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course PHY 231
9/26 1:23 pm
Class Notes 100920Experiments
Space 6 magnets, three on each side, symmetric ...
see how different you can make the symmetry and still get a balanced oscillation
For the magnets on the rotating strap:
`qx001. How fast was each of the magnets moving, on the average, during the second 180 degree interval? All the magnets had the same angular velocity (deg / second), but what was the average speed of each?
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My angular velocity was 180º/ 1 second which converts to π radian/ 1 second. Then I used the formula v_r= r_m*(π). where r= distance to center of magnet. My magnets were symmetrical so there were two magnets at the same distance from center on each side, therefore I will only refer to three magnets since all the values for one set are also the ones for the other. M1 is the magnet farthest from center, then M2 and M3 respectively.
v_M1= (14)π= 14π cm/s
v_M2 = 8π cm/s
v_M3 = 2π cm/s
you apparently composed this in something that used some special symbols; no big problem for me but you might find the posting difficult to read. It's a good idea to copy and paste your document into Notepad or some other pure-text editor to show you whether you're getting special symbols. If you save your document as a text-only document then open it, the special symbols might well be rendered in standard text-only format.
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`qx002. For each magnet, one of its ends was moving faster than the other. How fast was each end moving, and how fast was the center point moving?
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The above value are the velocities at center of the magnets so I wont repeat that, just the ends. The magnet are 2 cm wide so the radii are 1 cm ± the radii above.
v_M1_right = 15π cm/s
v_M1_left = 13π cm/s
v_M2_right = 9π cm/s
v_M2_left = 7π cm/s
v_M3_right = 3π cm/s
v_M3_left = π cm/s
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`qx003. What was the KE of 1 gram of each magnet at its center, at the end closest to the axis of rotation, and at the end furthest from the axis of rotation?
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KE_M1_center = 967
KE_M1_right = 1110
KE_M1_left = 834
KE_M2_center = 316
KE_M2_right = 400
KE_M2_left = 242
KE_M3_center = 19.7
KE_M3_right= 44.4
KE_M3_left = 4.93
KE = 1/2*m*v^2 and then I just substituted in the respective velocities from above. Units: mass is in grams and velocities are in cm/s so KE units here are g*cm^2/s^2.
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`qx004. Based on your results do you think the KE each magnet is greater or less than the KE of its center?
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I think greater than because I imagine the KE vs distance from center graph is not linear but likely quadratic at least, meaning that between 15 cm and 13 cm the kinetic energy is greater from 14 to 15 than 13 to 14 nonlinearly. Trapezoidaly this would mean the KE stays above the line segment connecting 13 to 15 more than below.
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`qx005. Give your best estimate of the KE of each magnet, assuming its mass is 50 grams.
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KE_M1 = 48525
KE_M2 = 15955
KE_M3 = 1151
I found this as an average of the kinetic energy at the three points on the magnet. It reads higher than the KE at magnet center for 50 grams and its just enough higher to make me think this is a pretty good estimate.
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`qx006. Assuming that the strap has a mass of 50 grams, estimate its average KE during this interval.
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Okay, the way that comes to mind for me to do this is with a Riemann sum. I know that might not be exactly the way you would expect everyone to do it but I don't see a clear way to do it otherwise. So a more direct reasoning route just isn't evident to me here. So this is what I did: KE_i= 1/2(m_i)(v_i^2) and here m_1 = (25/15)*∆r and v_i = π*r_i then the sum is ∑ 2.618(r_1)(∆r), after some simplification then the limit to find the integral that is: 2.618/2 r^2 and from 0 to 15 that is 295 g*cm^2/s^2. That is one side of the strap so I multiplied by 2 to get 590 g*cm^/s^2. Now this does treat the strip as being more like a rod than a flat rectangle since I didn't do the 50 grams per square area, or per cubic volume for that matter, but I think this is still a reasonably good, quick answer.
Outstanding analysis. The special-characters issue obscures some of the numbers and you haven't included units, so I can't tell why your result is off (should be more like 20 000 g cm^2 / s^2), but other than a little arithmetic you really nailed it. You should write this out yourself on the assumption of rod length L, mass m, angular velocity omega and see what the symbolic solution is. Having done that you'll be able to pretty much skip at least half a chapter when we come to this topic in the text.
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`qx007. (univ, gen invited) Do you think the KE of 1 gram at the center of a magnet is equal to, greater or less than 1/2 m v_Ave^2, where v_Ave is the average velocity on the interval?
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Well, put like this it seems that it would be equal to since the velocity seems to increase linearly, i.e. 7.5 cm is travelling 7.5π cm/s and 15 cm is travelling 15π cm/s. However, from looking at the KE values above it seems that it should be greater than the average velocity KE. So I may have something not quite right about one, or both, of those two parts.
v increases linearly with r, but KE increases as the square of v
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For the teetering balance
`qx008. Was the period of oscillation of your balance uniform?
