<p>The FRQ problems are up and I'm somewhat giddy to try to compile some solutions. </p>
<h1>3 Mechanics:</h1>
<p>Through this solution, I will be referring to ‘theta’ as ‘O’.
a) -BO = I*d2O/dt2 (differential equation)
b) T = 2pisqrt(I/B)
c) Just graph the points…no explanation required
d) Use point slope formula to find the equation
e) Since you are plotting T^2 vs I. The slope is going to be T^2/I, where T = 2pisqrt(I/B)</p>
<p>Substituting in, you get [4pi^2(I/B)]/I = 4pi^2/B</p>
<p>Since you are solving for B, you take reciprocal of the slope and multiply through my 4pi^2 to obtain B. </p>
<p>f) (not sure of this) When the line intersects the y-axis, at this point, the rotational inertia of the disk, I, is 0, but the period is not 0. This indicates some kind of error, since ideally, we would want there to be a y-intercept of ‘0’ in our LSRL (best-fit line). We can attribute this error to the fact that when there is no rotational inertia of the disk, there is still an ‘initial’ period that is caused by the inevitable swaying of the torsion pendulum from side to side. </p>
<p>Again, this is just my guess at #3. Feel free to add/correct anything. Hopefully someone can post up the other 2? If no one does, I can post up some parts that I know I did correctly.</p>
<p>EDIT: Actually, I’m fairly sure ‘f’ is right since I found this online: <a href=“http://www.space-electronics.com/Literature/Mass_Properties_Measurement_Handbook_part3.pdf[/url]”>http://www.space-electronics.com/Literature/Mass_Properties_Measurement_Handbook_part3.pdf</a></p>
<p>Read the first few sentences.</p>
<p>For 3f I put that the y intercept was the rotational inertia of the rod, itself, that the disks were being hung from.</p>
<p>I put that it’s impossible to have a rotational inertia of 0 for the object on part f… lol, I fail</p>
<p>Wow, I really messed up mechanics. I think I did a lot better on E&M…anybody going to post those solutions for us? :)</p>
<p>My answers that I can remember:</p>
<p>#1 Mechanics</p>
<p>a.) t = Jp/Favg</p>
<p>b.) m = Jp/vx</p>
<p>c.) Work = Initial Kinetic Energy = 1/2mv^2
Work = 1/2 (Jp/vx) (vx)^2
Work = (Jp)(vx)/2</p>
<p>d.) Work = Fb d
Fb = (Jp)(vx)/(2d)</p>
<p>e.) Conservation of energy
1/2 m v^2 = Work of force in block + Work of friction on table
(Jp)(vx)/2 = (Fb)(dn) + (fT)(D)
(Fb)(dn) = (Jp)(vx)/2 - (fT)(D)
dn = [(Jp)(vx)/2 - (fT)(D)]/Fb</p>
<p>f.) No idea</p>
<p>#2 Mechanics</p>
<p>a.) Normal force going towards the center of the quarter circle, Force of gravity vertically down.</p>
<p>b.) Fc = MgcosΘ</p>
<p>c.) Mg(7R/4) = 1/2 M v^2
v = √(7gR/2)</p>
<p>d.) μ = .583</p>
<p>e.) i.) M dv/dt = -kv
ii.) **v(t) = vde^(-kt/M) **
iii.) Graph begins at (0, kv/M), concave up, decreases asymptotically to zero.</p>
<p>#3 Mechanics</p>
<p>a.) -βΘ = I d^2Θ/dt^2</p>
<p>b.) T = 2π√(I/β)</p>
<p>c.) Graph I vs. T^2</p>
<p>d.) Find equation of your line using points on the line, NOT data points given</p>
<p>e.) slope = T^2/I and β = 4Iπ^2/T^2 = 4π^2/slope</p>
<p>f.) This represents the square of the period of the oscillation of the rod, alone.</p>
<p>Tell me what you think!</p>
<p>^That’s what I got for #2, except I messed up my graph since I started graphing at the wrong point. And, if those are right, I don’t think I got many points on #1…probably 5.</p>
<p>I have no idea if they’re right; what were your answers for #1?</p>
<p>Copied from the other Physics C thread are my answers:</p>
<p>My procedure/thought process for #1 Mech:</p>
<p>For (a) and (b), v is v_x</p>
<p>(a) J=Ft, so t=J/F
(b) J=mv, so m=J/v
(c) Work done in stopping the projectile is the change in kinetic energy. KE=.5mv^2, so it’s .5(J/v)v^2, which is .5Jv
(d) Work is equal to average force times distance. So .5Jv=Fd. So F_b=.5Jv/d
(e)/(f) I know that the impulse imparted by the projectile on the block is equal to something, which I wrote on the test for both parts. Hopefully I get points.</p>
<p>Same for #2 Mech:</p>
<p>(a) 2 forces, normal and gravity. Gravity straight down, Normal perpendicular to surface
(b) Screwed this one up, said Mgcos(theta). Maybe I’ll get something for writing normal=centripetal?
