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course Phy 241
10/5 11
1. Projecting point on CD onto paper on tabletop.`qx001. Your points will lie along (or close to) an x axis perpendicular to the line you sketched on your paper. With the origin at the center point, what were the positions of your points corresponding to theta = 0, 30, 60, 90, 120, 150 and 180 degrees? Report as 7 numbers separated by commas in the first line, with brief explanation starting in the second line.
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5.3, 4.5, 2.3, 0, -2.75, -4.8, -6.2
X axis is done in intervals of approx. 1cm.
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`qx002. What were the coordinates of your points corresponding to theta = 180, 210, 240, 270, 300, 330 and 360 degrees?
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-6.2, -5.3, -3.2, 0, 3.3, 4.5, 5.8
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`qx003. Suppose the disk rotated with a constant angular velocity, with an actual object moving along the tabletop just below the point on the disk. How would the velocity of that object change as the disk rotated through one complete revolution? Sketch (on your paper) and describe (below) a graph representing the velocity vs. clock time behavior of that point. Include an explanation connecting your results to your data.
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Using velocity = `dx/`dt, I found the differences in x between my time intervals. With t1 velocity = difference in x values between 0deg and 30deg. At v(t1) = -.8, v(t2) = -2.2, v(t3) = -2.3, etc….
This gave my graph the appearance to be concave up, that velocity is decreasing at increasing until crit pt(180deg, `dx becomes pos) where slope starts increasing at increasing rate before leveling back off towards final t interval.
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`qx004. For the same object as above, sketch a graph representing the acceleration vs. clock time behavior of that point. Desribe your graph and include an explanation connecting your results to your data.
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This graph appears to be concave down with crit pt at 120deg-150deg interval where a = 10.1(`dx)/5(t) = 2.02. After this point a decreases in a decreasing way
????We didn’t time this and I don’t think there is a way to so I left every interval in t, where it could be sec or whatever. Is this correct???????
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`qx005. For the same object, sketch a graph representing the net force on the object vs. clock time for one revolution of the disk. Describe your graph and include an explanation connecting your description to your data.
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This graph appears to be similar to accel graph , the greater the accel the greater the force needed to make the same mass accel faster.
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... describe r, v and a vectors ...
2. Quick collision experiment
`qx006. In the first line below give the landing positions of the 'straight drop', uninterrupted steel ball, the marble, and the steel ball after it collides with the marble, separated by commas.
In the second line, report the horizontal displacement of the uninterrupted steel ball, the marble, and the steel ball after it collides with the marble, separated by commas.
Starting in the third line give the units of your measurements and a brief explanation.
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1.5cm, 17.7cm, 26cm, 12.5cm
16.3cm, 24.5cm, 11cm
All measurments were done in cm, where horiz displacement was measured from straight down drop of steel ball.
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`qx007. Assuming that the time of fall was .4 seconds, what do you conclude was the velocity of each object at the instant it left the end of the last ramp? Report three numbers separated by commas in the first line, in the same order used in the preceding question. Units, explanation, etc. should start in the second line.
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40.75cm/s, 61.25cm/s, 27.5cm/s
Knowing horiz displacement and knowing `dt = .4sec, we can use `ds = vAve * `dt and solve for vAve. Where `ds is in cm and `dt is in sec then `ds/`dt will have to be in cm/s which is definitely a velocity measurement.
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`qx008. In the collision, the velocity of the steel ball changed, as did the velocity of the marble. What was the change in the velocity of each? Report number in the first line, brief explanation in the second.
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The steel ball went from having a velocity of 40.75cm/s when uninterrupted to 27.5cm/s when making contact with marble before leaving ramp. Because the marble was just sitting on the end of the ramp it could not have gained any velocity unless the steel ball transferred it to the marble in the contact made before both left the end of the ramp. This is obvious, another conclusion to be drawn is that the marble has a lighter mass because the steel ball lost around but not quite half its velocity during contact. The same energy that would have given the steel ball around 15cm/s velocity gave the marble a velocity of 61.25 which is almost 3 times as large. I am not sure exactly how this process works or what is going on but I am noticing the effects and have a slight understanding into how the system is working.
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3. Motion of unbalanced vertical strap
`qx009. The original vertical strap system oscillated about an equilibrium position with one end lower than the other. Why do you think the equilibrium position had that end lower?
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This would have to do with the mass of the lower side being greater than the mass of the elevated side
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`qx010. What changed about the behavior of the system when a couple of #8 nuts were added to the higher end? What is your explanation?
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At the equilibrium position it was the end that was lower, that is if enough #8 nuts were added to counteract the weight differences that were in place before.
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`qx011. Would it have been possible to balance the system at a position where the end with the #8 nuts was higher? Would it have been challenging to do so? Explain.
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It seems there would be some actions that could be taken to balance the system, but it would be fairly difficult to do due the fact that even after the #8 nuts were added the end was still higher so there must be a fairly significant mass difference.
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`qx012. Did the frequency of oscillation of the system appear to be constant?
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Of course the freq is a bit faster in the begeaging and a tad slower towards the end , I believe due to the fact that the force wears off as experiment continues. During the larger middle of the experiment freq. seems to be fairly constant.
