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phy121
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.
** CQ_1_26.1_labelMessages.txt **
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 diagram is a straight vertical line from the horizontal. The pendulum is drawn on the right side with a displacement of 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: ->->->->->->->->->->->-> :
If pulled in the X direction, the direction will be positive before the release of it, since it is pulled up to the right.
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The direction of the tension has to be opposite to the pull. Since I said above that it was pulled in the positive direction, therefore the tension force is in the negative direction.
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
F=m*a
5N= m*9.8m/s^2
M= 0.51kg
Arctan(10cm/200cm)= 2.86 degrees
Horizontal:
F= -m*g*cos(theta)
F= -0.51kg*9.8m/s^2*cos(2.86)
F= -5N
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
F=ma
5N=m*9.8m/s^2
M= 0.51kg
0.51kg*2.2= 1.12lbs
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
F=ma
5N= 0.51kg*a
A= 9.8m/s^2
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20 minutes
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Most of your responses are good, but you have a few misconceptions about the overall picture, especially regarding the direction of the tension force and the acceleration of the mass (string tension opposes most of the force of gravity so the acceleration is not 9.8 m/s^2; the acceleration is due just to the horizontal component of the tension).
Check out the link below. If you're not sure you understand everything, a revision is welcome, but is not otherwise required.
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&#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. &#