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course PHY 241
11/13/20118:37 pm
Only putting the data up for now. I have sent an email as to why the third experiment is missing. Also, the rest of this assignment will have to be put on hold, as I have a number of questions." "Balancing dominoes experiment:
Report all relevant data from the experiment, being sure to clearly identify the quantities you report and what they mean.
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left dominoe: 14 cm from center
right dominoe: 14 cm from center
second right dominoe: 8 cm from the center
weight on left: 15 grams
weight on right: 30 grams
distance from actual center to equilibrium: 2.5 cm
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Find the torque produced by the weight of each domino on the first beam, about the point of rotation. You may assume domino weights of 15 grams each.
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???
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Candy bar experiment:
How many oscillations did the candy bar complete in a minute?
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130
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What was the length of the rubber band chain when supporting 4 dominoes?
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35 extra mm
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What was the length of the rubber band chain when supporting 8 dominoes?
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72 extra mm
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Through how many radians did the reference point move during the 1-minute timing (it moved through a complete circle for every cycle you counted)?
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260pi
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What therefore was the angular velocity omega of the reference point?
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2.1670 oscillations per second
2pi per oscillation
4.334pi/second
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How far was it, on the average, from the lowest point to the highest point in the candy bar's oscillation (you didn't measure this; just visualize the motion and make an estimate)?
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5 cm
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The diameter of the reference circle is equal to the estimate you made for the preceding question. How fast was that reference point moving around the arc of that circle?
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velocity = 4.334pi/second
diameter = roughly 5 cm
21.67 pi cm/second
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What is the slope of the force vs. length graph for your rubber band chain? If you measured the domino stack then you can use the fact that for every millimeter of height the stack has mass 1.9 grams. If not just assume a mass of 15 grams.
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first length difference: 35 mm
second length difference: 75 mm
mass = 60 grams
mass = 120 grams
60/40 = 30/20 = 3/2
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Gravity exerts a force of about .6 Newtons on a mass of 60 grams. So the tension force corresponding to 60 grams is about .6 Newtons.
Your slope would thus be about
.6 N / (40 millimeters) = .0015 N / mm,
the same as about
1.5 N / m.
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Your last answer is your estimate of k, and your count resulted in your previous answer for omega. What therefore is m?
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omega = sqrt( k / m ).
You (nearly) found k, which is the slope of that graph.
You earlier found omega.
So you can find m.
Overall not bad so far. Check my notes.
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