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course Phy 121
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:
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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:
y-component=2m
x-component=.1m
vector direction= arctan(2m/.1m)=87.14m from x axis or 177.14 degrees depending on if it was pulled in the positive or negative x direction.
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What is the direction of the tension force exerted on the mass?
answer/question/discussion:
The force is exerted away from the pivot point of the pendulum towards the weight.
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The pullback force has this direction.
The tension in the pendulum string, however, has an x component equal and opposite to the pullback force, acting back toward equilibrium. (This is consistent with the orientation of the string).
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What therefore are the horizontal and vertical components of the tension?
answer/question/discussion:
Fy=5N*sin(87)=4.99N
Fx=5N*cos(87)=.26N
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What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion:
weight=4.99N+.26N=5.25N
F=M*a
5.25N=M*9.8m/s^2
M=.54kg
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What is its acceleration at this instant?
answer/question/discussion:
5.25N-5N=.25N= net force acting on system
F=M*a
.25N=.54kg*a
a=.46m/s^2
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Self-critique (if necessary):
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Self-critique rating:
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&#This looks good. See my notes. Let me know if you have any questions. &#