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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: ->->->->->->->->->->->-> :
I assume the pendulum is pulled back in the positive x direction and then the pendulum string goes from the origin up and to the left to point (-10 cm, 200 cm). The direction vector runs along the string and measured counterclockwise from pos. x
Angle = arctan (200 cm/-10cm) + 180 = 92.8 degrees
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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 vector runs along the string and measured counterclockwise from pos. x
Angle = arctan (200 cm/-10cm) + 180 = 92.8 degrees
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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 an angle of 92.8 degrees measured counterclockwise from the positive x axis
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
Tx=5 newtons * cos 92.8 degrees = - .25 newtons
Ty = 5N * sin 92.8 degrees = 4.9 Newtons
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
since the pendulum is in equilibrium its weight is the same as the vertical componenet of tension so 4.9 newtons. Its mass would be mass = weight/g = 4.9 newtons /9.8m/sec^2 = .5 kg
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
a = netforce / mass = -.25 newtons / .5 kg = -.5 m/sec^2
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&#Very good responses. Let me know if you have questions. &#