Lab 7 Modeling Friction Forces

Student: Xavier Lomeli
Partners: Matthew Ibarra, Billy Justin
Date of Lab: 21 September 2016 

Lab 7 Modeling Friction Forces

Mission Statement: To conduct five different experiments involving friction, with derivations and measurements included to explain each part of the lab.

Theory: Friction is the reciprocal force experienced by objects which either prevents or hampers their movement by countering any applied force. The kind of friction which prevents movement is known as static friction (SF), and the kind which hampers movement is known as kinetic friction (KF). Both kinds of frictional forces should be equal to the normal force experienced by the object multiplied by some coefficient of friction between 0 and 1, with the coefficient of SF typically being higher than the coefficient of KF. This experiment was divided into 5 different parts, detailed below
Experimental Procedure and Explanation Part 1:
. For Part 1, essentially, we were operating a hanging mass-pulley-sliding mass system, with the sliding mass being a block with an acrylic bottom which would resist sliding until the weight of the hanging mass proved to much. The weight of the hanging mass would be increased ever so slightly by adding more and more metal plates to it, thereby allowing us to pinpoint the precise weight required to overwhelm the SF being experienced by the object, shown below.

 We would then plot the two important quantities (SF and the object's normal force, itself equal to the weight of the object) and produce the graph shown below.

 The slope of the SF verses normal graph should be our particular coefficient of SF, which was 0.2693 in this case, very much a reasonable figure.
 Experimental Procedure and Explanation Part 2:
For the second part of the experiment, we calibrated a force sensor which, when tied to the object with a string, could detect the average force being exerted on the object if we pulled the object at constant speed (so as to have zero acceleration). Additionally, we collected the masses of the object and then the object with one, two, and three other objects mounted atop it. We proceeded to plot the force being detected by the sensor over the duration of the pull four different times, corresponding with the four different masses, shown below.

With the information provided by the above graph, we were now equipped to plot our KF verses normal force graph, shown below.

From this experiment, we determined that the coefficient of kinetic friction is equivalent to the slope of our KF verses normal force graph, which in turn resulted in a value of 0.3049, reasonable to be sure, but oddly higher than the SF value from Part 1, which we did not anticipate.

Experimental Procedure and Explanation Part 3:
For Part 3, we inclined the slab on which we conducted Parts 1 and 2, setting the angle to be 19 degrees, shown below. 

 Using this known angle, we proceeded to calculate the coefficient of KF, shown below.

 We obtained a value of the coefficient of KF to be 0.344, which is 0.0391 off of our experimental value for the coefficient of KF we determined in Part 2, and even higher than our experimental value for the coefficient of static friction we found in Part 1. 
Experimental Procedure and Explanation Part 4:
Next, for Part 4, we mounted a motion sensor at the top of the incline to measure the acceleration of the block as the block slide downward to determine the coefficient of KF between the block and the slab. Plotting the data being collected by the motion sensor as the block descended, we constructed the following graphs of position v time and velocity v time.

 We highlighted the region of the velocity verses time graph corresponding with the block's uninterrupted descent, the slope of which was the acceleration, possessing a value of 2.829 m/s/s.
With this value for the acceleration, we could now carry out our calculation to determine the particular value of the coefficient of KF for this particular trial, shown below. 

 The value of the coefficient of KF this time turned out to be 0.244, which is much more reasonable than the values collected during Parts 2 & 3 since it is below the SF value of 0.2693 we determined for Part 1. 
Experimental Procedure and Explanation Part 5:
For Part 5, we were finished with the hands-on component of the lab and proceeded to a purely computational component. Using the coefficient of KF we determined from Part 4, we proceeded to derive an expression for what the acceleration of the block would be if you used a hanging mass sufficiently heavy to accelerate the system, shown below.

Our calculation is shown below.

Following our calculation, we obtained a value for the acceleration of 2.96 m/s/s, which is slightly higher than the acceleration value of 2.829 we collected during Part 4.

