LabActivities927

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course PHY 231

10/12/10 2:29 pm

Lab activitiesNotes:

Include concise explanation: Whether specifically requested or not, all responses should include a brief explanation or description beginning on the line following other requested information.  Abbreviations and incomplete grammar are acceptable; if you go too far with this I'll let you know, but I don't want to keep typing demands reasonable.

Reporting data:

• Data consists of what you observed, not what you concluded or how you got from observations to conclusions. 

• Data should be presented in the specified format.  If no format is specified, give a succinct data report in the form of a table. 

• Unless otherwise specified, data should consist of numbers, with a subsequent note on the meanings and units of the numbers. 

• Basic rule:  Don't bury your data in a paragraph of explanation.  It's OK if it appears that way within your explanation, as long as there's a succinct data summary. 

Explaining your analysis:

• Typically the explanation of your analysis will include some combination of symbolic and numerical results in a sample calculation of one result.

• Other results should be reported in specified format; if no format is specified a list or a table would be appropriate.

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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0, 4.5, 4.5, 0, -4.5, -4.5, 0

The duplicated numbers indicate that the projection onto the x axis was the same for both angles, or all three angles for 0/90/180.

The projections wouldn't be the same for 0 deg and 90 deg, nor would they be the same for 30 deg and 60 deg.

Ask me next class and I'll clarify this.

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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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0, 4.5, 4.5, 0, -4.5, -4.5, 0

The numbers here follow the same pattern as before.

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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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Seems that this graph is sinusoidal, that is it's going from zero then increasing but as it reaches the other side of the rotation is slows down and stops for an instant and through 360º it would go through 2 of these cycles. This shows in the coordinates of the angles since the 30-60 interval is very small, distance wise, than the 0-30, 60-90, 90-120… intervals. This means that the object would slow down at the extremes and speed up trough the larger 'inner' intervals.

That's a good description. I'm still unsure of your data.

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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 has the same shape as the first graph however the larger values of a take place when the object is at the extremes rather than the inner angle area. This matches as you can see that the velocity changes the most in the 30-60 area. If the first graph were a sin function then this one would be a cos function, since a=dv/dt and since this graph corresponds to the first one as being the slope of the velocity graph.

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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 would be nearly identical to the acceleration graph since as a increases so to does F_net, when the mass in constant anyway, as it is here. The difference would be in the amplitude of the wave but it would peak when a peaks and be zero when a is zero.

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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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0, 9.25, 15.25, 2.75

My units here are cm. I 'zeroed' my measurements at where the dropped ball landed since that indicated, relatively accurately, where the edge of the ramp/table was.

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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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23.1, 38.1, 6.88

Units: cm/s. This was obtained by dividing the x distance by the time. One note: I seem to remember that these balls transferred to a flat ramp where they then impacted the marble. This velocity would not be correct if it was not done this way as there would have been an (v0)y component to the velocity.

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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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-16.22, 38.1

The units are again cm/s. This is ∆v and it was obtained in the steel balls case by subtracting the first solo velocity from the impact velocity. The marbles ∆v is just the velocity it had because its v0 was 0.

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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'm not remembering this experiment very well and can't seem to find any notes/data from it. I do remember the oscillating strap some but only what I saw you do. This has to do with center of mass though but can't say too much more than that.

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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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This would obviously change the center of mass in favor of the high end and it would then move to the bottom in order to obtain equilibrium as much as it could.

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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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Its possible but certainly difficult. This is due to the way that mass is distributed, with so much of the mass in such a small area even a very small inequality would allow the nuts end to spin to the bottom. There's very little room for error here, in other words.

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`qx012.  Did the frequency of oscillation of the system appear to be constant? 

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Relatively so, but this system has a lot more friction than a pendulum because of where there is meta/metal contact. The friction does not dissipate energy linearly and so the angular velocity would create more/less friction based on its value.

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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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This was easiest above below the axis as when there is more mass on one end than the other, that end will move to the lowest state possible (entropy at work maybe?).

good thought, but no mechanical work resulted from thermal energy changes, so while there were changes in entropy they did not explain the observed behavior of the system

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`qx014.  At which positions of the paperclip 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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Below the axis oscillated in the most constant manner, while along the axis of mass center the motion was more erratic, or so it seemed, and would easily 'topple' or flip over if the amplitude was to great. Above did not oscillate until it flipped over and became below.

good, but note that your description seems to describe the behavior in terms of the position of the center of mass with respect to the paperclip, not in terms of the position of the paperclip

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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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Lots of things at work here: mass, mass distribution, friction acting on the car from table, of the car on the table, and friction internal to the car, e.g. between the wheels and the axles., magnet positions, strength. Any/all of these would impact the car's distance depending on how much/many were depleting the kinetic energy.

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`qx016.  Which car do you think exerted the greater force on the other?

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Neither, the forces must be equal and opposite by Newton's Third Law of motion. Any 'observed' disparity has more to do with the forces at work above than the reaction.

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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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-20.7, -12.8, -5.4, - 1.32

1.9, 50, 3.75, 40

Clearly for such a small and jagged graph these are very rough approximations, but I did try to pick points whose values were more clear.

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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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1.9, 100 000, 3.75, 80 000

-20 000, 1.85

rise units = g cm^2/s^2, run units = cm, slope units= g cm/s^2

The rise is the ∆PE from one distance to another while the run is just ∆x. The slope is then ∆PE/∆x. The ∆PE is also the work done on the car by the magnets through the distance ∆PE. The slope is then related to something else here but I'm not sure what it is right at this moment. The units seem to be force units but I know the units in this area can be misleading.

Good. Since `dPE = F_ave * `dx, we get F_ave = `dPE / `dx, so your slopes should represent forces.

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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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41400, 25600, 10800, 2460

I just multiplied my previous slopes by 2000

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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'= -203.85x^-2.083

-30.3, -15.9, -5.85, -2.68

9.6, 3.1, 0.45, 1.36

The function sees to be messed up as it seems it should be 188.23 and -1.083 so that is what I used. Yeah its' easy to see that these varied from mine considerably.

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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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I don't know what this means. Is this the function from the graph? The derivative is still what I said above: ∆PE/∆x and I still am not sure about exactly what that implicates at the moment.

The PE vs. x function would just be 2000 * 188.23 x^-1.083, where 2000 represents the 2000 dynes per coasting centimeter. The 188.23 * x^-1.083 is just the coasting distance 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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&#This looks good. See my notes. Let me know if you have any questions. &#

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