tag:blogger.com,1999:blog-5088860151651047897.comments2017-06-23T05:35:34.263-04:00Jacobs PhysicsGreg Jacobshttp://www.blogger.com/profile/03854009948036330746noreply@blogger.comBlogger779125tag:blogger.com,1999:blog-5088860151651047897.post-2677991217131105142017-06-19T13:22:25.089-04:002017-06-19T13:22:25.089-04:00(1) From a force point of view, the two forces (gr...(1) From a force point of view, the two forces (gravity and normal) are equal and opposite and cancel out. When forces cancel this means there is no acceleration - which matches our scenario. <br /><br />(2) From a Work point of view, the two forces (gravity and normal) are doing the same amount of work, but Normal is doing positive work, and gravity is doing negative work. When Works cancel, there is no change in kinetic energy - which matches our scenario. <br /><br />(3) From a Potential Energy point of view, as the elevator rises, it gains gravitational potential energy. Also, Normal force is doing positive work. We see that all of the positive work that the Normal force does is stored as gravitational potential energy; so, the elevator's KE hasn't changed, but it has gained potential energy - which matches our scenario. <br /><br />----------------------------------------------------------------<br /><br />The key point is how gravity is considered with respect to energy. It can EITHER be considered as a conservative force (i.e., the work it does is CONSERVED as potential energy as in explanation 3 above), or a non-conservative force (i.e., the work it does is NOT conserved as potential energy as in explanation 2 above). In explanation 2, we don't consider PE related to gravity, and in explanation 3, we don't consider the work related to gravity. <br /><br />I hope that helps!Michaelhttp://www.blogger.com/profile/06967354327638923275noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-44565282893318166652017-06-16T18:58:53.545-04:002017-06-16T18:58:53.545-04:00Sorry I wasn't clear, but my context was learn...Sorry I wasn't clear, but my context was learning (or perhaps using without error) rather than the grading rubric or getting them to annotate their work. But I also think that equations and description sometimes come into conflict with one another, and that might lead to resistance by students. Let me try again. <br /><br />I take them directly from Delta E = W to Ef = Ei + W and then use the latter when doing examples of how to solve problems for only one reason: it mirrors Vf = Vi + a*t and is thus more consistently memorable. Teaching philosopy, or pedagogy if you prefer. Other orderings, whether Vi + a*t = Vf or Vf-Vi = a*t will convey the same physics, but those do not convey the idea that Vf is actually a function V(t), and functions are always defined on the left side of an equation. [I've never seen a textbook present Vi + a*t = Vf as The Equation.] I want them to think of momentum and energy the same way, as an alternative way of understanding the dynamical evolution of a system, and use that standard ordering to get work in the right place every time. <br /><br />Order is unimportant in Pf = Pi or Lf = Li problems, and that is about all you get at your level and mine. External impulse problems are rare in intro classes. I don't care how they write it, but I approach momentum consistently just for consistency. Pf = Pi + Favg*t, if you like, even if that is rarely used on a test. <br /><br />What complicates all of this is the descriptive process tends to be in temporal order. That is what leads to your statement (and mine) about what happens in that collision. We draw or describe motion diagrams from initial to final, and we should translate that into mathematics in the same sequence, Ei + Work = Ef, which is the exact opposite of how the equations are presented in textbooks! One of the two has to give if we are to push description first, before math, because students take The Book as gospel. CCPhysicisthttp://doctorpion.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-80521474871477697772017-06-15T08:37:50.111-04:002017-06-15T08:37:50.111-04:00Hi Greg! I would also love to get some information...Hi Greg! I would also love to get some information. I'll be a first year teacher this year and will be teaching some conceptual physics. I've just sent you an email as well I'd love to make it to one of your summer institutes, but I don't think that will be an option this summer. Brittany Neashttp://www.blogger.com/profile/07654225011343013750noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-75184981694846796572017-06-14T14:53:55.093-04:002017-06-14T14:53:55.093-04:00Edited for reasons of previous stupidity!
