<p>heh might be kinda impaired right now</p>
<p>Hah okay. Maybe the AAPT got a report of cheating or something so that’s why it’s “drastic”?</p>
<p>Ha ha you guys are cracking me up 
I doubt it has anything to do with cheating. Easier just to disqualify and release the qualifiers. I think the “drastic” scores imply that the scores do not follow the expected statistics. Another theory is that the answer key is incorrect :). The reason I say the scores are likely low is because all the predicted cutoffsso far are pretty low. Torus is the only one predicting 15, everyone else predict a cutoff between 11.5 to 12.5. </p>
<p>@togerbookmark Can you provide the link on Artofproblemsolving.com?</p>
<p>@woody123 - here you go: <a href=“http://www.artofproblemsolving.com/Forum/viewtopic.php?f=913&t=572987&start=20”>http://www.artofproblemsolving.com/Forum/viewtopic.php?f=913&t=572987&start=20</a></p>
<p>@tigerbookmark I’m predicting a 15 because I roflstomped the test. They might be reviewing their answer for the helicopter problem, that one has stirred up quite a bit of controversy…</p>
<p>Lolwut? roflstomped? as in … ?
The helicopter controversy might be why, but the AOPS post said that the AAPT was surprised …</p>
<p>@torusdonut - It’s plausible that their key is faulty. On Friday, AAPT was so sure that they would release the qualifiers by Tuesday at the latest. But once the key was released late Friday, someone may have expressed concerns with the key. That may have been valid enough to force AAPT to re-grade (explains why it takes them another week to get the results out).</p>
<p>@jx1729 - The “surprise” comment is the main reason why I believe that the cutoff is very low (maybe below 10 even). AAPT would never expressed “surprised” if the scores were high because that would be insulting to the students (i.e. that they have such low expectation of the student pool). Given that the level of difficulty is comparable to last year’s exam, a cutoff around 12ish would not have surprised them given the a priori likelihood. Based on these reasons, I believe the cutoff may have fallen below 10 How’s that for making everyone feel better.</p>
<p>i can guarantee that the cutoff will be low this year. please don’t worry guys. the cutoff. will be low.</p>
<p>99.99999% sure it’s because they answered the helicopter wrong.
^ cutoff will stay low, it’s probably because the cutoff ± 1.25 points had so many people in the range that they had to reevaluate the entire cutoff</p>
<p>I agree that the cutoff will probably be very low (12ish), but why is their answer for the helicopter problem wrong? Everyone I spoke to got the same answer, and my physics teacher agreed with the key.</p>
<p>@pk5144, what was your physics teacher’s solution?</p>
<p>Which helicopter problem? The one I analyzed and my son got it wrong?</p>
<p>Yes, the one and only legendary helicopter problem</p>
<p>There are other 2 problems which mentions the word “helicopter”</p>
<p>Then we mean the rope problem</p>
<p>Are we still on the helicopter rope problem? I missed this one too, but here’s an explanation after the fact that justifies the answer in the key: the point where the string is attached to the helicopter is a fixed point. Immediately below the fixed point, the string has the maximum leverage to move because the pivot is fixed. The farther down the flexible string, the string would have to move against a moving pivot, and the leverage you have decreases the farther down you go. This is because the farther down the string would have more degrees of freedom to move. The tip at the bottom would therefore be closest to having the same direction as gravity. So the rope is close to vertical at the free end.</p>
<p>Anyways, there is probably a more rigorous solution to this using a calculus of variation solution based on optimizing the string shape to minimize potential energy and some other parameter like entropy :)</p>
<p>there is no such thing as a moving pivot in this problem because we are talking about the case when the rope is in STATIC EQUILIBRIUM and therefore the “pivot” stays put. Also I don’t know what you mean by “leverage to move” and “degrees of freedom”. I don’t know if you mean torque by leverage to move, but that’s not a physical term. And degrees of freedom implies that the rope can move in different dimensions, which in case is false.</p>
<p>Also we have established many times that the rope is slanted and straight, with zero counterarguments for THREE RIGOROUS solutions, while every single argument for other answer choices has been refuted (without counter-refutations).</p>
<p>Here are two solutions that majorineverythin and I posted beforehand:</p>
<p>Mine:
When the helicopter is in its steady state, each rope element is moving at the same speed. This means that the force of air resistance on each mass element will be identical, and therefore we can consider the air resistance to act as a gravitational field. This situation is now equivalent to a rope under the Earth’s gravitational field in conjunction with a virtual gravitational field acting in the opposite direction of motion. The net field is a straight slanted line. The rope will take on the shape of this constant gravitational field. Thank you for your attention. </p>
<p>Majorineverything:
Every differential segment of the string will have the same net tension force acting on it, since there are effectively two tension forces on it - one from above pulling it up, and one from below pulling it down. The vertical component of the tension balances gravity, resulting in a forward-directed net force, which depends on the angle. A simple free-body diagram and calculations will give you T = mg/tanθ, where θ is the angle from the vertical. Air resistance is constant and proportional to the cross-sectional area (and thus angle) and the speed (which is constant, as stipulated in the problem). This gives you something like F = CLcosθ*v^2, where C is a constant and L is the length of the segment of string. Both of these are functions of θ. For the string to be in equilibrium - still moving relative to the ground, but with zero net force, the two forces must be equal. This can only occur at one value of θ, meaning a curve cannot be a configuration in equilibrium - therefore it must be incorrect.</p>