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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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SEED Question 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: ->->->->->->->->->->->-> :
Given:
L = 2 m, ds = 10 cm = 0.10 m, T = 5 N
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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: ->->->->->->->->->->->-> :
sin (theta) = 0.10 / 2 = 0.05
theta = 2.86 deg
tan (2.86) = 0.05 deg
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• What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The direction will be the right angle (90 deg) + or - (angle theta) 2.86 from the x axis. The pendulum could be pulled back in either the left or right direction.
90 deg + 2.86 deg = 92.86 deg from the + X-axis or 90 deg - 2.86 deg = 87.14 deg from the - X-axis.
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• What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
w * sin (theta) is the horizontal component.
0.044 * w * cos (theta) is the vertical component.
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• What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
w = 5 N
w = mg
m = w / g = 5 N / 9.8 m/s^2 = 0.5 kg
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• What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
F = -mg * sin (theta)
F = - (0.51 kg) * (-9.8 m/s^2) * sin (0.05)
F = 0.25 N
F = m * a
a = F / m
a = 0.25 N / 0.51 kg
a = 0.5 m/s^2
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Solution
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