Conclusion- This laboratory was admittedly rife with uncertainty, predominately random uncertainty. In terms of systemic error, we followed the instructions and set up everything accordingly, so I am led to assume that the professor's initial remarks about how friction is very difficult to precisely model could best explain why our coefficients of KF for Parts 2 and 3 were actually higher than the coefficient of SF for Part 1. The best explanation I could provide would appeal to the fact that despite seeming smooth, both the white slab and the acrylic bottom of the block possess countless microscopic imperfections which compounded  to contribute significantly to the dependencies noted above as the block continued with its motion.

Lab 5 Modeling the fall of an object falling with air resistance

Student: Xavier Lomeli
Partners: Matthew Ibarra, Billy Justin
Date of Lab: 14 September 2016 

Lab 5 Modeling the fall of an object falling with air resistance

Mission Statement:To determine the relationship between air resistance force and speed.

Theory: Objects falling or otherwise moving through the atmosphere don't simply experience the force of gravity downwards, but also some kind of air resistance opposite their direction of motion as the object plows through the molecules which comprise the atmosphere.
Experimental Procedure:
For this problem, we reasoned that the force of air resistance is proportional to the velocity of the falling/moving object and would work in the direction opposite the motion of the object. We determined that to best experiment with this concept, we would walk together to a nearby building with a large internal cavity within which we could drop coffee filters without worry of crosswinds. The professor walked up to the balcony and dropped the filters whilst everybody video recorded the descent. There were 5 of these drops, the first being only one coffee filter, the second being two filters cupped together for added weight, continuing up to the 5th drop, which involved 5 filters cupped together, falling against a black cloth backdrop.
When we returned to the classroom, we used LoggerPro to place dots correlating with the position of the falling filter at each frame of its fall, and when finished, the dots were referenced in the creation of a graph showing position v time of the filter. We did the same for the other four videos in our archive, recording the slope of the curve on each graph, the slope of which should be the terminal velocity once we did a curve fit of the later points. Below, I show the aforementioned graphs, together with a fifth graph showing terminal velocity v air resistance.

Lists/Tables/Graphs of Collected Data with Explanation:

PART 1

Below is the position v time graph for our  falling mass system of (1) coffee filters, with the slope of the highlighted, linearly-fit region of the curve being the terminal velocity of the falling mass. For this drop, the terminal velocity reached 2.104 m/s.

Below is the position v time graph for our  falling mass system of (2) coffee filters, which fell faster, the terminal velocity being 2.393 m/s.

Below is the position v time graph for our  falling mass system of (3) coffee filters, which fell even faster, the terminal velocity being 2.512 m/s.

Below is the position v time graph for our  falling mass system of (4) coffee filters, which fell still faster, the terminal velocity being 3.792 m/s.

Below is the position v time graph for our  falling mass system of (5) coffee filters, which fell fastest, the terminal velocity being 4.022 m/s.

Below is the (terminal) velocity v air resistance graph 

From the above graph showing the relationship between air resistance and terminal velocity, we determined our values for k and n to be 0.007173 plus/minus 0.001746 and 1.263 plus/minus 0.1933, respectively.  

PART 2 

Now we applied the mathematical model we developed in Part 1 to predict the terminal velocity of any given number of coffee filters. We set up  a spreadsheet with various condition columns which would let us adjust values of k, n, the time interval size between points of the object's fall, and the mass of the falling mass. 

To model the descent and terminal velocity of our falling mass system containing (1) coffee filter, we set up the spreadsheet shown below.

Notice how m is the mass of  just one coffee filter. With the spreadsheet fully fleshed-out, we kept scrolling down until we noticed the acceleration fell to approximately zero, suggesting constant speed henceforth, shown below.

The terminal velocity shown here is approximately 1.172 m/s, which is not quite the 2.104 m/s which we observed. 

We then proceeded to double, triple, quadruple, and quintuple the mass to model each of our tested falling mass systems, with the results shown below. 

For the (2) coffee filter system, we collected the following information

*Notice I wrote 'weight x2' in D2 for clarity purposes

Here we see the terminal velocity (the velocity when the acceleration is approximately zero) is shown to be 2.029 m/s, which is yet again lower than our observed value of 2.393 m/s.

