<p>Do you guys want to compile a list of multiple choice answers, so we get an approximation of how we did on the test as a whole?</p>
<p>yes! 
please</p>
<p>polar bear: m/(m+M) d or something
uhhhhhh LOL do not remember any other problems</p>
<p>@tmiddlet:</p>
<p>No I see what you’re saying, I completely get how they are independent of each other. That was the point I was trying to make though: that the net Force in F=ma and the net Torque in T=I*(alpha) is not uMg and uMgR (where u is the coeficient of friction, mew). </p>
<p>THINK ABOUT IT!!! If we said F=uMg=ma then our equation would be saying that the force was acting purely against the ring’s linear velocity. The M’s cancel, so we could also then say ug=a. So the frictional force, which “dislikes” the rubbing, would “want” the ring to decelerate at a rate of a=ug. To my understanding, only bottom part of the ring that is touching the ground decelerates at a rate of a=ug: So in other words the friction doesn’t know that the ball is rolling, it thinks it actually slowed down at a rate of a=ug but in reality it decelerated at a much lesser rate and just started turning since that was “easier” than slowing down. </p>
<p>So in reality, yes the linear velocity of the portion of the ring in contact with the ground with respect to the ground slowed down at a rate of a=ug, but the actual linear velocity of the ring’s center with respect to the ground did not slow down at a rate of a=ug. </p>
<p>^This makes me think that solving for anything using F=Ma=uMg (or the torque equivalent) would yield an incorrect answer in this problem (Mech FRQ.3), since that equation over-represents the net force ‘F’ that is SPECIFICALLY and WHOLESOMELY working to decelerate the velocity of the center of the ring with respect to the ground.
[ uMg > F = Ma ];[ uMgR > T = I(alpha) ]</p>
<p>Does anyone else see what I’m saying? Or am I just wrong… :(</p>
<p>I do see what you’re saying, and agree with you completely on an intuitive basis, but newton’s second law f=ma is referring to the NET force ANYwhere on the object and the acceleration is ALWAYS the acceleration of the center of mass.</p>
<p>It seems completely wrong, but the whole right will decelerate at a rate of ug while the angular velocity is increasing up until the point that static friction takes over and the ring rolls without slipping. I had a hard time with this concept too because it doesn’t make sense, but that’s the way it is. Gyroscopes don’t make any intuitive sense either.</p>
<p>"…newton’s second law f=ma is referring to the NET force ANYwhere on the object and the acceleration is ALWAYS the acceleration of the center of mass."
^I have to disagree with that. If you were rolling a log and were pushing on the middle of it, you’d expect it to go straight. If you were pushing with the same force but from the left side of it, it would go straight but also start to turn. But comparing the component of its velocity going straigt, wouldn’t you expect it to be faster when you pushed from the center? Rather than when it was also turning? What I’m suggesting with this, is that the fact that the ring starts turning tells me the value of the net force in F=ma is LESS than uMg, since some of the uMG is turning the ring instead of slowing it down. Does this make ane more sense? To anyone?</p>
<p>BTW I love this discussion, I wish school was like this lol.</p>
<p>Did you look at the video link I posted? If you trust the world’s best professor from MIT then that should be convincing. </p>
<p>I can’t think of any intuitive way to explain this </p>
<p>Now that the solution guidelines are up here are my answers for electricity and magnetism FR 3:</p>
<p>A. Flux= BhL (ignored the sub-0)
B. clockwise, decreasing magnetic flux
C. I=E/R = (BLv)/R
D. ma= Fg - Fb
ma= mg - BIL = mg - (B^2 L^2 v)/R
m(dv/dt)= mg - (B^2 L^2 v)/R
E. a= 0 so mg = (B^2 L^2 v)/R
v= (mgR)/(B^2 L^2)</p>
<p>F. Increases, as shown in the above equation, v and R are directly proportional</p>
<p>the FRQ’s are on college-board if anyone wants to look</p>
<p>@tmiddlet
