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course Phy 241
12/7 11
TIMER-related experiment:Everyone should do the exercise at
http://vhcc2.vhcc.edu/dsmith/forms/ph1_2_testing_hypothesis_regarding_time_intervals.htm
The exercise uses the TIMER program.
Distance students generally report little difficulty completing this exercise in 30-60 minutes.
Slope at which domino tips:
At what slope does your thickest domino tip?
You might well have dominoes of two or even three different thicknesses. How many different thicknesses do you have?
Test a domino having each of these thicknesses similarly.
Briefly report your results and how they were obtained:
!!!I only have 2 domenios and they have same thickness, but they tiped at different slopes!!!!!
1st = .1443(slope
2nd = .1199(slope)
Started off with level plane and then elevated until domino tipped over. Then to find slope I noted the ht I had raised the plane domino was sitting on and measured dist from one end of plane to the opposite end. Using Pythagorean theorem and knowing hypo side and opposite side I found run. Knowing slope = rise/run, I then divided the rise(or ht) by the run to get the slope
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Refer to the analysis in Class Notes of the cart-suspended mass-rubberbandChain system we've been observing in class.
The cart system in was assumed in that analysis to have mass .5 kg, which resulted in the conclusion that the domino should tip when the amplitude of the oscillation is greater than 5 cm. This was in contrast to the observation that the tipping point occurred at an amplitude of about 3.8 cm. If the actual mass of the cart system is .6 kg, how does this affect our analysis and our comparison with the observed tipping amplitude?
Observing that domino tips at accel = 2 m/s^2, where we found omega = sqrt( k / m ) = 6.4 rad/s we would have to recalculate omega using info provided
omega = `sqrt(k/m) = `sqrt(20N/m/.6kg) = approx. 5.7735rad
We then would recalculate omega^2*A >= 2m/s^2
omega^2*A >= 2m/s^2
A >= 2m/s^2/omega^2 = 2m/s^2/(5.7735rad)^2 = approx. .06m
So if A >= 6cm then tipping of the domino would occur. So by changing the mass by .1kg this would in turn move the dist that the cart needed to placed in order for domino to tip back another cm so that domino shouldn’t tip unless A >= 6cm.
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We estimated the 2 m/s^2 'tipping acceleration' based on data that had an uncertainty of +- 10%, and the estimated mass of the system, which we might consider to be uncertain by +-20%. Are these uncertainties sufficient to explain the discrepancy between the observed tipping amplitude (3.8 cm) and the predicted tipping amplitude (5.0 cm)?
University Physics students should do a symbolic solution, finding the expression for A in terms of the 'tipping acceleration', k, and m then calculating and applying the differential of this expression. For the moment consider our value of k to be accurate, with negligible uncertainty (ain't so, but for simplicity assume it).
omega = `sqrt(k/m) and the centripetal acceleration vector has magnitude = omega^2*A.
a_max = a_cent_ = v^2/r = (omega*A)^2/r = omega^2*A
So for tipping point
a_max >= 2m/s^2
omega^2*A >= 2m/s^2, now we solve for A
A>= 2m/s^2/omega^2
A = 2m/s^2/ omega^2
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General College Physics students may simply use the appropriate maximum and/or minimum values of 'tipping acceleration' and mass, along with the given value of k, to calculate the predicted 'tipping amplitude'. If you're unsure what this means, start by assuming 10% greater 'tipping acceleration' and 20% greater mass, and calculate the resulting amplitude necessary to tip the domino; determine by what % this differs from the result .05 m obtained in the notes.
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&#Your work looks good. Let me know if you have any questions. &#
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