For the (3) coffee filter system, we collected the following information

Here we see the terminal velocity is shown to be 2.795 m/s, which is actually higher than our observed value of 2.512 m/s.

For the (4) coffee filter system, we collected the following information

Here we see the terminal velocity is shown to be 3.511 m/s, which is lower than our observed value of 3.792 m/s 

. For the (5) coffee filter system, we collected the following information

 Here we see the terminal velocity is shown to be 4.189 m/s, which is slightly higher than our observed value of 4.022 m/s.

Conclusion-
 While our observed values were still within the same order of magnitude of difference from the calculated values, the difference is still quite significant, with our observed values for the (1), (2), and (4) coffee filter systems being higher than their calculated counterparts while the observed values for the (3) and (5) coffee filter systems were actually lower than their calculated counterparts. These discrepancies can best be attributed to our admittedly arbitrary dot-placement when we were trying to capture the descent of our five different systems of falling masses. While we were as precise as possible with our placements of dots, the combination of blurry video and an inadequate backdrop definitely hindered our ability to properly place the dots along the trail of the falling mass.
In terms of systematic error, LoggerPro malfunctioned only once luckily, and we were able to recover our data. Otherwise, there were no pieces of equipment which were used, thus the aforementioned discrepencies, while not severe in scope, could nonetheless be the result of simple human error.

With regard to the premise of the experiment  itself, we confirmed that the air resistance rises with increasing terminal velocity, suggesting that heavier objects experience more air resistance, since the air will become increasingly unable to get out of the way of the descending object quickly enough to allow for continuously-increasing velocity of descent.

Lab 4 Non-Constant acceleration problem/activity

Student: Xavier Lomeli
Partners: Matthew Ibarra, Billy Justin
Date of Lab: 12 September 2016 

Lab 4 Non-Constant acceleration problem/activity

Mission Statement:To determine the distance covered by a rocket-equipped elephant on frictionless roller skates going down a hill, igniting the rocket the moment the elephant reaches level ground, which provides thrust in the direction opposite the motion of the elephant and whose mass is diminishing with time.
 The exact phrasing of the problem is the following:
A 5000-kg elephant on frictionless roller skates is going 25 m/s when it gets to the bottom of a hill and arrives on level ground. At that point a 1500-kg rocket mounted on the elephant's back generates a constant 8000 N thrust opposite the elephant's direction of motion. 
The mass of the rocket changes with time (due to burning the fuel at a  rate of 20 kg/s) so that the mass of the rocket is depicted by the function m(t) = 1500 - 20t.

Theory: Previously, kinematic problems we've done were done under the precondition that the acceleration, if involved, be held constant. Now we must apply the principles of calculus in order to solve kinematic problems like the one described for this lab which involves non-constant acceleration. Furthermore, this problem also involves the mass of both the elephant and the rocket, with the mass of the rocket being a function of time, since fuel is being burned.
Experimental Procedure:
For this problem, we actually did not attempt the calculus-based method of solving the problem since the integration techniques required would have been too time-consuming for most of the class. Instead, we opted for a numerical approximation using Excel. For this problem, we knew the initial velocity of the elephant-rocket system, the (constant) mass of the elephant, the diminishing mass of the rocket, which was a function of time due to burning the fuel at a given rate, and lastly the (constant) thrust produced by the rocket when ignited.

Lists/Tables/Graphs of Collected Data with Explanation:
Upon opening the spreadsheet for Excel, we devoted columns to time, acceleration, average acceleration, change in velocity, speed, average speed, change in position, and the position of the elephant. Furthermore, we inputted the initial conditions provided, shown below, with the time changing in increments of 1.

We see that the 'v' column begins at 25 m/s but eventually diminishes to 0 m/s somewhere between
t = 19 and t = 20 seconds. During that time period, the distance traversed by the elephant before the rocket begins to reverse its motion is numerically shown to be 248.379 meters.

Next we changed the time increment size from 1 to 0.1 and obtained the data shown below.

