of the angle as a function of time would show that this is no-longer perfectly sinusoidal, and a careful investigation of the period of the oscillation would show that it is no-longer be completely independent of the amplitude of the oscillation.
[this part is optional – perhaps it should be in the book rather than in the prelecture]
One way to understand why this approximation works so well is to perform a Taylor expansion of sine theta about theta = 0, for theta expressed in radians. When we do this we get an infinite series whose first few terms are shown. We can immediately see that as long as theta is less than one radian (which is about 57 degrees), then the first term will be much bigger than all the others. 25

26. 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) B) C) R D

27. 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?
pivot (nail) A) B) C) D

28. Use parallel axis theorem: I = ICM + mR2
The angular frequency of oscillation of the hoop for small displacements will be given by
Use parallel axis theorem: I = ICM + mR2
= mR2 + mR2 = 2mR2
pivot (nail) R X CM So m