Hi Micha...Edited for reasons of previous stupidity!<br />Hi Michael. Thank you for your explanation but I'm still struggling; I must be missing something but please indulge me! Surely in your first system, gravity and normal forces are "conservative" and cancel, not as you describe. In your second, gravity is conservative and the elevator force (generated by the motor) is non-conservative. By this I understand that the elevator motor must perform more work than the increase in energy of the system because some is lost as heat, noise, vibration etc. I don’t understand what you mean by “gravity’s energy”. At the end of the ascent the increase in energy, mgh Joules, is the same work performed, F*d, also Joules (less losses in the second case) but I can’t see how F is derived. It still seems like no net work has been done at the top of the elevator journey, but it apparently has. I can't quite picture why just because one force is conservative and the other is not they can't be calculated in the ordinary way for vectors, and therefore sometimes cancelled. In this case I would find it hard to draw a vector diagram showing a force which has moved. Would you be able to do that?Hamish griffithshttp://www.blogger.com/profile/09336250080012848887noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-47828681878448076082017-06-14T11:06:03.962-04:002017-06-14T11:06:03.962-04:00This comment has been removed by the author.Hamish griffithshttp://www.blogger.com/profile/09336250080012848887noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-35640534342442117872017-06-13T22:02:21.880-04:002017-06-13T22:02:21.880-04:00Hamish, great question! I believe the issue lies w...Hamish, great question! I believe the issue lies with "double-counting" the force of gravity. The general formula for the conservation of energy is:<br /><br />PE+KE+Wnc=PE+KE<br />In words that would be: potential energy initial + kinetic energy initial + work done by non-conservative forces equals potential energy final + kinetic energy final<br /><br />So, where do you consider gravity's energy? As Work =F*d=mg*h? Or as PE=mgh?<br /><br />If the earth IS NOT part of the system, then there are two non-conservative forces: gravity and normal. Since they point in opposite directions, they cancel (or since one adds energy and the other removes energy, they cancel). Since the earth is not part of the system, then no PE can be stored. The equation boils down to 0=0.<br /><br />If the earth IS part of the system, then there is one conservative force (gravity from the earth) and one non-conservative force (normal force from elevator). The normal force does positive work as it lifts the person up, and on the other side of the equation, positive PE is gained since the person has been lifted off the ground. The equation boils down to N*d=mgh (with N=mg due to equilibrium and the distance moved is h, so mg*h=mgh). <br /><br />I hope that makes sense!<br /><br />Thanks,<br /><br />-Michael<br /><br />Michaelhttp://www.blogger.com/profile/06967354327638923275noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-12062840783937832862017-06-13T10:29:24.807-04:002017-06-13T10:29:24.807-04:00I hope you don't mind me joining this rather l...I hope you don't mind me joining this rather late. I used to fly aeroplanes and intellectual difficulty arises there also. My problem and, it seems that of others, is this; in the "constant velocity" ascending elevator the normal force exactly equals the "contact force" which is exactly opposite. There is no "net force" because the vector quantities exactly cancel. Even if you allow that the elevator does "positive" work and gravity does "negative" work the two also appear to exactly cancel. How is any net work accomplished? (It obviously is because the system ends up with more potential energy but I can't see that a net force has been moved through a distance, which was always my understanding of the definition of work. I have ignored the acceleration/deceleration which also cancel in a friction-less system). Interestingly (I think!) in the aeroplane example there is an analogous situation in flight at constant velocity; there thrust and drag are equal and opposite and similarly cancel but despite movement of the system there is no change in energy.<br /><br />Hamish Griffiths Hamish griffithshttp://www.blogger.com/profile/09336250080012848887noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-4916827473704524772017-06-13T09:03:37.034-04:002017-06-13T09:03:37.034-04:00Great point, CC... I certainly teach the mathemati...Great point, CC... I certainly teach the mathematics of energy conservation by starting with the change in the system energy, setting that equal to the sum of changes in each particular form of energy. Your approach is a good one.<br /><br />So how does that square with my discussion above? Perfectly. A typical student would start the derivation with L = L', and go from there. Your student would start with 0 = L'-L. Works for me! <br /><br />The AP exam is graded by humans, not robots, for a reason. We look to see if the student clearly communicated the starting point of the derivation -- that is, didn't just plug variables into equations, didn't just write a final answer, but said clearly what fundamental principle was used as a springboard.