<p>what did you guys get for the mcehanics mc about the period of the pendulum with the hanging mass and the period of the rod?</p>
<p>it could be either
depends on length l</p>
<p>(assuming the rod behaved like a pendulum)
If it behaved like a physical pendulum then it would be 2pi<em>sqrroot</em>(MGD/I)</p>
<p>but if i recall the question stated that it behaved like a simple pendulum so thats why i picked it depends on length l
(the equation for period of simple pendulum for small angles is only gravity or length dependent)</p>
<p>wait but they wre asking for the ratio between the pendulums
so it wouldnt depend on l right?</p>
<p>What do you guys think the curve will be like on the E&M? I got maybe 30/45 on the FRQs and a low estimate of 20/35 on the MC…is this enough for a 5?</p>
<p>Was a and b in FRQ3 asking about their velocity/angular velocity during “pure sliding”? Also, can we use “F” in expression?</p>
<p>No you couldn’t have used F.</p>
<p>Llazar that should be enough for a 5 but it is a raw score of 56 which is pretty close to a 4</p>
<p>For E&M Frq with the charged spheres, did anyone else get…</p>
<p>A. E=(1/4pi(epsilon 0))<em>(Qi/r^2)
B. E=(1/4pi(epsilon 0))</em>((Qi+Qo)/r^2)
C. V=E*r…so you essentially just get 1/r instead of 1/r^2
D. QT=2.22 nC
E. I think this was the graph of E, it starts at 0 until it reaches r=0.1. From there the curve starts at E<a href=“0.1”>part A</a> and is concave up (just the graph of 1/r^2) until r=0.2. From there the curve jumps up to E<a href=“0.2”>part B</a> and is concave up to infinity…
F. Very similar to the graph in E, except the curve is flipped below the x-axis for r=0.1 to r=0.2 as the voltage goes from -100 V to 100 V, and the curves are 1/r instead of 1/r^2</p>
<p>Let me know if you got anything similar to this</p>
<p>D should be one half of that, 1.11 x 10^-9. Qo was positive and qi is negative</p>
<p>You’re right, my silly mistake.</p>
<p>Woah woah woah… Mech FRQ 3:
I don’t see how the net force AND net torque equations can be applied. Or at least not with friction being the NET force of both. I would think the frictional force could be wholesomely applied to slowing down the linear velocity OR wholesomely applied as torque to increase the angular velocity OR partitioned between the two BUT not wholesomely applied to both. The object begins to turn because it would rather start to turn than straight up slow down. But the fact that it turns tells me its redirecting some of that force on the velocity and applying that portion to turning it. So I don’t see how F=ma=(mew)mg or the torque equivelant could be correct for any parts of the problemN including part A and B, as they would both be over acounting for the net force and net torque applied.</p>
<p>How do you distribute the friction between a resisting force and a torquing force? The only thing I can think is maybe do something with energy, so that W=Fx=Fvt, but I’m lost from there.</p>
<p>On second graph of the E&M FRQ #2, was the curve for V(R4) the same as the one for V(R5)? I drew one e^(-x) line and labeled it as both.</p>
<p>A force is a force - it is applied, not distributed as energy is.</p>
<p>Yes, it should be the same graph for both V<em>R4 and V</em>R5 as the voltage across parallel resistors is the same.</p>
<p>@tmiddlet:
I get what you mean, but please look further into this. Conceptually, imagine what it would look like if friction was infinate: the ball would instantanously start rolling. But if I plugged negative infiniti (or just a really large negattive quantity) into the F=ma equation, the math would say the ball’s linear motion would suddenly stop, however we can both see this to not be true as it would just keep on rolling. Yes, forces are forces, but they can be applied in different ways. I suggested possibly looking at it in terms of energy, since that would be a way to relate both translational and rotational kinetic energies to the force seen as work done collectively to both of them. And actually, thinking about it now… Hmmm… The P=dW/dt equation might be a good start at making a differential equation asked for in parts A and B… I’m not giving up on figuring this out lol.</p>
<p>Hey for question #2 of Mech, can I set up an inclined plan (they give us track, don’t they) and release the cart? Then using the stopwatch, measure the time and thus velocity of cart => PE = KE??? Anyone does it that way?</p>
<p>For 2c (or d, I’m not sure), they ask for one possible explanation of why E(after)>E(before). Isn’t that impossible though?</p>
<p>@llazar: Congrats, I think you’re looking at a 5:)</p>
<p>Idonotknow- haha I did exactly that. I had no idea what kind of writeup we were supposed to do though an the extent of detail we were supposed to show because I have NEVER seen an frq like that. I think it was a good idea but I probably lost some points in explaining.</p>
<p>@CherubEngr</p>
<p>Yes, if you applied an large negative force to the hoop, anywhere, it would quickly stop and change direction at the same speed regardless of where the force is applied. However, depending on where the force is applied, rotational motion may begin also. These are two completely independent things.</p>
<p>If you need some concept development on this, I recommend this video [Lec</a> 21 | 8.01 Physics I: Classical Mechanics, Fall 1999 - YouTube](<a href=“- YouTube”>- YouTube) from 14:45-20:00. This rotation stuff is very non-intuitive so I see why you could be confused.</p>
<p>I used FR=I(alpha) to solve parts A and B of the third free response where friction was the torquing force at a radius R. Turned alpha into both an angular velocity and velocity differential.</p>
<p>Which questions on the MC part for mechanics did you guys find hard or had trouble with?</p>