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Phy 231
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: ->->->->->->->->->->->-> :
There is a force which is the tension in the string. Because it is pulled 10 cm back in the negative x direction, the tension is upward and to the right. There is a downward force resulting from the weight of the pendulum.
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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: ->->->->->->->->->->->-> :
The pendulum string is upward and to the left of the pendulum. That would mean that the vector is in the same direction as tension, going up the length of the string.
To find the direction, you need to find the angle. To find the angle you would use: actan (opp/ adj)
The length of the string is 2 meters, or 200 cm, and the distance along the x axis is -10 cm. Therefore, actan(200/-10) = -87 degrees
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The angle is
actan(200/-10) + 180 degrees.
When the x component is negative the angle is in the quadrant directly opposite the arcTan.
The correct angle of the string, and therefore the tension, relative to the positive x axis is therefore 92.9 degrees.
This agrees with your subsequent result, but is obtained somewhat more directly and possibly with less chance of error.
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To find it from the positive axis, 90-87.14 = 2.86
2.86 + 90 = 92.9 from the positive x axis.
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What is the direction of the tension force exerted on the mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The direction of the tension force is the same as the vector. The angle would b 92.9 degrees.
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What therefore are the horizontal and vertical components of the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
The x comp of T: Tx = 5 * cos(92.9) = -.25N
The y compof T: Ty = 5* sin(92.9) = 4.55 = 5 N
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What therefore is the weight of the pendulum, and what it its mass?
answer/question/discussion: ->->->->->->->->->->->-> :
The pendulum is in equilibrium. Therefore, the weight must be equal to Ty. So, the weight is 5N
The mass of the pendulum is F = m * g
5 = m * 9.8
m = .510 kg
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What is its acceleration at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Fnet = m * a
We need to use the x comp of tension.
-.25 = .510 * a
a = -.490 m/s^2
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50 min
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Very good.
Note that the pullback is 1/20 the length of the pendulum, the net force is very close to 1/20 the weight of the pendulum and the acceleration is very close to 1/20 the acceleration of gravity.
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