<br /><br />Thus, even the words "conservation of angular momentum" would be sufficient. So would "the rod-disk system's angular momentum can't change." <br /><br />And, note the point I made about annotation. I did see a number of papers with odd groupings of terms... often these were clearly labeled with the words "before collision" and "after collision". My overwhelming preference is for a student to communicate in both words and mathematics, in order to make discussions of mathematical order unimportant. :-)<br /><br />GCJ<br /><br />Greg Jacobshttp://www.blogger.com/profile/12121422726610824760noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-53310501639801088392017-06-13T08:50:57.487-04:002017-06-13T08:50:57.487-04:00I have a question about teaching philosophy. You ...I have a question about teaching philosophy. You wrote "with before the collision left of the = sign". All of dynamics is structured as change (final minus initial) equals whatever drives the dynamical change. That naturally leads to final velocity equals initial velocity plus a*t, or final energy equals initial energy plus net work. I encourage my students to stick with that ordering even when the change is zero. Do you have a reason for writing inital plus physics equals final? I have toyed with, and found success with, that reordering for collision and energy problems but would not dare try it with the kinematics equations. As I see it, the conflict is between consistency (which tends to eliminate fundamental sign errors) and linking the equation to a pictograph or textual description of what is happening. The latter tends to run from past to present to future. <br /><br />Your thoughts? CCPhysicisthttp://doctorpion.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-38084616822713659692017-06-12T10:46:53.367-04:002017-06-12T10:46:53.367-04:00Steve, I've often referred students straight t...Steve, I've often referred students straight to the source - to the official course and exam description. You can have students read the first 6 pages or so and the topic list. https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap-physics-1-course-and-exam-description.pdf. <br /><br />(I've also heard of teachers whose summer assignment is "procure the 5 Steps book" :-) )<br /><br />Take a look at the post linked here for a letter I wrote to my class in August outlining what to expect: https://jacobsphysics.blogspot.com/2016/08/school-starts-for-me-in-couple-of-days_29.html<br /><br />gregGreg Jacobshttp://www.blogger.com/profile/12121422726610824760noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-71971022846890817762017-06-12T07:39:40.801-04:002017-06-12T07:39:40.801-04:00Hi Greg,
Big fan of your blog, your AP1 prep book...Hi Greg,<br /><br />Big fan of your blog, your AP1 prep book, and all of your hard work more generally. Apologies for the slightly off-topic comment, but curious if you could help with something. I'm in the process of generating a summer assignment for my 2017-2018 AP Physics 1 students and want to include a reading component, giving them the opportunity to read and learn more about the AP Physics 1 curriculum. I really love your prep book's chapter 1 and 2, but for obvious reasons I cannot simply "post" those chapters to my personal class website. Do you have a favorite resource you draw from to give students a basic, but accurate and informative, idea of what the AP Physics 1 curriculum is all about? Thanks! Stevenoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-27677995319019910632017-06-09T08:07:07.588-04:002017-06-09T08:07:07.588-04:00Hey, Will... the rubrics will be out in September....Hey, Will... the rubrics will be out in September. I wasn't on that question. I can't imagine that the actual free body diagram would be required, as the question did not ask for it. However, I would expect that at least clear mathematical or verbal statements of equilibrium horizontally and vertically (or parallel and perpendicular to an incline) would be necessary. The first principle on that problem is simply Newton's second law with a=0; start there and communicate, and you should be fine.Greg Jacobshttp://www.blogger.com/profile/03854009948036330746noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-32666217326202037342017-06-08T21:41:52.123-04:002017-06-08T21:41:52.123-04:00Clint Alexander is an amazing man who went out of ...Clint Alexander is an amazing man who went out of his way to take my troubled son under his wing during his time at WFS. His leadership and genuine care for the WFS boys and community was unlike anything I've ever seen. His departure is a sad, sad loss for WFS as a community, the current students, and the future students, who won't even realize how much they are missing out on.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-81778274454126497122017-06-04T10:40:51.761-04:002017-06-04T10:40:51.761-04:00One question: on Question 2, part b), of this year...One question: on Question 2, part b), of this year's AP Physics 1 exam, the students have to derive an equation for the coefficient of static friction. Now, there are a few ways to measure that, but one could use, say, a spring scale and measure some forces directly. I've always tried to hammer in that my students must draw a free body diagram whenever writing equations involving forces, but that wasn't specifically asked for in the problem here. Will it be required for full credit?Willnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-39717826825311143262017-05-20T21:51:27.367-04:002017-05-20T21:51:27.367-04:00Two thoughts.