The highlighted row is the row of interest. Now we see that when time is just before 19.7 seconds is when the speed of the elephant-rocket system is zero, and the distance traversed is 248.698 meters.

Finally, we changed the increment from 0.1 to 0.05 seconds and obtained the following data.

The relevant row here is highlighted and shows that when the time is somewhere between 19.65 and 19.7, the distance traversed is 248.698, which did not change from when the increment size has 0.1 and not 0.05, suggesting that calculus analytics would achieve the same number.

Conclusion- In fact, through calculus, we would have achieved the approximately same number of 248.7 meters, with the computational work shown on the first page of the lab, therefore confirming that our numerical technique is valid. Sources of error in this lab would likely only be due to human error during the calculations since there were no physically moving parts involved, but it would be very curious if there were.

Lab 3 Propogated Uncertainty in Measurements

Student: Xavier Lomeli
Partners: Matthew Ibarra, Billy Justin
Date of Lab: 7 September 2016 

Propagated Uncertainty in Measurements

Mission Statement:To determine the propagated error of our measured density of two different cylinders.

Theory: Every kind of calculation is in truth an approximation, and the tolerance of uncertainty depends on the application the calculation will be for. When multiple calculations are involved, then the uncertainty ripples, or propagates, through to the final result. Learning the method by which this kind of propagated uncertainty is determined is the essence of this lab.
Experimental Procedure:

 In order to determine the propagated uncertainty of our measurements, we first needed to familiarize ourselves with the calipers provided to us. Before beginning the experiment ourselves, the professor showed us a clever technique for accurately reading measurements up to the hundredths place. Then we proceeded to measure the height and diameter of two different cylinders with the aforementioned calipers, together with weighing the cylinders on a digital scale. Now equipped with these measurements, we proceeded to carry out the computations, shown below, to obtain the propagated uncertainty of the calculated density of the aluminum and tin cylinders, respectively.

Lists/Tables/Graphs of Collected Data with Explanation:

From the calculations above, I determined that the density of aluminum is 2.87 plus/minus 0.058 g/cm^3 and the density of tin to be 11.73 plus/minus 0.058 g/cm^3. The density I determined for aluminum is roughly close to the scientific value of 2.70 g/cm^3, but the density I determined for tin is considerably off of the accepted value of 7.26 g/cm^3.

Conclusion- The difference between our calculated values and the accepted value, particularly for the tin cylinder, is most likely due to errors in weighing the cylinders, since our calipers were mechanical in nature. I suspect the oddly higher-than-expected density of tin is more than anything else due to human error in calculations, although reviewing my written work above, I honestly do not know where I went off.

Lab 2 Determination of g (and learning a bit about Excel) and some statistics for analysing data.

Student: Xavier Lomeli
Partners: Matthew Ibarra, Billy Justin
Date of Lab: 31 August 2016 

Determination of g (and learning a bit about Excel) and some statistics for analyzing data.

Mission Statement:To determine whether, in the absence of all other external forces except gravity, a falling body will accelerate at 9.8 m/s^2. 

Theory: Our theory is that in a controlled environment,the value of g can be determined through free-fall mass experimentation.
Experimental Procedure: In order to determine g, we used a strip of spark paper, marked by electric chars dispensed at equal time intervals by a falling body initially held in place by electromagnets to produce the strip of spark paper showing a dotted line with the dots increasing in distance with each other. This now became our permanent record of the fall corresponding to the position of the falling mass every 1/60th of a second, and we then proceeded to measure the distances between the char dots (in cm) and type them into our Excel spreadsheet, together with other columns for time transpired, change in distances between times, mid-interval times, and mid-interval speeds. We then selected the columns for mid-interval time and speed and made a XY scatter graph with the points not connected. Finally, we installed a linear fit from which we could get the equation of the line. We repeated this procedure for the columns of time and distance.


Lists/Tables/Graphs of Collected Data with Explanation:
Below are both of my graphs, showing distance v time and mid-interval speed v mid-interval time.

