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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.
** CQ_1_26.1_labelMessages.txt **
Samantha Rogers
PHY 201
Seed 26.1
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
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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: ->->->->->->->->->->->-> :
angle = arcTan(200 cm / (-10 cm) ) = 2.9 deg + 90 deg = 92.9 deg.
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension force on the mass is directed along the string, at angle of 92.9 deg as measured counter clockwise from the positive x-axis.
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
the tension is 5 N and the components are
T_x = 5N * cos(92.9 deg) = .25 N
T_y = 5N * sin(92.9 deg) = 5 N
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
the pendulum is in equilibrium, the weight is the same as the vertical component of the tension (5N).
Mass is m = weight / g = 5 N / (9.8m/s^2) = .51 kg
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
a = Fnet / m = /25 N / (.51 kg) = .49m/s^2
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&#Very good responses. Let me know if you have questions. &#