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course PHY 201
The only thing I've changed is question 3--I've tried to denote it better this time. It is under your previous comment. 11/16 10 pm
Assigment 927 redone
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course Phy 201
11/15 3 pmBelow is the 9/27 lab. The only portion I've redone is the question asking about the velocity vs time graph, question 3
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Assignment 927
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course PHY 201
10/27 10 am1. 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, 3.5, 2.1, 0, 3.6, 4.8, 6
Starting with 0 degrees and rotating through, these numbers correspond to the number of cm away from the axis at the given degrees.
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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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5.3, 4.4, 2.6, 0, 3, 4.5, 5.7
These numbers start at 180-360 and correspond to the number of cm away from the axis.
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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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I am still lost on this question and the next, but I wanted to go ahead and submit
Every 30 degrees of rotation corresponds to the same time interval. How far does the object travel during the first time interval? How far during the second? etc. ...
So when it it moving most quickly, when most slowly?
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The angular velocity would be moving at the same speed through each interval. The velocity vs clock time graph would be that of a straight line parallel to the x axis, assuming x=clock time and y= velocity.
That describes the point on the disk.
However the object is moving directly below the disk, and its velocity is not constant. It moves a lot further between the 60 degree point and the 90 degree point than between the 150 degree point and the 180 degree point. According to your data, for example, that the first of those distances would be 2.1 cm (from 2.1 cm to 0) and the second would be 1.2 cm (from 4.8 cm to 6 cm).
See also the class notes. I've posted some note on this situation that parallel the discussion in the last class.
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`qx004. For the same object as above, sketch a graph representing the acceleration vs. clock time behavior of that point. Describe your graph and include an explanation connecting your results to your data.
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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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... 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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2.5, 18, 13.5, 9.5
16, 11.5, 7
The measurements are in cm, and the horizontal is the extra distance it went using the free drop as baseline.
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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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80, 57.5, 35
The units are cm/s. I arrived at these answers by using `dx horizontal and the `dt to find Vave. Knowing that vf=0, V0=2*Vave
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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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57.5, -45
cm/s The change in velocities is the difference between the initial and the final. The marble was at rest when hit by the steel ball, therefore the change in velocity is 57.5-0=57.5 cm/s. According to the velocity in the previous question as the ball left the ramp, it was 80 cm/s, however, after it collides with the marble, the velocity is 35 cm/s, the difference being 45 cm/s.
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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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I would assume that end weighed slightly more.
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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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The system rotated until the nuts were at the lower end of the system. My explanation is that the 2 nuts weighed more than the opposite end.
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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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I am not sure it would have been possible because that end wanted to stay lower.
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`qx012. Did the frequency of oscillation of the system appear to be constant?
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Yes, after the system decided which end was lower and which was higher.
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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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At a point above the center of its mass. I think it was because the majority of the mass was below the paperclip therefore making it more stable and easier to balance.
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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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The system oscillated when it was in the center of its mass with the best frequency and when the paper clip was above the center of its mass. When the paper clip was below the center of its mass, it oscillated non-constantly and often ended with the side portion facing upward.
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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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I think the cars traveled different distances when released because of the differences of magnetic force related to the size of the cars and potentially to any drag caused by rubber bands or friction occurring.
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`qx016. Which car do you think exerted the greater force on the other?
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I think the car that was held steady exerted a greater force on the one being moved closer.
When you hold two magnets together, which exerts the greater force?
The answer is that the forces are equal and opposite.
The same is so when the magnets are attached to toy cars.
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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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-13, -11, -8,- 1
5.5, 30, 6.5 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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5.5, 60 K, 6.5., 44 K
rise= 60,000– 44,000=16,000 run=5.5-6.5=1 slope =16000
g cm^2 / s^2, cm, g* cm/s^2
The rise in the amount of energy, PE lost, coming from the magnets, the run is the distance coasted, and the slope is the amount of PE lost over the distance.
Good.
When you divide `dPE by the displacement `ds over which it is lost, then since `dPE = F_ave_cons * `ds, you get F_ave_cons, the average conservative force.
So in this case you get the average magnetic force.
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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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26000, 22 000, 16000, 2000
I multiplied the beginning slopes by 2000 g cm^2 / s^2
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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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`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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... 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 ...
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Very good overall.
See my notes on those few questions related to the motion of the point beneath the CD. Submit just a copy of those questions, with your best answers to my new questions.
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We talked about that situation in class today. You might have gained some insight from that discussion.
I can't tell what's new in the submission. Can you submit another copy and mark what is different from before, using &&&& before and after the new information? You are of course welcome to make more changes, if necessary.
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See my note, which relates your question to your data and also gives you a reference to class notes. Let me know if you don't understand, but I think you will.