I looked at some parts of it, but not all 47 minutes lol. Are there any parts he talks about dealing with rotation AND translation at the same time? I get how “F=ma” can be rewritten as T=I(alpha). BUT this does not apply since the translation and rotation are not synchronized!!! </p>
<p>NORMALLY: “a=(alpha)R” and “v=wR”</p>
<p>BUT IN THIS CASE tangental velocity “wR” IS NOT the ring’s linear velocity “v”. If it were, the ring would not be sliding. </p>
<p>‘linear velocity’ = ‘rotational tangent velocity’ + ‘scraping velocity’</p>
<p>^that last line alone should explain why F=ma AND T=I(alpha) would be wrong in this problem if we said “uMg” was force and “uMgR” was torque.</p>
<p>Does this get us anywhere? lol Can you try to explain why it WOULD be equal, specifically with respect to this problem?</p>
<p>^does that mean my putting t=Ialpha and v=wr and thats it not even get me 2 points?</p>
<p>The polar bear question…</p>
<p>I went with (m/M)d, because</p>
<p>F(tension 1)=F(tension 2)</p>
<p>m<em>a(1)=M</em>a(2)</p>
<p>since d=(a/2)(t^2), then a=2d/(t^2)</p>
<p>so: m<em>2d/(t^2)=M</em>2D/(t^2)</p>
<p>they have the same time, so the t’s cancel</p>
<p>and depending on which distance (d or D, i don’t remember) we solve for, it ends up being (m/M)d or (M/m)D</p>
<p>anyone have solutions to frqs? also, there were two forms and the problems collegeboard posted for e&m aren’t the ones i got :(</p>
<p>And for number 3, this is what I did for part D.</p>
<p>K(slide0)=K(roll)+K(slide)+Work(friction)
having V=v(0)
(1/2)mV^2=(1/2)mv^2+(1/2)Iw^2-F<em>L
(1/2)mV^2=(1/2)mv^2+(1/2)mr^2</em>(v/r)^2-umgL
(1/2)mV^2=mv^2-umgL</p>
<p>V=sqrt[2(v^2-ugL)]</p>
<p>…only problem is it didn’t say to express answers in terms of L…despite it being a given and necessary quantity (because velocity has to depend on length if there is work being done by friction. Consider the scenario where L=0…)</p>
<p>@idonotknow
</p>
<p>This would work, except for ONE major thing. You can only measure the carts velocity by measuring the time and displacement when the track is level. While the cart is going downhill, velocity is increasing. Measuring time would only give an average velocity, and it would not allow us to find the KE at the end.</p>
<p>Essentially, you should do exactly your experiment, plus have a level portion of the track a length L. Start the stopwatch when the cart reaches the bottom of the slant and end when it reaches the end of the track. This will be the velocity of the cart at the end of the slant, when PE=0 (we disregard friction and air resistance)</p>
<p>@pastorek
I talked to my teacher today and she said you would say that the linear velocity Vf would be equal to WR at the point where it is purely rolling–use this with the standard equations of motion to solve for d and t</p>
<p>@pastorek: yup that’s what I did. I start the stopwatch when the cart reaches the bottom of the ramp. I also measured the height of the inclined plane, so basically I have PE(top) + KE(top) = PE(bottom) + KE(bottom), with KE(top) and PE(bottom) = 0. But well I can’t remember if I was that detailed in my response - we have 15 mins after all…</p>
<p>I love reading threads for subjects the year before I study them. It’s cool to think that I have literally no idea what you guys are saying, but in a few months I’ll completely understand it</p>
<p>@CherubEngr</p>
<p>F=ma is intependent of T=Ialpha. They do not have to be synchronized to apply individually.</p>
<p>For the video, there is a somewhat similar problem that he describes from 14:45 to 20:00. Maybe that will help things make sense</p>
<p>@pastorek</p>
<p>This is actually a very complicated problem to analyze in terms of energy. The work done on an object’s linear energy is force times distance, and the work done on an object’s rotational energy is torque times angle, so there is more to the work done than just umgL. You are better off using the differential equation solutions found in part B.</p>
<p>hey guys i dunno what happened but my mind just blanked out and couldn’t see the checkbox for the cart on the second FRQ…so i just chose “a set of objects” as my object to roll down the track…i explained everything but it sounds so stupid do u think they’ll give me a zero…?</p>