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Yes, I had 9 counts per oscillation for as long as I counted. Note: I have adjusted my counts to where I count about twice as fast as I used to because my count was too slow to measure some of these systems and intervals.
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`qx009. Was the period of the unbalanced vertical strap uniform?
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With this it seems to be related to how hard you spin it. At slower speeds the periods were erratic and it didn't spin well but spin it harder and it behaved more like the balanced strap. So yes and no to uniform period.
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`qx010. What is the evidence that the average magnitude of the rate of change of the angular velocity decreased with each cycle, even when the frequency of the cycles was not changing much?
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This is troubling me some as I can quite picture this in my head but I want to say that its related to the distance travelled, that is less distance is travelled but at a slower rate and then greater distance travelled has a greater rate with it.
you're getting there; we'll leave this question open for awhile
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For the experiment with toy cars and paperclips:
`qx011. Assume uniform acceleration for the trial with the greatest acceleration. Using your data find the final velocity for each (you probably already did this in the process of finding the acceleration for the 09/15 class). Assuming total mass 100 grams, find the change in KE from release to the end of the uniform-acceleration interval.
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∆KE= KE_f- KE_s ----> KE_f = 1/2*(.1)*(0.95)^2= 0.045 J, KE_s = 1/2(.1)(0)^2-----> ∆KE = 0.045 - 0 = 0.045 J
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For the experiment with toy cars and magnets:
`qx012. For the experiment with toy cars and magnets, assume uniform acceleration for the coasting part of each trial, and assume that the total mass of car and magnet is 100 grams. If the car has 40 milliJoules of kinetic energy, then how fast must it be moving? Hint: write down the definition of KE, and note it contains three quantities, two of which are given. It's not difficult to solve for the third.
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v= √(2KE/m) = √(2*40*100)= 0.89 m/s
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`qx013. Based on the energy calculations you did in response to 09/15 question, what do you think should have been the maximum velocity of the car on each of your trials? You should be able to make a good first-order approximation, which assumes that the PE of the magnets converts totally into the KE of the car and magnet.
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0.2379, 0.173, 0.01
For magnet separation of 3 cm, 6 cm, 9 cm respectively in m/s. Using the same formula as above.
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`qx014. How is your result for KE modified if you take account of the work done against friction, up to the point where the magnetic force decreases to the magnitude of the (presumably constant) frictional force? You will likely be asked to measure this, but for the moment assume that the frictional force and magnetic force are equal and opposite when the magnets are 12 cm apart.
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There would be less KE available for motion up until the 12 cm after which the remaining KE would be spent even more to overcome friction and is eventually 'drained' as a result of this.
this is something you can actually calculate with the given information, and with reasonable accuracy (assuming the accuracy of your observations)
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`qx015. If frictional forces assumed in the 9/15 document were in fact underestimated by a factor of 4, then how will this affect your results for the last two questions?
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It would considerably reduce the maximum velocity attainable by the car since even more KE would go to overcome friction since it is so much greater.
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`qx016. What did you get previously for the acceleration of the car, when you measured acceleration in two directions along the tabletop by giving the car a push in each direction and allowing it to coast to rest?
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26.9 cm/s^2 was my average acceleration, with a little expected variance but still all in all close to this in both directions. Actually one way was just a little lower than and the other just above.
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`qx017. Using the acceleration you obtained find the frictional force on the car, assuming mass 100 grams, and assuming also a constant frictional force.
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F= m*a------> F = 0.1 Kg(0.269 m/s^2)= 0.0269 N = 26.9 mN. This is the force required to accelerate the car and it is opposite in direction to the frictional force and since the car stops friction must also generate this force. So frictional force is 26.9 mN in the opposite direction of motion.
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`qx018. Based on this frictional force
How long should your car coast on each trial, given the max velocity just estimated and the position data from your experiment?
... this could be done with an inclined air track ...
... collision: release two cars simultaneously, one carrying two magnets and the other carrying one
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I'm not sure what this is asking exactly. For the questions just above, the cars coasted as long as they could since they all came to a stop and the magnet/air track statement just confuses me more. What does this have to do with the magnet cars? Since I don' know what it means by 'coasting' I don't know how to use the max velocity and position data together.
For example if the car coasts to rest from a speed of 40 cm/s, how long should it take to come to rest?
The lines with ... are just notes to myself that I neglected to edit out. They probably provide fair warning about lab exercises yet to come.
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The following questions are for university physics students, though all but one are accessible to general college physics students, who are invited but by no means required to attempt them. Questions of this nature will be denoted by (univ; gen invited). Questions which actually require calculus are denoted (univ; calculus required). General College Physics students with a calculus background are invited to attempt these questions.
