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PHY 201
Your 'question form' 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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Regarding the following problem, did I complete it in the correct way?
A washer of mass 21 grams is rotated in a circle at the end of a string of length 2.9 meters. If the breaking strength of the string is 4 Newtons, what is the maximum rate at which the washer can be rotated in this manner?
.tau=I*a
4N=.21 kg(2.9m^2)*a
4N=1.8 kg m^2*a
2.222 m/s^2=a
The string exerts the centripetal force required to hold the washer in its circular path.
The centripetal force is m v^2 / r.
So when m v^2 / r exceeds 4 N, the string will break.
You would thus solve the inequality
m v^2 / r >= 4 N,
obtaining
v >= sqrt( 4 N * r / m) = sqrt(4 N * 2.9 m / (.021 kg) ).
Additionally, I have a couple of more questions regarding my test prep. There are several questions that have popped up that begin coasting from rest down a certain hill, whose slope is variable, I reach a speed.... Given that it is not uniform acceleration, can you point me in the right direction to begin that problem?
Use energy conservation. `dPE_ac = `dPE_ab + `dPE_bc, where a and c are the top and bottom of the hill and b is the intermediate point.
Any `dPE is equal and opposite to the corresponding `dKE. So any combination of PE and KE changes for two phases gives you `dPE and `dKE for the third. For example if you know `dKE for the second half and `dPE for the entire hill, then you know `dPE for the entire hill and the second half. You can easily find `dPE for the first half.
If you are given height and/or velocity information, this can immediately be translated into PE and/or KE information. And if you have the PE and/or KE information, you can easily translate this into information about height and/or velocity.
Also, the infamous rollercoaster question: A roller coaster is at the top of a circular loop of radius 20 meters. The roller coaster is turned upside down. How fast must it be moving for a passenger sitting on a scale to register her usual weight?
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I asked about this one on my in class questions at one point. Again, can you potentially point me in the right direction on how to begin this question?
Thanks,
Hope you had a great Thanksgiving!!
The centripetal force required to hold her in the circle is m v^2 / r. At the top, gravity contributes m g to her centripetal force. If the scale reads her correct weight, then it is pushing down on her with force m g. So her centripetal force is 2 m g.
Solving m v^2 / r = 2 m g for v, we get v = sqrt( 2 g r).
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Good questions. See my notes. We'll also look at this in class tomorrow.