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PHY 201
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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: ->->->->->->->->->->->-> :
OK
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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=.1
y=sqrt(2m^2-.1m^2)
y=1.99m
The angle will be tan^-1(1.99/.1)=87, 180-87=93
degress.
The direction of the of the pendulum string is
93 degrees from horizontal.
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What is the direction of the tension force
exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The y components are 93 degrees from horizontal
and 270 degress. The x component are 0 degrees
and 180 degrees.
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The tension force is in the direction of the rubber band, at 93 degrees.
The gravitational force is at 270 deg.
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What therefore are the horizontal and vertical
components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
x=T(cos93)=5N(cos93)=-.26N
y=T(sin93)=5N(sin93)=4.99N
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What therefore is the weight of the pendulum,
and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
(4.99kg*m/s^2)/9.8m/s^2=.5kg approx
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What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
5N/.5kg=10m/s^2 approx
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The y component of the tension is 4.99 N upward, and the weight is 4.99 N downward. The net force in the y direction is zero. The weight balances the y component of the tension.
There is not a net force of 5 N acting on the pendulum.
Only the x component of the tension is unbalanced, so the net force is the x component -.26 N of the tension force.
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*#&!
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