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phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Final exam/final retake
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Mr Smith,
I am going to be taking the final tomorrow, on Monday. I will be making sure that I have as much of harmonic motion mastered by that point as I can. My question to you is this: if I take the final early enough on Monday, would it be possible to have the results on Tuesday so I could retake it? I understand if this is not possible.
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I should be able to post that grade by 2:00. I can't guarantee that, but it should be possible.
I believe the Learning Lab is open Tuesday evening, so that would give you time.
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#$&*
phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Pendulum confusion
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A pendulum of mass 1.5 kg and length 3.44 meters is initially displaced .4472 cm in the horizontal direction from its equilibrium position. The pendulum is released from rest at this position.
How much work does gravity therefore do on the pendulum during its fall to the equilibrium position, and how much work does the pendulum do against gravity?
What therefore will be the change in the kinetic energy of the pendulum as it falls?
Using energy considerations determine the velocity of pendulum at its equilibrium position.
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Mr Smith,
I have repeatedly gone through final exams loaded from the website, in an effort to be sure that I will be ready for any problem that could be on it. On every single one, without fail, there is a pendulum problem similar to this. It gives us the mass, length, and HORIZONTAL displacement of the pendulum, then asks about things like work, force, or the period of motion. Now, in QA 25 I remember dealing with something like this, but we were given ANGLES not DISPLACEMENT. As such, I am completely lost whenever I see such a problem, and have no idea what to do. I have some equations with regards to period of motion, but they are much less ingrained then with a spring or other such device. How do I solve these kinds of problems? Quick feedback would be very much appreciated
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There are two ways to solve this problem.
One is to solve the right triangle with hypotenuse 3.44 meters and one leg equal to .4472 cm. The second leg will be slightly shorter than the hypotenuse.
That leg is the distance of the pendulum mass below the point of attachment of the pendulum, when the pendulum is at the .4472 cm position.
When hanging from equilibrium the pendulum mass is 3.44 meters below the point of attachment.
The difference between the two distances is how much higher the pendulum is when displaced .4472 cm.
The difference is very slight, so you have to be sure your calculator is accurate enough to represent the difference, and you have to be very careful with significant figures.
The other option is to understand that, for displacement x from equlibrium,
T_x / T = x / L,
where T is the magnitude of the tension and T_x its x component, so that
T_x = T * x / L.
For small displacement T is very close to m g, where m is the mass of the pendulum, so that
T_x = m g * x / L.
The net force on the pendulum is T_x, so we could write
F_net = m g * x / L.
This is the net force when displacement is x.
At maximum displacement x = A the force is thus
F_net_max = m g * A / L
In this case, as the pendulum returns to equilibrium, x changes from .4472 cm to 0.
Since the force is proportional to x, then, the average force is the average of the x = 0 force, which is zero, and the force m g * A / L at maximum displacement.
The average force is this
F_net_Ave = (0 + m g A / L) = 1/2 m g A / L.
The distance from position x = A to equilibrium at x = 0 is just A, so the work done by the net force is
`dW = F_net_Ave * `ds = 1/2 m g A / L * A = 1/2 m g A^2 / L.
Or you could use the fact that the restoring force for the pendulum is
F = - k x,
with
k = m g x / L
(this for the reasons outlined above).
The potential energy of the pendulum at position x is 1/2 k x^2, for reasons with which you should be familiar.
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