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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 Newtons.
• Sketch the system with the pendulum mass at the origin and the x axis horizontal.
answer/question/discussion: ->->->->->->->->->->->-> :
I sketched the system with the pendulum hanging downward and being pulled to the right site 10cm.
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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 direction would be in the positive x direction .
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The direction of the tension would be opposite of the direction it is pulled so the tension would be in the negative direction.
The tension force acts in the direction of the string.
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
The x component is : Tsin (10)
The y component is Tcos(10)
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The weight of the pendulum would be the x and y added together so, 5.77N and then multiply this by 9.8 for gravity would be the mass of: 56.55N.
To get the magnitude of a vector you don't add its horizontal and vertical components.
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
The acceleration would be with gravity so it would be 9.8m/s/s
The net force on the weight is not m g, so the acceleration is not g.
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10 minutes
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You should submit a revision on this one.
Note that T sin(10 deg) = T cos(80 deg), and T cos(10 deg) = T sin(80 deg). If the angle of the pendulum is measured relative to positive x axis, the angle is 80 deg and we can use the general rule that T_x = T cos(theta) and T_y = T sin(theta). If you know right-angle trigonometry well enough to use it here, then your components are correct, provided the string extends from the mass into the first quadrant.
&#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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