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course phy-231
ExperimentsSpace 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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angular velocity was 24 degrees per second.
Outer magnets had an average speed of 5.4 cm/s
middle magnets had an average of 3.4 cm/s
inner magnets had an average of 2.1 cm/s
the outer magnet was 13 cm away from the center rod, so I took the distance the magnet traveled 13pi, and divided it by the time interval, 7.5 seconds.
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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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outer magnets, outer edge=6.3
outer, center=5.4
outer, inner=4.4
middle, outer=4.2
middle, center=3.4
middle, inner= 2.3
inner, outer=3.1
inner, middle= 2.1
inner, inner= 1.0 cm/s
the distance from center of the rod to the magnet is question number 1 was measured from the center of the magnet, so the data in number 2 for the center is the same as in number 1. the distance used for outer and inner of each magnet is that distance either plus or minus 2.5 cm.
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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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outer magnets, outer edge=19.8 N
1 gram at 5.4 cm/s would have kinetic energy 19.8 gram * (cm/sec)^2 = 19.8 gram cm^2 / sec^2, or 19.8 dynes
outer, center=14.6
outer, inner=9.7
middle, outer=8.9
middle, center=5.8
middle, inner= 2.6
inner, outer=4.8
inner, middle= 2.2
inner, inner=1
I used f=.5mv^2 to obtain my results
good, except for your units
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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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seems real equal to me. I averaged the outer and inner edge and they were pretty much equal to the center value.
v is a linear function of position, so v^2 is not; the two can't be equal
from the shape of 1/2 m v^2 vs. v you should be able to figure out whether the KE of the entire magnet is greater or less than the KE based on center speed.
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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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outer=729
middle= 289
inner= 110
good, but include correct units at some point (perhaps in a note at the end)
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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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14400 (deg*gram)/(sec)
for this result I used the angular velocity as v
since KE = 1/2 m v^2, you need to use v, not the angular velocity omega
the complication is that different parts of the strap have different velocities
the best you can do is make an estimate of whatever velocity v could serve as an average
the velocity halfway between the center of the strap and an end would be a reasonable first candidate; then you would want to modify, considering the nonlinearity of v^2
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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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equal to because the center KE is pretty much equal to the average KE of the outer, middle, and inner parts of the magnet.
nonlinearity (and consistent direction of concavity) of v^2 with respect to position shows that it's not the same
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For the teetering balance
`qx008. Was the period of oscillation of your balance uniform?
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no. it would take longer for one side to go up then it would for it to go back down.
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`qx009. Was the period of the unbalanced vertical strap uniform?
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no. it would sway to the left longer than it would to the right.
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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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It would slow down a lot faster than a balanced strap.
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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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39.6
139
154
180.5
220.5
264.5
364.5
450
joules, I used the formula 50v^2
those units don't come out to be Joules
there is no general formula 50 v^2; of course from 1/2 m v^2 you would get this result, but the formula is KE = 1/2 m v^2, and your specific formula (which should include units) would result from this
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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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.89 cm/s
.4=.5v^2
with appropriate use of units your calculation would yield a velocity in m/s, not cm/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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should be double the numbers for velocity on question “qx011”
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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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the frictional force will deplete the KE and once the car reaches 12 cm past the magnet, the acceleration of the car will become negative and it will slow down and come to a stop.
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`qx015. If frictional forces assume 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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the car will not travel as far and the frictional force and magnetic force will be equal sooner, therefore causing the car to start slowing down sooner.
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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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| 11.7 | cm/s^2 in the right direction
| 15.8 | cm/s^2 in the left direction
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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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N=.981 newtons
Magnetic force+coefficient of friction*.981=1.17
(1.17 is acceleration in the right direction times the mass of 100 grams)
it's not clear what you were using for the magnetic force or how you obtained that
nor is it clear what you used for the coefficient of friction
given your accelerations it appears that the acceleration due to friction is about 14 cm/s^2, so the frictional force is 14 cm/s^2 * 100 g = 1400 dynes
the weight of the car is about 98 000 dynes, so the coefficient of friction corresponding to this force is about .014
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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?
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you know vf, `ds and a so you should be able to find `dt
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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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vf^2=v0^2+.5a’ds
It's 2 a `ds, not .5 a `ds
vf^2=v0^2+480
for v0=0, vf=21.9 cm/s
for v0=5, vf=22.47
for v0=10, vf=24.1
for v0=15, vf=26.6
for v0=20, vf= 29.7
I got a graph that has the equation y=|.02x^2+21|, where y is vf and where x is v0.
the graph of that function would be concave up, but the graph of vf vs. v0 is concave down
vf = sqrt(v0^2 + 2 a `ds), and for constant a and `ds you get a graph which is increasing but concave down, asymptotic to the straight line vf = v0.
It looks like that because if v0 changes, but the other factors stay the same, then the final velocity is going to increase.
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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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vf=|.02v0^2+21|
For the given quantities 2 a `ds = 480 cm^2 / s^2, so the graph would be vf = sqrt(v0^2 + 480 cm^s / s^2), which could be rearranged to give you vf = 22 cm/s * sqrt( .002 v0^2 sec^2 / cm^2 + 1).
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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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positive, the final velocity increases if the starting velocity increases and all of the other factors remain constant. Adding more velocity to the system, will increase the ending velocity.
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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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y’=.04x, where x is v0.
That would not follow from your equation, nor would it follow from the correct equation y = 22 sqrt( x^2 + 1). The derivative of that equation would be 22 x / sqrt(x^2 + 1)
Vf increases with v0.
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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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yes. I substituted delta s, in the equation for vf^2 with (delta s)/2. I then solved for vf^2 and then replaced vf with vmidx. Then I took the constant acceleration equation vf=v0+at and substituted it in for vmidx of my other equation and solved for t. I came up with t=(sqrt(v0^2+2ads)-v0)/a
Good.
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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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yes,. I took the equation vf=v0+at, and substituted t for t/2. then I substituted vf for vmidt. I then solved the equation ds=v0t+.5at^2 for t using the quadratic formula, then substituted my equation for t into the above equation that I used t/2 as t. I then solved for vmidt and got vmidt=(sqrt(v0^2+2ads))/2
close, but I believe it should be
vmidt=(sqrt(v0^2+2ads) + v0)/2.
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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?
The time it takes for vmidx and the expression for vmidt are similar, but I don’t know how to relate them.
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In general your work is good. However these are challenging questions so nobody is going to get every detail. I've inserted a number of notes.
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