Does anyone know when the calculator policy changed? All the videos I’ve been watching say that a calculator is NOT allowed, but I looked it up and it is. Is that new this year?
@Woandering I believe the force vector asked is the force from the support/axle on the bar, not the string on the bar.
@itsmyusername It changed just this year, but as I’ve said a few times in this thread, the change should not be an issue at all.
Hi guys, does someone have the answers to the 2014 Exam that was posted here earlier?
@sleepdeprived4
Iw for disk vs Iw for ring: (I/2)(2w) for disk equals Iw of ring (angular momentum is same) and then for rotational kinetic energy… 1/2(l/2)(2w)^2 = lw^2 which is greater than that of the ring 1/2Iw^2
I plugged in 1/2 and 2 just as an example but any number when multiplied equals one would work
1/2 times 2 equals 1 btw.
if kinetic energy is greater then the potential energy would be greater: thus a greater height.
A question for everyone else:
on the scoring rubric for physics it says 1 point added for correct units on all answers. If I put a random number with the correct units for everything with no work would i get this point?
also, any tips from people who have previously taken the exam? I’m kinda freaking out now as I was supposed to self study but I haven’t all year. I know basic concepts but I’m scared shitless for this… any words of advice for someone like me who is trying to cram? I’m taking both Mechanics and Electricity and Magnetism tomorrow 
@punctiliouseye lol that’s a good catch! I wonder too…
And also does someone happen to have the 2014 Electricity and Magnetism exam, by chance? Sorry for so many posts, too
@Kyuutoryuu Right, but your explanation mentioned something about making the net force 0. How does that explain the question?
Does anybody know the answer to my question above about the sign of the answer for 2c on the 2014 practice and why the work-energy theorem doesn’t seem to be working. If anybody does, I would really appreciate it as I’d really like to understand what sign the force should be and where I’m going wrong if the answer key is right in its sign.
Good luck for tomorrow!
@purplepiano I saw your post and decided to look at the practice test 1 MC… Did half of them and its definitely harder D: luckily Princeton Review doesn’t make the AP exam
Are we supposed to design an experiment for the laboratory FRQ or are we answering questions about an experiment?
@Sajidur4 Look at 2012 FR 2, so that means there could possibly be one
Are we supposed to memorise all the moment of inertias for objects such a solid spheres, horizontal spheres, solid cylinders, discs and whatnot?
@Woandering You know there are two forces: tension directed up-left, and weight directed down. Adding those two vectors in that order, you will get a right triangle missing its base. Therefore, the only other force vector, from the axle, must point to the right in order to complete that right triangle and make the net force zero. (I’m adding the three forces geometrically).
@punctiliouseye Yes. The units points is literally for the correct units, even if your numerical answer is wrong.
@themusicman123 No. Problems will always give you moments of inertia. However, I have ran into situations where memorizing the moments of inertia for rod/sphere/hoop/disk saves time. Instead of having to integrate for a moment of inertia, for example, I was able to use a memorized MoI combined with Parallel Axis Theorem to solve it.
@themusicman123 , the only ones you really need to know are:
Disc: 1/2MR^2
Bar about Center: ML^2/12 (note: either memorize bar about end to be ML^2/3 or rederive it via the parallel axis theorem)
Solid Sphere: 2/5 MR^2
If it was some kind of other shape they’d give you the moment of inertia (unless it’s a collection of point masses in which case you can add up the MR^2 terms for each point mass).
@Kyuutoryuu Why do you need to make the net force zero? The question is asking for the force exerted on the point.
@themusicman123 memorize?? Why not just put all of them into your calculator
@Woandering Right, and in order to find the force exerted on the point, you must use the knowledge that all forces add to zero (net force=zero). Nothing is moving, after all. Because nothing is moving, net force=zero.
@stoopidfoose Same logic with the formula sheet they give you (which I do not touch at all). Playing Where’s Waldo, looking for your key formula, simply drains time. The more you memorize, the more you can blaze through the test.
I took the 2014 MC and got 29/35… Then I took 2008/2009 and they were much harder, got like 23/35 and 22/35 respectively. For FRQs I’m getting like half the points for each question. I guess I better aim for a 4. Anyone else in a similar situation?