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PHY 121
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** 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: ->->->->->->->->->->->-> n:
The pendulum when at rest is aligned with the vertical axis. After it is pulled back, the pendulum is at 10cm along the positive horizontal axis.
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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: ->->->->->->->->->->->-> n:
pendulum is pulled .1m on positive x so x component is -.1m
theta = arctan(2m/-.1m) + 180
=92.9degrees
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What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> n:
92.9degrees
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What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> n:
Tension_x = 5N * (cos(92.9))
=-.26N
Tension_y =5N * (sin(92.9))
=5N
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What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> n:
Pendulum is at equilibrium
Weight = tension_y
=5N
M = weight/gravity
=5N/9.8m/sec^2
=.51kg
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What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> n:
a = F_net/m
a = .25N/.51kg
=.49m/sec^2
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&#Very good responses. Let me know if you have questions. &#