Questions/Analysis
1)
2) In order to get the acceleration due to gravity from the velocity/time graph, we simply look at the slope of our fit line, which turns out to be 954 cm/s^2, which corresponds with 9.54 m/s^2, which is fairly close to the accepted value of 9.81 m/s^2.
3) In order to determine the acceleration due to gravity from the position/time graph, the second derivative of any particular point must be found. Using Excel, I identified the position function to be y = 475.91x^2 + 74.29x + 0.02 and thus the first derivative to be 951.82x + 74.29 and the second derivative to be simply 951.82. Since our units of length were centimeters, this value corresponds to 9.52 m/s^2, which is within 0.02 of the value for g we got from our velocity/time graph, and fairly close to the accepted value of 9.81 m/s^2.

Conclusion- The difference between our calculated values and the accepted value is most likely due to measurement uncertainties when we determined the distances between char marks on the spark paper and also systematic error (i.e. less-than-cutting-edge equipment). Nevertheless, our value fell well within the range of values other groups collected.

29-Aug-2016: Deriving a power law for an inertial pendulum.

Student: Xavier Lomeli
Partners: Matthew Ibarra, Billy Justin
Date of Lab: 29 August 2016 

Deriving a power law for an inertial pendulum

Mission Statement:To measure the inertial mass of an object by comparing the resistances experienced by the object when subject to changes in its motion. 

Theory: Due to mass not being dependent on gravity, gravity is not required within the derivation to find the inertial mass of the object, since mass is ultimately a quantitative measure of an object's inertia, or resistance to motion. So if the force of gravity were unknown, such as on another planet, the mass of any given object could still be identified by utilizing a mechanism that exploits this characteristic of constant inertial mass.
Experimental Procedure: For this experiment, we were supposed to determine the mass of any given object (stapler, pencil, etc.) by clamping an inertial balance to the edge of the worktable, placing a thin piece of masking tape on the end of said balance, and then setting up a photogate so that when the balance oscillated the tape completely passed through the beam of the photogate. Every two times the strip of masking tape blocked the beam, the computer program Logger Pro recorded the period duration via a connected LabPro. We recorded 9 different periods corresponding to 9 different masses atop the inertial balance. With this data we then proceeded to find the relationship between mass and period via some power-law type of equation described below.
Lists/Tables/Graphs of Collected Data with Explanation:
The table below shows the different masses we tested together with their respective periods (T)-

From the data on the table above, we could now use the equation T = A(m+Mtray)^n to determine the mass (m) of any given object. However, first we needed to solve for the unknown variables A and n, which correspond to the y-intercept and slope, respectively, of the above equation when written in the equivalent form Ln(T) = n*Ln(m + Mtray) + Ln(A). Furthermore, we needed to arbitrarily determine the appropriate value for Mtray such that the plot of the equation would be linear with a correlation coefficient of 0.9998. We discovered that there were a range of values for Mtray, which we discovered to be 210 +/- 20 grams. Beyond this range, the plot lost linearity. Below, I took pictures of three graphs, one depicting the lowest possible value, one the highest, and the third depicting the median value for Mtray.
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Lowest value for Mtray (190 g)

    n = slope = 0.5351
   A = e^(-4.083) = 0.0168

Highest value for Mtray (230 g)

    n = slope = 0.5950
   A = e^(-4.505) = 0.0110

Middle value for Mtray (210 g)

    n = slope = 0.5497
   A = e^(-4.186) = 0.0152
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Now that we found the proper values for A, n, and Mtray, we could put our equation into practice and calculate the inertial mass of any given object upon knowing the period of said object. Since we would be inputting three different values of Mtray per object, we expected three different values for the inertial mass per object. We decided that these two objects would be my partner's phone and the classroom stapler. Below is my work.
*Keep in mind Ln(A) = Y-intercept, thus A = e^(Y-intercept)

Next, we went and weighed both the phone and the stapler to record the gravitational mass. Below are pictures of the scale.

Conclusion- Fittingly, these observed gravitational masses corresponded very well with their respective inertial counterparts, falling well within the range of calculated values for the inertial masses. Thus, the end purpose of this lab, to find a mathematical relationship between mass and period for an inertial balance, has been proven to be satisfied by the equation described above.

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