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: ->->->->->->->->->->->-> :
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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: ->->->->->->->->->->->-> :
x = 0.1 m
0.1m^2 + y^2 = 2m^2
y = 1.99 m
tan-1(0.1 m / 1.99 m) = 2.88 + 90 degrees
92.88 degrees
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
92.88 degrees
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
x = 5 N * cos(92.88)
x = 0.25 N
y = 5 N * sin(92.88)
y = 4.99 N
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
w = 5 N
the weight would be equal and opposite to the y component you found above; it round to 5 N but you correctly expressed it above to 3 significant figures as 4.99 N
m = w/g
m = 5 N / 9.8 m/s/s
m = 0.51 kg
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
a = 0.25 N / 0.51 kg
a = 0.49 m/s/s
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30 mins
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Good responses. See my notes and let me know if you have questions.