Jillian,+Ani,+Ariel,+Nicole

=Lab 5: Acceleration Down An Incline=
 * Group Members:** Ariel Katz, Jillian Laub, Ani Papazian, and Nicole Margulis
 * Class:** Period 2
 * Date Completed:** December 20, 2010
 * Date Due:** December 21, 2010

Purpose:
The purpose of this lab is to determine the relationship between acceleration and incline angle.

Hypothesis:
If an object is moving down an incline, then acceleration should be equal to gravity (9.8) times sin of the incline angle minus µg (9.8µ) times cos of the incline angle. If an object is moving up an incline, then acceleration should equal µ (the coefficient of friction) times mass 1 times g times sin of the incline angle plus mass 1 times gravity sin of the incline angle minus mass 2 times gravity all divided by the sum of mass 2 and mass 1. These hypotheses are based on Newton's second law, which states that force equals mass times acceleration along with the equation to determine friction, which states that µ times normal force equals friction force.

Materials:
wooden block, aluminum track, ring stand, clamp, meter stick, photogates and data studio, picket fence, tape, pulley string, masses, masshanger

Procedure:
see prelab

Pre Lab Derivations
1)

2) 3)

Video of Data Collection:
media type="file" key="ariel is the best mermaid lol physics.mov" width="300" height="300"

Data:




Class Data:




Graph:
The slope of the line is half of what it should be because when we collected the data, Data Studio calculated the band spacing to be the space in between the black bands. It should really have been calculating from the leading edge of one black band to the leading edge of the next. Therefore since there is double the amount of distance, there is double amount of acceleration as shown in our adjusted graph.



Data/Graph Analysis:
>> % error = |Actual value- Theoretical value| / Theoretical value X 100 >> >> |9.3093- 9.8| / 9.8X 100 >> |-0.4097| / 9.8 X 100 >> 5% error
 * Find the coefficient of friction between your incline and the block using the equation of your trend-line.
 * [[image:Pictaegaergznvaergea.png width="167" height="220"]]
 * Calculate the percent error between the slope and g(earth). Show this calculation.
 * Compare the value of the coefficient of friction between your incline and the block to that from last week's lab.
 * our coefficient of friction last week was .204 and this week was .2068. These are very similar, which it should be because we used the same surfaces as before.

Percent Difference with Class:



 * Part B Calculations**



Analysis Questions:

 * 1) Discuss your graph. What does the slope mean? What is the meaning of the y-intercept?
 * The slope of our adjusted graph is close to the theoretical slope, which should be the force of gravity times the sine of the angle theta (angle of incline). The y-intercept equals the product of -µ, gravity, and cosine theta. These meanings come based on the equation for acceleration derived from the equations for the system (as seen in pre-lab derivations).
 * 1) If the mass of the cart were doubled, how would the results be affected?
 * If the mass of the cart were doubled, the acceleration would, based on the equation of the line of best fit, be slightly smaller. Changing the mass doesn't change gravity, the sine or cosine of the incline angle, it only would change µ. Because of this, µcosø would be larger, and since it is subtracted from gsinø, the total value of acceleration at that angle would be smaller than if the mass were smaller.
 * 1) Consider the difference between your measured value of g and the true value of 9.80 m/s^2. Could friction be the cause of the observed difference? Why or why not?
 * The theoretical value for g should be 9.8. However, we found that g is equal to 9.3. Friction on the x-axis could have caused our value to seem be lower than the theoretical value because there is friction between the block and the track, which could have been a source of error in the calculations.
 * 1) How were your results in Part B? Why was the expectation that your results be within 2% considered to be reasonable when in other labs we allow much larger margins of error?
 * We did not complete Part B of this lab.

Conclusion:
Our lab was pretty successful in finding the equation of acceleration based on the sine of the angle of incline. The graph should theoretically have been a straight line with a positive slope of 9.8 and a negative y-intercept. Because each angle changes what 9.8*sineø and µcosø equal, it was difficult to come up with a theoretical y-intercept and slope. Even though our r^2 value of 9.78 is relatively high, it might have been even higher had we redone the data for the angle with the smallest sine. This point visibly throws off the line of best fit on our graph. Still, we had a percent error of only 5%, a relatively small value. Some main sources of error in this lab include human error and using the incorrect coefficient of friction in our calculations. Even though we used the same surface, the coefficient changed slightly, which may have changed our results slightly. In addition, we started by only changing the angle by minuscule amounts, so the differences in the values of acceleration were difficult to see. However, the equation for the line of best fit of our data is still very close to the derived theoretical equation. Also, we can use the graph to estimate the acceleration given a value for incline angle. Theoretically, the larger the sine of the angle, the larger the acceleration should be for any mass (masses cancel and are not in the equation for acceleration, only the coefficient of friction is subject to change).

