Is This Free Body Diagram of a Pendulum Correct?

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SUMMARY

The forum discussion centers on the accuracy of a free body diagram for a pendulum with a maximum amplitude of 30 degrees. Participants confirm that the diagram correctly reflects the angle and amplitude, clarifying that amplitude measures the maximum displacement from the vertical. The period of the pendulum is defined as the time taken for one complete cycle, which includes passing through the equilibrium position twice. The final consensus is that the diagram is correct, with no further adjustments needed.

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-EquinoX-
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Homework Statement


I am asked to draw a free body diagram of a pendulum and a bob with it's maximum
amplitude of 30 degrees. Below is my attempt, I just forgot to say that theta is equal to
30 degrees, other than that is it right?

Homework Equations


none

The Attempt at a Solution


attached as a picture

http://img246.imageshack.us/img246/7276/pendulumqa2.jpg
 
Last edited by a moderator:
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Looks good to me.
 
reason I am not so sure is because of the amplitude of 30 degrees, what does that mean?
 
The amplitude is just a measure of how far the pendulum bob swings away from equilibrium (the vertical). An amplitude of 30 degrees means that the pendulum swings until it's 30 degrees from the vertical before swinging back.

You've reflected the angle in your diagram, so you're OK.
 
ok, and the period of this pendulum is basically the time it takes the position I move the pendulum 30 degrees away from the equilibrium then I release it and it goes back again to the same position right?
 
I think your theta should be flipped to the bottom. Currently, your amplitude is 150^{o}.
 
Last edited:
-EquinoX- said:
ok, and the period of this pendulum is basically the time it takes the position I move the pendulum 30 degrees away from the equilibrium then I release it and it goes back again to the same position right?
Right. The period is the time it takes for one complete cycle. If it starts out to the right at 30 degrees at t=0, then in one period it will have gone to equilibrium, moved to the left at 30 degrees, come back to equilibrium and then back to the starting point at 30 degrees to the right.
 
kevtimc said:
I think your theta should be flipped to the bottom. Currently, your amplitude is 150^{o}.
No, the angle shown is the correct amplitude.
 
Doc Al said:
Right. The period is the time it takes for one complete cycle. If it starts out to the right at 30 degrees at t=0, then in one period it will have gone to equilibrium, moved to the left at 30 degrees, come back to equilibrium and then back to the starting point at 30 degrees to the right.

ok, thanks that helped a lot... so in one period it passes the equilibrium point twice then
 
  • #10
Doc Al said:
No, the angle shown is the correct amplitude.

EDIT: Nevermind, I see the image now, and I missed the T part. It's correct.
 

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