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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 drew a dotted line coming down 2 m to represent the pendulum length. I then drew a solid line on the x-axis going 0.10 m to the right to represent the total distance pulled from equilibrium. I then drew another solid line from the end of the x-axis line to the beginning of the y-axis dotted line that represented the pendulum length.
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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 vector on the x-axis would start at origin and go 0.10 m towards the east.
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This vector has magnitude 0.10 m. The string is much longer than that.
The vector representing the direction of the string has both an x component of length 0.10 m and a length of 2.0 meters; its y component is very close to its length.
You need to specify the direction of this vector. Its direction will be an angle measured from the positive x axis.
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
tan^-1 (2 m / 0.10 m) = 87 deg. However, I would say the point from where the mass is pulled back 0.10 m to the top of the string would be 180 deg - 87 deg = 92.8 deg.
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Your answer here covers the direction nicely. I agree with the 92.8 degrees as well as your reasoning.
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
horizontal component would be 5 N cos 92.8 = -0.244 N
vertical component would be 5 N sin 92.8 = 4.99 = 5 N
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
Not sure where to begin.
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The pendulum doesn't fall straight down, so its weight is pretty much supported by the y component of the tension. At an angle of 92.6 degrees, the y component of the tension won't differ significantly from the vertical component.
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Unable to answer this question until I can answer the previous.
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15 mins
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Good up to the very end.
See if my note helps.
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