It is just as short as question n...Two thoughts. <br /><br />It is just as short as question number 1. The suggested time implies to me that the students should be as careful with details in the sketch as they are with the paragraph. Good writers get more time on the drawing, and good graphers get more time on the paragraph. <br /><br />What is the velocity AT t = 1? (I've never bothered to get access to the secure answer sets to see if Greg cares about this detail, but why not worry about that instead of worrying about the inertia of the disk?) CCPhysicisthttp://doctorpion.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-92231420537423786342017-05-20T21:29:06.864-04:002017-05-20T21:29:06.864-04:00Regarding the last observation that you didn't...Regarding the last observation that you didn't notice the statement about the rod mass until after you had worked on the problem for awhile, you just explained why it is important to read the entire problem before answering it. Tell your students about your experience so they can learn from it! The sentence before the first question is as important as the question itself, as is the instruction "without manipulating equations" that follows the possible answers. <br /><br />When you ask at the start "Is there something that I'm missing with this question?", I would say you missed that they never said the disk was uniform! Your formula should have I_disk in it. You can't assume more, although you might be allowed to assume that the disk's radius of gyration is << d based on the picture. <br /><br />Finally, I think it is fair to assume (and perhaps state the assumption) that the statement about the masses is tantamount to a statement about their respective moments of inertia about the axis. Changing the radius of gyration of the rod merely shifts the threshold for what "much more" means quantitatively. If they had merely said "more", your concerns would be valid BUT you couldn't answer the problem without resorting to equations. And if you have to use equations, you are reading it wrong. CCPhysicisthttp://doctorpion.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-65254565434471627442017-05-20T20:37:03.473-04:002017-05-20T20:37:03.473-04:00I'm a bit late commenting, but I was reading t...I'm a bit late commenting, but I was reading this all from the beginning. I'll first say that I came to appreciate "easy" questions after being told that some were used on the graduate comprehensive exam we took because they tested clarity of mind over mindless use of the coolest technique available. They are also easy to grade. Fewer arguments over what partial credit to assign in the rubric. <br /><br />What I liked about 3 (c) was that the units DO work out, so a student might waste time on checking that rather than looking at how it fits with the qualitative reasoning in 3 (a). Why worry about whether the exponent on d should be 4, which requires using equations that you were specifically told not to use, rather than question whether particular exponents should be positive or negative? Direct and reciprocal dependences are the first thing that someone should look at when considering if an expression is physical or not. CCPhysicisthttp://doctorpion.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-90224005468032518252017-05-18T15:38:11.640-04:002017-05-18T15:38:11.640-04:00the sum is a little tricky and I had forgotten eve...the sum is a little tricky and I had forgotten everything so it was a mind boostAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-84381215785824675122017-05-16T09:16:33.748-04:002017-05-16T09:16:33.748-04:00Jason, you pretty much nailed it. "The first...Jason, you pretty much nailed it. "The first resistor's voltage increases because I calculated 4 V the first time and 8 V the second time" is at least in the right direction. But "The first resistor's voltage increases because it always takes the same fraction of the battery's voltage, and the battery's voltage increased" is excellent physics reasoning. Then, when the student can show in variables the calculation showing that the first resistor the same fraction of the battery's voltage, his training is complete. Greg Jacobshttp://www.blogger.com/profile/12121422726610824760noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-66282284660106484262017-05-16T07:36:12.653-04:002017-05-16T07:36:12.653-04:00Greg,