=Lab 4: Coefficient of Friction=
 * Group Members:** Ariel Katz, Jillian Laub, and Ani Papazian, Nicole Margulis
 * Class:** Period 2
 * Date Completed:** December 13, 2010
 * Date Due:** December 14, 2010

Purpose:
The purpose of this lab is to determine the coefficient of static and kinetic friction between a wooden block and an aluminum track.

Hypothesis:
If we drag a wooden block with mass across an aluminum track, and use data studio to calculate the tension and assume tension equals friction, then we can find out the coefficient of friction. If we raise the track to the highest angle without the block moving, and we find the average of 8 different trials, then that is static friction. If we raise the track to the lowest angle that the block would move, and we find the average of 8 different trials, then that is kinetic friction. Based on the values of the coefficient of friction between wood and cleaned metal found on the internet, this value should be between 0.2 and 0.6.

Materials:
track block mass string data studio connector protractor

Procedure:
see website

Video of Data Collection:
media type="file" key="ariel physics ani nicole jillian.mov" width="300" height="300"

Data:
0.5 kg

1.0 kg

1.5 kg

2.0 kg

2.5 kg

[|excel december 13th.xls] Part A: Calculations used for part A:

Maximum tension - used to calculate coefficient of static friction

mean tension - used to calculate coefficient of kinetic friction

Part B: maximum angle without sliding - used to calculate coefficient of static friction Calculations used: minimum angle to slide at constant speed - used to calculate coefficient of kinetic friction Calculations used:

Data Analysis, Graph:
· From measured data in Part A, use Microsoft Excel to plot a graph of Static Friction vs. Normal Force. Add a second line for the kinetic friction data, on the SAME graph. Set the y-intercepts to zero, and show equations of the lines with the regression coefficient.

·slope for µs = .2498 -slope for µk = .2159 Compare the slope of line with calculated m s average (% difference). The slope of the line and the calculated µs average are very close. The calculated average for the coefficient of static friction is 0.246. The slope of the line is 0.2498 for the coefficient of static friction. The percent difference between the two is 1.52%. Graph vs. Calculated percent difference:

· Compare your values of m found in Part A with those found in Part B.

Data Analysis, Percent Difference:
Class information:



Our results vs. the class's results for coefficient of status friction: Our results vs. the class's results for coefficient of kinetic friction:

1. Why does the slope of the line equal the coefficient of friction? Show this derivation. The slope of the line is equal to the coefficient of friction because the equation is given, where f = friction, M = the coefficient of friction, and N = normal force. When graphing the friction force vs. the normal force from our data collection, and making f=y and n=x, then m=the slope based on the equation y=mx+b, where m=slope. Therefore, the slope of the line equals the coefficient of friction on our graph.

2. Look up the coefficient of friction between wood and alum inum. Discuss if your measured results fall within the range of theoretical values. Be sure to cite your source! According to three websites (http://www.engineeringtoolbox.com/friction-coefficients-d_778.html, http://www.engineershandbook.com/Tables/frictioncoefficients.htm, and http://www.school-for-champions.com/science/friction_coefficient.htm) the coefficient of wood on clean metal (aluminum is a metal and we cleaned it with Windex) is in the range 0.2-0.6. The values we calculated for the coefficients of static friction (0.26) and kinetic friction (0.22) both fall within this range. This supports that our values are correct.

3. What variables affected the magnitude of the force of friction? What variables affected the magnitude of the coefficient of friction? The magnitude of the force of friction was affected by the amount of mass and, therefore, weight we placed on the wooden block that we slid. The greater the mass on the block, the larger the force of friction was. The magnitude of the coefficient for friction was affected by the amount of friction between the block and the track (which was affected by the mass). If the friction force is larger and the normal force stays the same, then the coefficient of friction must increase. In this case, the coefficient of friction was a constant because as the mass increased so did the normal force (acceleration = 0 so w=N).

4. How does the value of coefficient of kinetic friction compare to the value for the same material’s coefficient of static friction? The value of coefficient of kinetic friction is lower than the value of coefficient of static friction for the same material. This is because when an object is already in motion (kinetic friction) their is less "hold" on the surface and the material. When an object is at rest, it will take more of a force to get it to start moving.