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4. Balancing the styrofoam rectangle
`qx013. Was the styrofoam rectangle easiest to balance when the paperclip was inserted along an axis through the point below its center of mass, at a point above its center of mass, or at the point of its center of mass? Why do you think it was so?
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I don’t recall this experiment, but I believe that it would be easiest to balance with the paperclip inserted above its center of mass. I would think this was so because force of gravity helping to stabilize the mass when the paperclip is at the top of the rectangle.
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`qx014. At which positions of the paperclip did the system did the system oscillate? At which positions did it appear to oscillate with constant frequency? At which positions did it appear to oscillate with nonconstant frequency?
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I would believe that the most constant freq. would be for the paperclip inserted into the center of the rectangle. The other two positions would oscillate but not constantly
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5. Two cars with repelling magnets
`qx015. Why do you think the two cars traveled different distances when released?
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The one car seemed to have a smaller mass that traveled the furthest while the larger car traveled the smaller distance.
????I believe the cars had approx. the same magnetic force on top of them, is that correct??????
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`qx016. Which car do you think exerted the greater force on the other?
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I think they exerted approx. the same force on each other only the car with the smaller mass required less force to travel further.
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6. Graphs
`qx017. The graphs you were given in class depict coasting distance, in centimeters, vs. separation in centimeters, for a 120-gram toy car whose acceleration due to friction is 15 cm/s^2 (plus or minus an uncertainty of 10%). Sketch four tangent lines to the first curve, spaced equally from near one end of the curve and the other. Find, with reasonable accuracy, the coordinates of two points on each tangent line, and use these coordinates to find the approximate slope of each tangent line. In the first line below, report your four slopes. In the second line report the x and y coordinates of the two points used to find the slope of the third tangent line, reporting x and y coordinates of the first point then x and y coordinates of the second, using four numbers separated by commas.
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-15cm/cm, -12.5cm/cm, -4.4444cm/cm, -2cm/cm
(5, 30) (6.8, 22)
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`qx018. Each centimeter of coasting distance corresponds to very roughly 2000 g cm^2 / s^2 of energy lost to friction. That energy came from the potential energy of the magnets at the given separation. So the vertical axis of your graph can be relabeled to represent the energy lost to friction, and hence the potential energy of the magnet system. For example, 20 cm on the vertical axis corresponds to 20 cm of coasting distance, each cm corresponds to 2000 g cm^2 / s^2 of potential energy, so the 20 cm coasting distance corresponds to 20 * 2000 g cm^2 / s^2 = 40 000 g cm^2 / s^2 of potential energy. The number 20 on the vertical axis can therefore be cross-labeled as 40 000 or 40 k, representing 40 000 g cm^2 / s^2 of PE. You should be able to quickly relabel the vertical axis of your graph.
Using the relabeled vertical coordinates, find the y coordinates of the two points you used to find the slope of your third tangent line, then report the x and y coordinates of those two points as four numbers in the first line below. In the second line report the rise and run between these points, and the slope. In the third line report the units of the rise, the units of the run and the resulting units of the slope. Starting in the fourth line explain what you think the rise means, what the run means, and what you think the slope means.
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(5cm, 60000 g cm^2 / s^2), (6.8cm, 44000 g cm^2 / s^2)
rise = -16,000 g cm^2 / s^2, run = 1.8cm, slope = (-16,000 g cm^2 / s^2)/1.8cm = -88888.8889 g cm/s^2
rise is in grams per cm^2/sec^2, run is in cm, slope would be grams per cm/sec^2
rise is energy gained by magnets but lost during coasting due to friction, run is separation of magnets in cm, slope is the rate that the energy gained from the magnet and lost to friction is used per cm of initial sep. between magnets or another way to look at it is slope is how fast the energy is being used so it also tells you how much more energy the car is getting the closer the magnets are together.
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`qx019. Report the slopes of all four of your tangent lines, in terms of your relabeled coordinates, as four numbers in the first line below. You can easily and quickly find these four slopes from the slopes you previously reported for the four tangent lines. Starting in the second line report very briefly how you found your slopes.
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-30,000 g cm/s^2, -25,000 g cm/s^2, -8,888.8889 g cm/s^2, -4,000 g cm/s^2
I just mult. rise by 20,000g cm^2/s^2 and divided by original run.
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`qx020. University Physics Students: Find the derivative of the given y vs. x function y = 88 x^1.083 (this is a simple power function with a simple rule for its derivative) and evaluate at each of the four tangent points. Give the derivative function in the first line, in the second line the values you got at the four points, and in the third line compare your values to the slopes you obtained previously.
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y’ = 95.304x^.083
103.4938, 107.7749, 110,4312, 112.6526
First major thing you notice is when using derivative all slopes are positive (I think it was supposed to be x^-1.083), also my slopes were much higher meaning I believe I did not be as conservative enough when approx. the coordn. for my x and y points for my tang line.
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`qx021. University Physics Students: What is the specific function that describes PE vs. separation for the magnet system? What is the meaning of the derivative of this function?
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For PE vs. sep. , the resulting slope is `dPE/`ds or ave rate of change of PE wrt sep is ave force, therefore Force is derive of PE wrt position.
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... questions related to class notes ...
balancing demo
vertical strap demo
opposing cars demo, question of balancing paper clips
... vf ' = v0 / vf
... ref circle in complex plane ...
I believe I got it the first time. Let me know if there's any difference between the two submisisons.
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