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phy 231
Your 'cq_1_26.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A simple pendulum has length 2 meters. It is pulled back 10 cm from its equilibrium position and released. The tension in the string is 5 Newtons.
Sketch the system with the pendulum mass at the origin and the x axis horizontal.
answer/question/discussion: ->->->->->->->->->->->-> :
The sketch would show a force vector vertically downward from the weight of the pendulum and a force vector vertically upward from the tension in the string and equilibrium.
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Sketch a vector representing the direction of the pendulum string at this instant. As measured from a horizontal x axis, what is the direction of this vector? (Hint: The y component of this vector is practically the same as the length; you are given distance of the pullback in the x direction. So you know the x and y components of the vector.)
answer/question/discussion: ->->->->->->->->->->->-> :
The pendulum pulled back would have a horizontal and vertical component. Since the length it’s pulled back is small compared to the length of the pendulum, the y-component is approximately the length of the pendulum.
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What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
Make an x, y axis origin at the point where the string is pulled back to, then the string’s top is located at approximately the point (-10, 200) on the axis. This forms a right triangle in which we can find the angle arctan(200/-10) = -87.1 degrees below horizontal.
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@& The x component of the tension force would be negative, so you would add 180 deg to this angle, obtaining 92.1 deg. This should be consistent with your sketch, which would indicate a tension force directed into the second quadrant.*@
What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
To find the components of the tension you need the angle 180 - (87.1 + 90) = 2.9 degrees.
Ty = 5N cos 2.9 degrees = 4.99 N
Tx = 5N sin 2.9 degrees = .25 N
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@& Using the 92.1 degree angle you would get the correct signs on your tensions.
The x component of the tension would be 5 N cos(92.1 deg) = -.25 N.*@
What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The vertical component of the tension should give you the mass: m = F/a = 4.99 N/9.8 m/s/s = .51 kg
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What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
a= Fnet/mass = .25 N/.51 kg = .49 m/s/s
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*#&!
@& Very good but see my notes. No revision is necessary but check the discussion at the link below.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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