<p>I was thinking gravitation, kinematics, weird momentum/center of mass. I cannot deal with springs…</p>
<p>Okay, I’m having major problems on this MC question without using a calculator. I can get the answer fine with a calculator but seeing as how we’re not allowed to use one on the MC…</p>
<p>Go to page 37, #5:
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>I know work is W=Fdcos(angle) = 5N<em>13m</em>cos (angle). I know the angle between the x-axis and F is 53 degrees but the angle between F and d is 53 PLUS the angle between the d vector and the x-axis. The onyl way I could do it without a calculator was witht he double angle formula cos (53 + x) but that seems way too impractical…</p>
<p>A ball of mass m is attached to the end of a string of length Q as shown above. The ball is released from rest from position P. where the string is hori¬zontal. It swings through position Q. where the string is vertical, and then to position R. where the string is again horizontal. What are the directions of the acceleration vectors of the ball at positions Q and R?
Position Q , Position R
(A) Downward , Downward
(B) Downward , To the right
(C) Upward , Downward
(D) Upward , To the left
(E) To the right , To the left</p>
<p>Picture is a pendulum starting at point P, which is on the left side. Q is on the bottom of the pendulum and R is on the right side.</p>
<p>Answer: C… Why? I know the centripetal acceleration is always pointing toward the center (pivot point in this case) and that gravity exists…</p>
<p>If I’m not mistaken, isn’t the acceleration vector always to the center of the circle the pendulum creates since it’s centripetal motion?</p>
<p>@JDong for the calculator issue: you’re supposed to solve the problem by splitting the force in x, y components, which is conveniently done for you :)</p>
<p>Bump 10char… I need help :(</p>
<p>I think you answered your own question… The tension at Q causes the acceleration to be up, and the gravity causes acceleration to be down at R</p>
<p>u can miss 2 whole free responses and get a 5 haha</p>
<p>@aznjunior: but at R, both tension and gravity are present, so shouldn’t it be D, to the left?</p>
<p>@math
IS tension present? (Tension is centripetal force, centripetal force = mv^2/r, and v equals…)</p>
<p>
You get a formula sheet for the free response section. For the multiple choice, you don’t get formulas, but you do get a pretty useless sheet of constants and random stuff.</p>
<p>^Yes, in correspondence with ChemE14, at position R, there is no velocity. Hence, there is no centripetal acceleration, so the only acceleration is due to gravity. At Q, the centripetal acceleration points upwards, so the direction of acceleration is upwards.</p>
<p>can anyone explain #2 part b on the 2008 frqs?
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>It’s in static equilibrium, so torque<em>net = 0. The pivot point is where the hinge is at, so we are able to essentially ignore the force of the hinge. The reading on the spring scale is going to be the force of tension, so just set up an equation, like “t</em>counter-clockwise = t_clockwise,” and solve for the force of tension.</p>
<p>I just finished doing the FRQ from 2007-2009… and it wasn’t fun (specifically 2009 #2, I still haven’t graded it). As long as there is no more than one extremely hard question, I’ll be fine, but with this current trend, it does not look good for us!</p>
<p>I hate torque because I never can decide on the length of the lever arm!</p>
<p>o wait i mean part c</p>
<p>For I_rod, use parallel-axis theorem:</p>
<p>I<em>rod = I</em>center + MR^2, where R is distance from center to new pivot point. They give us I_center</p>
<p>You also have to add I_block, and the block is just treated as a point-mass.</p>
<p>use the parallel axis theorm to find the moment of inertia of the rod and then add on the moment of inertia of the block (.5*.6^2) to get .42</p>
<p>I tried doing what u said ChemE14 but when i did it, i ended up with</p>
<p>I_rod = (2 x .6^2)/12 + (2.5)(.3^2) = .285</p>
<p>am i using the formula wrong or something?</p>
<p>I<em>total = I</em>rod + I_block</p>
<p>I<em>rod = ML^2 / 12 + M(L/2)^2 <— parallel-axis theorem
I</em>block = M_block*L^2 <— regular “I=MR^2”</p>
<p>Add those together, should get 0.42. For I<em>rod, for your “M” in MR^2 you used “M</em>rod + M<em>block” instead of just “M</em>rod.” I think I might have confused you saying the block was a point-mass. What I meant by that was that you just use your standard I = MR^2 to find the moment of inertia of the block.</p>