`qx019. (univ; gen invited) Looking at how v0 affects vf, with numbers: On a series of trials, a car begins motion on a 30 cm track with initial velocities 0, 5 cm/s, 10 cm/s, 15 cm/s and 20 cm/s. By analyzing the first trial in the standard way, the acceleration is found to be 8 cm/s^2. Using the equations of uniformly accelerated motion, find the symbolic form of the final velocity in terms of the symbols v0, a and `dx. Then plug the information common to all trials into this equation (i.e., plug in the values of `ds and a) to get an expression whose only unknown quantity is v0. Finally plug your values of v0 for the various trials into your expression, and obtain your values for vf. Sketch a graph of vf vs. v0 and explain as best you can, in terms of your direct experience with these systems, why the graph has the shape it does.
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My graph here is a concave up curve that is increasing at an increasing rate. This is not a velocity vs time graph and so does not have to be linear, which it would be since acceleration is constant. vf increases greater and greater with increased v0 meaning that it is not a linear relationship and this is reinforced by the equation I used here, which is: vf^2 = v^2 + 2*a*∆s. I used this since it has only the four quantities which we are dealing with. The squared shows quadratic relation between vf and v0.
On a graph of vf vs. v0, the change in vf should on a given graph interval should be less than the change in v0, and the slope should decrease with increasing v0. If you solve the equation for vf you get vf = sqrt(v0^2 + 2 a `ds).
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`qx020. (univ; gen invited) Use your calculators to graph vf vs. v0, using the expressions into which you plugged your values of v0, and verify your graph.
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Yeah since I used the same equation in both they are the same.
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`qx021. (univ, calculus required) Should the derivative of vf with respect to v0 be positive or negative? Don't answer in terms of your function, your graph or your results. There is a good common-sense answer based on the behavior of the system and the nature of uniform acceleration.
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It should be positive since as vo grows so does vf and thus the change in vf wrt v0 will be positive. More of v0 leads to more of the vf.
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`qx022. (univ; calculus required) What is the derivative of vf with respect to v0? What does this derivative function tell you about the behavior of the system?
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Here I have: vf' = v0/(√(v0^2+2*a*∆s) which is exactly vf' = vf/v0 since I used the denominator as my equation for vf.
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`qx023. (univ; gen invited) Using the velocity and position functions for uniform acceleration, and the resulting equations of uniform acceleration, can you get an expression for the time required to achieve velocity v_mid_x in terms of v0, a and `dx?
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For this I arrived at: t= [(√(v0^2 + 2*a*∆x)-v0]/a. I was able to eliminate vf, which here is v_mid_x, to just have the three quantities on one side.
you appear to have used the velocity function, but not the position function; you probably need to use both
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`qx024. (univ; gen invited) Using the velocity and position functions for uniform acceleration, and the resulting equations of uniform acceleration, can you get an expression for the velocity v_mid_t in terms of v0, a and `dx?
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This I could not do. I worked every way I could think of and never could get an expression with just v_mid_ t on one side by itself. I got one that did only v_mid_t on the left but there was also one in my right side. The big thing was not be able to eliminate t without having v_mid_t on both sides. I think that here you don't need ∆x but ∆t instead since I could eliminate ∆x and just have v0, a , and ∆t on one side and v_mid_t on the other.
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`qx025. (univ; gen invited) Can you interpret the expressions for v_mid_x and v_mid_t to answer at least some of the open questions associated with the ordering of v0, vf, `dv, v_mid_x, v_mid_t and v_ave? Can you develop expressions that can be interpreted in order to answer the remaining questions?
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Well I know now that v_mid_x will always be greater than v_mid_t since it takes more time to get to v_mid_x than t/2. All of my equations have vf growing larger as v0 grows larger, which is expected and well known. I have suspected all along that v_mid_t = v_Ave and here are my equations that prove it: v_Ave = ∆x/t = (v0*t+1/2*a*t^2)/t = v0+1/2*a*t and, since mid_t = t/2 I have: v_mid_t = v0+ a(t/2)= v0 +1/2*a*t which are clearly the same. Another idea that had early on is that as v0 grows larger ∆v will grow smaller and eventually be at the bottom of the list and here is proof of that: ∆v = vf-v0, vf= √(v0^2+2*a*∆x) then substitution gives: ∆v = (√(v0^2+2*a*∆x))-v0. This shows that for different trials, with the same a and ∆x but increasing v0, the second half of the square root becomes negligible as compared to v0 and so we get something very close to √(v0^2)-v0 which is 0, but ∆v will never be 0 if a and/or ∆x are always > 0. So the only real change in my first assertion about the ordering of these is that v_mid_x and v_Ave/v_mid_t are not equal but v_Ave and v_mid_t still are. So as v0 increases from 0 the only quantity that moves in the ordering is ∆v which moves down the ranks from being next to vf at the top, when v0 = 0, to less than v0 at the bottom when v0 gets very large, and the rest are v0 < v_Ave = v_mid_t
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Your work is excellent. Not without some errors or omissions, but you are working at a very high level.
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