(c) Conservation of energy from point A (all gravity potential) to point D (all kinetic). h for mgh is 7R/4.
(d) I only remember getting 7/12 for my coefficient of friction. Which might not even be what I put. 
(e) i. -kv=m(dv/dt)
ii. Integrate
iii. Initial acceleration (i.e. at point D) is kv/m (even though I only put v). Concave up, decreasing, horz. asy. of acceleration equals 0.</p>
<p>For Mech. #3 (a=alpha, B=beta, O=theta):</p>
<p>(a) Ia=-BO. Replace a with double derivative of theta with respect to time and divide I over.
(b) Using small angle approximation, alpha=-(angular velocity)^2 times theta. Angular velocity is sqrt(B/I). Period=2pi/angular velocity. Could also use mass on spring analogy, with m=I and k=B.
(c) Graph.
(d) My slope was like 160, y-int .20. y=160x+.2 or T^2=160I+.2
(e) Checked my calculator, got .247. Units are same as that of spring constant.
(f) No clue.</p>
<p>Will do E&M later.</p>
<p>I’m not sure your e is correct…that’s the one that seems the most “wrong” in my mind, as in I had something different. I do remember getting a,b,c that was the same as yours, and d looks familiar as well. #1 is all a blur to me since that’s the one I had the least time to do. But, #2 and #3 are fresh in my mind. Anyway, I think I got around 25 points on the FRQ…which means I need around 19 questions correct on the MC for a 5.</p>
<p>
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<p>The answer was Fc = mgcos(theta) since the normal force was mgcos(theta) and that provided the centripetal force.</p>
<p>It can’t be Mgcos theta because as the angle increases that indicated that the normal force decreases which isn’t true</p>
<p>
</p>
<p>Of #1, #2, or #3?</p>
<p>
</p>
<p>Someone else on CC (I forget who) said that normal isn’t providing all the centripetal or something and it made sense. I’m hoping I’m right, though.</p>
<p>^I was referring to NYEM’s post, and it was for #1.</p>
<p>And no, I’m pretty darn sure the only force providing the centripetal force is the normal force. If the curved portion of the track had friction, then one component of the frictional force would be part of centripetal force, but since the problem said to assume that the curved part of the track was frictionless, the answer was mgcos(theta) = Fc.</p>
<p>^I sure hope so!</p>
<p>Well, since nobody else is stepping up, I guess I’ll kick off the E&M attempts.</p>
<p>(I’m using ~ as dot product…because I can’t figure out how to type it otherwise…)</p>
<p>1.</p>
<p>a. ∫E~dA=Qenc/epsilon, so pick any/every spherical surface within the shell and integrate, finding that Qenc is always zero, so E is also always 0 (0 * 2piR = 0).</p>
<p>b. Yes, the field is still zero everywhere in the sphere. Basically just repeat ∫E~dA=Qenc/epsilon for the exact same Gaussian surface(s) and you find that Qenc is still zero, so E is still zero.</p>
<p>c. Flux = 0 for ABCD, ABGH, ADEH. Flux = ∫E~dA, so when the angle between E and dA is 90 degrees, the dot product is 0 (cos90=0). This works because the sphere can be treated as a point charge located directly at point A.</p>
<p>d/e. The field has the least magnitude at corner A, because again, ∫E~dA=Qenc/epsilon (Gauss’ Law), so pick an infinitely small spherical surface that only surrounds the point at corner A. All of the sphere’s charge will be outside of this surface because charge moves to the outside of a conducting object. Hence, Qenc = 0 so E = 0 as well.</p>
