@punctiliouseye , what exactly did you do to get an answer without applying Newton’s Second Law for Rotation? I’m not 100% sure, but I think the problem with just using 1 and then multiplying or dividing by the radius is that it’s slipping as it rolls. All of those relationships where you multiply or divide by the radius only apply to problems with rolling without slipping. If you have rolling with slipping or sliding as in this case, you have to use both.
@punctiliouseye , your other question on the 2014 practice with the sine of theta is to do with the fact that torque is the cross product of force and distance at which the force is applied. The sine theta gives just the component of gravity perpendicular to the radius which is the component causing the torque. However, we can’t solve a second order differential equation with sine in it (without using computer software), so that’s why we approximate sin (theta) to just theta (which, if you’ve had Calc. BC, makes sense as the first term of the taylor series for sine centered about x=0 is just x. Hope that helps.
@Kyuutoryuu -thanks for letting me know. I find it odd that they did once allow for regression on the calculator: https://secure-media.collegeboard.org/secure/ap/pdf/physics-c-mechanics/ap-2008-physics-c-mechanics-scoring-guidelines.pdf?gda=1431209888_9d3309def642662956d4649d5bb49eea (if you scroll down to the last one on the 2008 scoring guidelines). However, I haven’t seen them allow for the regression option since. I think it’s still something I’ll try to do just to check the accuracy of my slope (as even with the intercept the calculator gives, its slope should be reasonably close to the real slope).
Does anybody have or want to make a calculator program with all of the possible formulas on it? Thanks and much appreciated.
Q)11 15 29 of http://mralsterscience.■■■■■■■■■■/uploads/3/8/0/0/3800138/ap_physicsc_m_practice_exam_2014.pdf
please. I got a 31/35 in the mcq section and 26/45 in the frq. Is this a safe 5? (if I get 30 mcq what is the minimum frq score for a 5?)
Also how do I prepare for the lab questions? (I don’t know most of the apparatus and have not done most of the lab activities)
@punctiliouseye mgsintheta is the component of the weight of the body which contributes to the torque (resolve mg)
Also for the frq it specifically asks for the method involving newtons second law (net force/net torque) I don’t think you can do it differently (you will need to know the net restoring force/torque)
@bantzking - here is an AP score calculator based upon a real curve: http://appass.com/calculators/physicscmechanics . If you plug in your scores, it’s a 66/90, and you only need a 50/90 for a 5, so that should definitely be a 5. in fact if you get 30 MC right, at least for that year, you could have gotten all 4’s on the FR’s and still have gotten a 5.
So, for 11) I got C… Since the point is in the midpoint of the two spheres, it is located a distance of (3R) for each planet. Therefore, the masses of the planets have to be the same so Fg1 - Fg2 = 0. The gravitational force from one planet will be (GM)/(3R)^2.
For 15) I got B… so I logically deducted it couldn’t be A, C, E. But what I did is used the conservation of momentum and plugged in the values to see if they would equal mv. So, B gave me the correct answer.
Honestly, I have no idea for 29). Does anyone know how to do it? I keep getting 12N, which isn’t even an answer choice :((
Hi,
@sscharter and @bantzking , I can help with 29. The fact that the block doesn’t slip on the cart means that both can be treated as one object. Thus, you have 15N=3a by Newton’s 2nd Law; consequently, the acceleration of the cart and block system is 5 m/s^2. Now, writing Newton’s 2nd Law for just the block, you have Ff=1a (as friction is the only force that acts on it). It’s accelerating at the same rate as the block (both move together), so Ff=15=5N in magnitude, which is answer choice B. The only reason we care about the coefficient of static friction is that a 5N frictional force is possible as it doesn’t exceed the maximum of 100.7=7N.
Oh that makes sense… I set up my equation as F -Ff = 5N and got Ff=7N which kept giving me 12N. 
Is everyone ready for tomorrow? I have to take e and m too, so I’m nervous.
Thanks !
@stoopidfoose. Oh my bad… a 23/35 is still very good. You’d probably need to average around 50% for all the FRQs just to be safe. They give out points for some trivial things on the FRQs too since they know it’s a hard test. You should be fine.
can anyone explain 20-22 on mech 2014
http://talk.collegeconfidential.com/profile/100324497/Chalmydia Here are the answers to 20-22:
20 - the two equations using Newton’s second law are as follows: T2 - uMg ( the frictional force because of mass M) = 0 ( because of constant speed) and uMg - T1 = 0, if you solve these you will see that T1=T2. - choice E
21 - since the friction is only between the top block and the bottom block, the resulting frictional force will be u*Fn of the top block. Since the friction is only dependent on the u and the weight of the top block, the frictional forces must be equal. - choice E
22 - so, if we name a force along the beam as L, the vertical component of L is 1/2 mgsin theta. The 1/2 comes because the mass is concentrated in the middle of the mass. Then T = L cos theta. If you solve for L in the first equation and plug it into this equation, you get choice A.
Can someone explain FRQ d) iii for me. Why would the KE increase if you do PE = KE?
@sscharter It is because the mass of the object is increased while velocity remains same (mv^2/2)
can someone please explain #7
@rockinman1 Just add the new rotational inertia to the one given. The formula for rotational inertia is sum(MR^2) and it’s just adding a piece of clay with mass m. And the clay is put onto the turntable at radius R/2, so add m(R/2)^2 to mR^2/2 (given), so that’s mR^2/2 + 1/4(mR^2) = 3/4mR^2
@MrWiggles thanks for the answer. 2 questions:1)why do you add the interias 2)could you possibly explain 8, 18, and 19 thankssss!
Sure, for 7), you use the parallel theorem which is that I= Iold +md^2. So this would be MR^2/2 + M (r/2)^2, so you get 3/4MR^2. The R/2 comes from the distance from the old pivot point to the new one.
bantzking - thanks!
Nevermind, I figured out how to do #18!
@rockinman1 For #8 just remember that angular momentum is always the same because it’s conserved. L = Iw is the formula, so if you’re increasing the Moment of inertia (I), the angular velocity (w) must decrease. Therefore the answer is D.
For 18, I think i got it. The total energy for an object in simple harmonic motion is TE = 1/2kA^2 (A = amplitude) while the potential energy is U = 1/2kx^2. So U = 1/2(TE) = 1/2(1/2kA^2), so it’s basically asking, “What value of A gives me an answer of 1/4kA^2?” I am looking for 1/4kA^2 because this is equal to half of the total energy (which is split between potential and kinetic). If you plug in A/sqrt(2), it gives you 1/4kA^2.