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PHY 201
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 Newton’s.
• Sketch the system with the pendulum mass at the origin and the x axis horizontal.
answer/question/discussion: ->->->->->->->->->->->-> :
The pendulum at the origin is 10 cm’s from the vertical axis, and the length of the pendulum is labeled 2 m with a tension of 5 N.
The pendulum sits right above the horizontal positive x-axis 10 cm’s from it, but the angle now has 90 degrees added to it, because we moved it from the vertical to horizontal axis. The tension of the string is 5 N, which is labeled right above the line vector, along with the length of the pendulum 5 N.
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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: ->->->->->->->->->->->-> :
I think that the direction of the pendulum string is going downwards towards the mass on the end of the string.
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension is going up the string from the mass on the end of the string.
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
The x and y components:
x-comp: 5 N cos(angle)
y-comp: 5 N sin(angle)
After I sketched my picture, I now two sides of the triangle, which is the length of the pendulum and the length at which it is pulled back straight from the x-axis. Since, I know two sides I can find the angle: tan-1 = (10 / 200 cm) = 2.87 degrees. Since, I moved the angle to the positive x axis, the angle is now: 2.87 + 90 deg = 92.87 degrees. Now, I can find the x and y components of the tension:
x-comp: 5 N cos(92.87) = -.25 N
y-comp: 5 N sin(92.87) = 4.99 N
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
I am not sure how to get the weight of the pendulum, and in order to get the mass I would need the weight of the pendulum, right??
The weight of the pendulum is directed in the negative y direction. The net force in the y direction is zero, since the pendulum is neither rising nor falling (at least not very much). So the weight is equal and opposite to the y component 4.99 N of the tension.
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
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45 minutes
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You won't need to revise, but be sure you understand:
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your 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. &#