I noticed the phrasing "briefly ex...Greg,<br /> I noticed the phrasing "briefly explain reasoning without manipulating the equation" showed up a few times on the AP 1 FR. I have interpreted this phrasing to mean "do not fabricate values and solve for an unknown" (which is many of my student's favorite end around moves for conceptual/algebraic problems). Was this added to keep solutions more in line with analyzing and applying the concepts instead of a plug and chug approach?<br /><br />Thank you for your timeJason W.http://www.blogger.com/profile/14085318774231202211noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-5083279153173035242017-05-14T17:14:19.161-04:002017-05-14T17:14:19.161-04:00Sorry, I'm not teaching AP2 this year, so I...Sorry, I'm not teaching AP2 this year, so I'm not writing out my solutions. They'll be published on the College Board site in September.<br /><br />But, any teacher is welcome to write up solutions and post them to PGP-secure. If you need help with that process, please email me.Greg Jacobshttp://www.blogger.com/profile/03854009948036330746noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-40090570008911965802017-05-14T04:32:14.325-04:002017-05-14T04:32:14.325-04:00Do you plan to upload the ap physics 2 2017 free r...Do you plan to upload the ap physics 2 2017 free response solutions too?<br />It would really help if you do coz I'm gving the exam in late testing<br />Thank youUnknownhttp://www.blogger.com/profile/02144929599387269215noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-52831529582272253172017-05-10T20:32:10.744-04:002017-05-10T20:32:10.744-04:00To me, I would think that the possibility of an un...To me, I would think that the possibility of an uneven mass distribution is exactly the type of logic that we would expect the students who get 5's to employ, but not the students who get 4's. (Note that I'm not saying that I'd expect them to assume an uneven mass distribution--just that they would consider this possibility.) I'm thinking specifically back to that ridiculous (IMHO) problem last year where students were supposed to somehow realize that they were dealing with a superelastic collision. I had one student last year who assumed a superelastic collision on that problem, and he was one of my 5's; everyone else I spoke with assumed that there was some breakdown in the ball becoming less elastic with increasing impact velocity, i.e. the change in slope going the other way. Similarly, while I haven't asked him, I bet that my aforementioned student from last year would not assume an even mass distribution here--that's just his thinking style. As for me, it was quite easy and early to assume an uneven mass distribution for a very simple reason: I noticed the part about the rod mass being much greater than the disk mass after I had already spent a bit of time on the problem. :-) Thus, I already had it in my mind that the answer would definitely depend on the parameters of the bar.<br /><br />That's a good note on the equation for omega. As you say, the problem asks for students not to manipulate equations, but if they start to become troubled by the logical possibilities regarding the bar, they might derive an equation on separate paper "just to be sure", and then come to incorrect conclusions.Jasonnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-68968451931023525442017-05-09T17:06:43.765-04:002017-05-09T17:06:43.765-04:00I actually also explored this and pondered it for ...I actually also explored this and pondered it for a bit when my students asked about this problem. In the end, I think it's obvious to us teachers that we should the assume the mass to be evenly distributed--whereas a student likely would not have seen mass unevenly distributed and would not think of this anyways.<br /> <br />What does trouble me is that some students may derive the equation for omega and see the x^2 term in the denominator somehow rationalize this to be an inverse relationship (naively). However, the problem asks the students not manipulate equations, so in a way they are guided toward a more torque-related response.Joshnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-6665766173858095202017-05-09T15:54:07.185-04:002017-05-09T15:54:07.185-04:00I considered that possibility when I was exploring...I considered that possibility when I was exploring the problem at the beginning, and whether or not the maximum is beyond the length of the rod depends on the dimensionless moment of inertia (which I'll just call K): the question is what is K*m_rod/m_disk. While m_rod >> m_disk, it is still possible for K to approach zero. As K-->0, x-->0. It's fun to check this out on desmos.com/calculator using the various sliders--one for I_rod, another for m_disk. The equation for omega is omega = vx/(I_rod/m_disk + x^2) in the limit where we assume the disk radius is small. (The value of v doesn't affect the location of the peak.) (If not, this adds another term that is also variable and can approach zero.) When playing around with these sliders, it's important to note that m_rod >> m_disk does not necessarily imply that I_rod >> I_disk for a non-uniform rod. (However, it *would* imply this if they called point C the center of mass--this would have saved a lot of discussion!)Jasonnoreply@blogger.com