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course Phy 201
Oscillating Candy Bar
You suspended a candy bar from a rubber band chain and counted its oscillations for a minute. You did the same for a stack of 2 dominoes, and for a stack of four dominoes.
Insert a copy of your data here, along with any previously submitted work you wish to include:
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ROLOS
140 bounces in 1 min
4 dom = 112 in 1 min
8 dom = 67 in 1 min
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For each number of dominoes, determine the force constant k, based on your observations of the number of oscillations in a minute (which you can use to find the angular frequency omega) and the mass of the domino stack.
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Rx = R cos theta = 1 cos 45 =.71
Ry = R sin theta = 1 sin 45 = .71
omega= sqrt k/m
k= .4 with rubberband
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Are your values of the force constant reasonably consistent?
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45 degrees isn't relevant here. The system oscillates in one direction, up and down, so the entire motion is along a single vertical straight-line axis.
What is omega for 4 dominoes?
What is the mass of 4 dominoes?
Using omega = sqrt(k/m), what do you get for j?
Go through the same questions fo r 8 dominoes.
You will get two different values of k. They will be different, but reasonably consistent with one another.
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yes
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Based on your best estimate of the force constant and your observation of the frequency of the oscillations of the candy bar, what is the mass of the candy bar?
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Randians = 2pi/time = 6.28/60 sec = .104 rad/sec
m= .4/(.104^2)
m= 37 grams
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That's a reasonable mass, but it doesn't follow from your data.
In 60 seconds that oscillator goes through 112 cycles, so you would have 112 * 2 pi / (60 seconds).
m = k / omega^2, so you're on the right track with that.
But k isn't .4. Remember also that k has units.
Use the average of the k values you got when you corrected the 4-domino and 8-domino cases.
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