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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 the mass at the origina with a length of the string of 2 m and point of top of the string at (-.10 m, 2 m). I also represented T as 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: ->->->->->->->->->->->-> :
theta = arctan(2m / -0.10 m) = -87 deg. +180 deg. = 93 deg from the positive x axis.
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What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
towards the top of the pendulum.
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More specifically the direction is at 93 degrees, which is understood to be relative to the positive x axis as measured in the counterclockwise direction.
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What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
T_y = 5N * sin 93 deg. = 4.99 N = 5 N
T_x = 5N * cos 93 deg. = -0.3 N
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What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
T_y = F_g = m*g
5N = mass * 9.8 m/s^2
Mass = 5 N / 9.8 m/s^2 = 0.51 kg = 0.5 kg
F_g = 0.5 kg * 9.8 m/s^2 = 4.9 N
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What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
F_net_x will be 0.3 N
A=F_net_x / m = 0.3 N / 0.5 kg = 0.6 m/s^2
This may not be right but I took a shot at it. I am not sure if the F_net_y will account for some acceleration or not. I did figure the mag. of the F_net to be 0.3 N at 162 deg. F_net approx. = F_net_x approx, so I guess this is correct.
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The net force in the y direction is very nearly zero, the upward component of the tension exerting very nearly 5 N of force and very nearly 5 N of gravitational force.
As the pendulum moves the tension actually increases a bit to exert the centripetal force necessary to keep the motion of the mass circular, but for small oscillations this is pretty much negligible.
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Clearly also the mass does descend a bit, so there is some acceleration in the vertical direction, but once again for a small oscillation this is pretty much negligible.
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&#Good responses. See my notes and let me know if you have questions. &#