2010 AP Physics C Study Thread

Test Preparation AP Test Preparation
41

<p>No fluid mech and thermo are not part of the AP Physics C: Mechanics curriculum. However, if you take physics courses at university, those topics are discussed. Don’t worry they won’t show up on the AP exam. :)</p>

42

<p>Fluids and Thermo show up on the Physics B exam, but not the C:Mech. C:Mech is just newtonian physics with energy and some other things thrown in.</p>

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<p>btw, how does College Board want us to round off the answers for Physics C?</p>

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<p>I assume this is where significant figures come into play. Remember sig figs? There are different rules for add/subtract and multiplication/division.</p>

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<p>my teacher didn’t really tell us anything about sig figs. I remember doing them in chemistry, but now I completely forgot the rules when adding/multiplying.</p>

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<p>do we need to know damped and driven oscillations for the AP test?</p>

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<p>I don’t think so, but someone correct me if I’m wrong. If damped oscillations are on the ap, it would be pretty hard. I’ve been working thru past ap exams and i haven’t seen anything like that.</p>

48

<p>bumping to keep the thread alive</p>

49

<p>can someone give me some hints on this problem (1974 #1b)</p>

<p>

</p>

<p>The answer is the square root of the quantity gl, but that’s not what I’m getting. What I’m trying to do is setting up a centripetal force equation.</p>

<p>Sum of centripetal forces = ma(centripetal) = Tension cos 30
(mv^2)/l = T cos 30</p>

<p>I did T cos 30 because that’s the component (horizontal) that contributes to the centripetal acceleration. I don’t see any other force (or components of force) that contributes to the centripetal acceleration.</p>

<p>After doing all the necessary math, I can’t derive the answer that they had (root of the quantity gl). What am I doing wrong?</p>

50

<p>what you can do is use conservation of energy</p>

<p>U(start)+K(start)=U(end)+K(end)</p>

<p>Since it starts from rest, you can cancel out K(start), and you can define point P as h=0, so you can cancel U(end). So you’re left with</p>

<p>U(start) = K(end)</p>

<p>mgh = (1/2)mv^2</p>

<p>h = l sin 30</p>

<p>mgl sin 30 = (1/2)mv^2
2mg l sin 30 = mv^2</p>

<p>sin 30 = 1/2, cancel out the m’s</p>

<p>v^2 = gl</p>

<p>v= root (gl)</p>

51

<p>oooh i see i couldve done it that way…thank you ^_^</p>

<p>but i still wanna know why my solution is wrong :(</p>

52

<p>The sum of the centripetal forces is T - mg sin 30, not T cos 30. Draw a free-body diagram when the mass is at 30 degrees … the centripetal forces are all those in the direction of the pendulum arm.</p>

<p>So, you would get T - mg sin 30 = m v^2 / l , but this is one equation and two unknowns, so it doesn’t help you very much here.</p>

53

<p>r u sure? because this is what i have (diagram)</p>

<p><a href=“http://img594.imageshack.us/img594/732/pendulum.jpg[/url]”>http://img594.imageshack.us/img594/732/pendulum.jpg&lt;/a&gt;&lt;/p&gt;

<p>so the component that would contribute to the centripetal force is the horizontal component (which is the cos)…or do i have it all wrong??</p>

<p>(sorry for the horrible paint skills btw :p)</p>

<p>to solve for T, i set up the sum of forces in the y-direction ma(y)=T sin 30 - mg
since a(y)=0, T sin 30=mg, T = mg/(sin 30)</p>

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<p>okay guys, how’s it coming? how many of guys think you’re going to pull off a 5?</p>

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<p>hopefully 10char</p>

56

<p>@atotc17: remember that the centripetal force is always the net force directed toward the center of the circle, which in this case is along the arm of the pendulum pointing back to the pendulum pivot point, i.e., not horizontal unless the pendulum is horizontal.</p>

57

<p>hey guys is potential gradient in the curriculum of AP physics C (E&M)</p>

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<p>I’m going to start studying Physics C after aps are over this year, does anyone have any suggestions for a good textbook that isn’t ridiculously abstruse? And preferably comes with a detailed solution manual? </p>

<p>Thanks</p>

59

<p>the palisade charter school thing it is not working anymore does anyone have copies?</p>

60

<p>Here’s a good practice question. Took me pretty long to get it.
Note: Do not use Energy Conservation method.
A sphere starting from rest rolls without slipping down a 22 degree 12 m long inclined ramp. What is its velocity at the bottom? Do not use Energy Conservation methods to solve. </p>

<p>Answer: Has a 9 in the tenths place.</p>