<p>f. Ummm I don’t know/think I got this right, but I set ∫E~dA=Q/8epsilon, because only one eighth of the charge was within the cube. Then I said that each of the 3 sides with flux would receive the same flux, so ∫E~dA for side CDEF would just be 1/3 of the total flux from that portion of charge, so I divided Q/8epsilon by 3 to get my answer of Q/24epsilon. Again, I doubt what I did here was right, but maybe someone else can tell me how it’s supposed to be done.</p>
<p>Phew, anyone else care to post #2 or #3? Please? :D</p>
<p>I took #1d for E&M as which corner of the cube, other than A, has the least magnitude, so I chose F. But I think you could be right since the field is zero at corner A. WongTongTong, do you remember how you approached 1e? I’m going to ask my physics teacher from last year for his opinions on everything tomorrow (hopefully, if I get my green insert back).</p>
<p>
</p>
<p>Wait, I think I can prove it.</p>
<p>So, since force=m*a, the force is only going to change when the vector, acceleration changes.</p>
<p>The magnitude of the acceleration comprises the tangential component of acceleration and the normal component of acceleration, so a = aT + aN</p>
<p>and aT = v’ (where v is the speed in this case) and aN = kv^2 (where k is the curvature and v is the speed again), and since the speed is constant throughout uniform circular motion, neither the tangential component of acceleration nor the normal component of acceleration change. Also, since the curved portion of the track is a circle with radius R, the curvature is just 1/R, which is constant. So since curvature and speed are constant from B to D, then aT and aN constant and so the magnitude of the acceleration is also constant. </p>
<p>
</p>
<p>So, no. The normal force does not decrease. </p>
<p>I think. Actually, idk w tf I’m even doing. I think I’m over complicating this, but the answer is def mgcos(theta) lol.</p>
<p>Any NYEM, no, I do not. But, I will say I had a lot of trouble with parts e and f of that problem, so I think whatever you wrote is more reasonable than anything I would have written.</p>
<p>Yeah I felt weird putting A and E=0 because it said to give answers in L, Q, etc. What did you put for 1f on E&M, NYEM?</p>
<p>(and btw I would post my #2 answers, but seeing as we never learned/mentioned/heard of LC circuits, my response to that question was something like ‘lol…lol?..lol :(’ )</p>
<p>I had what you put for 1f, APcollege - I had no idea how to do it otherwise.</p>
<p>For the E&M, question 1
part b is not zero everywhere. Remember that although there is no electric flux through the Gaussian sphere you construct within the shell, you cannot bring the electric field out of the closed integral because it is not uniform throughout the surface (due to non-uniform charge distribution). Gauss’ Law can only calculate the electric field for shells that have uniform charge. This is an awkward question in any case, given that nature always forces charges to uniformly distribute themselves.</p>
<p>In part d the answer is simply F because it is farthest away from the point charge. For point charges you don’t need Gauss’ law, using the idea that the electric field is an inverse square field is sufficient to conclude that due to it being the farthest, it has the minimum field. The field can be calculated using the formula E = kq/r^2. In this case r appears to be equal to the square root of (l^2+l^2+l^2) because its a cube.</p>