5. Did putting the track on an incline significantly change the coefficient of friction? Why or why not? The coefficient of friction would be slightly lower when the track is on an incline. The force of gravity would also be acting on the block of wood, enabling it to slide down the incline easier then when it is flat. The equation for incline friction uses the components of gravity as opposed to just weight on a flat plane. The components of gravity are cos theta mg for the y-axis and sin theta mg for the x-axis.

Conclusion:
The goal of this lab was to find the coefficient of friction between aluminum and wood. Also, the purpose was to find the relationship between normal force and friction force. We tested what an incline would do to the coefficient of friction and found that it decreases because it is broken down into x and y components. Our results from the experiment supported our hypothesis. We hypothesized that the coefficient for static friction would be larger than kinetic friction. We resulted with a larger static friction of .2498 and a smaller kinetic friction of .2159. The class averages were .2170 and .1876 showing that our results are reasonably accurate. We used Newton’s 2nd Law to show that tension is equal to friction acting on the system.

In this lab there are many possible sources of error. One potential source of error could have happened while pulling the block. The lab only works by assuming the acceleration of the block is zero. Although we tried to pull the block at constant speed, there could have been some acceleration or deceleration. This error could have been improved by having a machine move the block to make it go at an uniform speed. Another source of error could have been from the track. If there were any dust particles on it, it would make the results change. Although we wiped down the track in the beginning of the experiment, this error could have be minimized if we wiped it down before every trial. In the second part of the lab when lifting the track, we had to find the angle at which the block was moving at a constant speed. It was difficult to conclude when it was moving at constant speed because it is subjective. To decrease this error we could have done more trials.

=Lab 3: Newton's Second Law, Atwood's Machine= = =
 * Group Members:** Ariel Katz, Jillian Laub, and Ani Papazian, Nicole Margulis (Absent)
 * Class:** Period 2
 * Date Completed:** December 6, 2010
 * Date Due:** December 7, 2010

= =

Purpose:
The purpose of this lab is to again determine if Newton's Second Law can be supported. We want to observe graphs made using data obtained from Data Studio recordings of acceleration in different trials while changing force to see if the equation F=ma can be supported by data.

Hypothesis:
According to Newton's Second Law, if the force is changed and the total mass is kept the same, then the acceleration should change proportionately according to the equation F=ma. The graph should be linear and as the difference in the masses is increased, the acceleration should increase because of their inverse relationship.

Materials:

 * Atwood's Machine
 * Pulley
 * Clamp
 * Rod
 * Stand
 * 2 mass hangers
 * Masking tape
 * Masses: 100 g, 200 g, 10 g, 5 g
 * Photo-gate
 * String
 * Data studio connection

Procedure:
Procedure
 * 1) Set up the Atwood Machine
 * 2) clamp on a Pulley to the top of the machine
 * 3) plug in the usb link and open data studio
 * 4) set up balances on either side of the pulley
 * 5) place masses on mass hanger 1 and mass hanger 2 ; always keep the net mass between the mass hangers constant
 * 6) For each of the following masses, perform 3 trials:
 * 7) MH1 = 350 g MH2 =300 g
 * 8) MH1= 345 g MH2 305 g
 * 9) MH1 = 340g MH2 = 310 g
 * 10) MH1= 335 g MH2 = 315 g
 * 11) MH1 = 330 g MH2 = 320 g
 * 12) Record data using data studio
 * 13) figure out acceleration using the slope of the velocity – time graph
 * 14) place data points in excel spreadsheet
 * 15) while placing data points in, populate a graph to determine the relationship between net force and acceleration
 * 16) analyze the data
 * 17) analyze the graphs to determine the relationships

Video of Data collection:
media type="file" key="ATWOOD ARIEL JILLIAN ANI.m4v" width="300" height="300"

Data:










Graph:





Analysis Questions:
1. Explain your graph(s) thoroughly! a. If linear: What is the slope of the trend line? To what actual observed value does the slope correspond? How does it compare to this actual observed value (find the percent error between the two)? Show why the slope should be equal to this quantity. What is the meaning of the y-intercept value?

Our graph has a linear slope, as predicted. The slope is 0.666, which should correspond with the total mass of the system. The actual total mass of the system was 0.650; giving us the extremely small percent error of 2.4%, which could have been caused by not taking friction and the mass of the pulley into account. The slope should be equal to the mass because based on the equation, F=ma. This equation matches slope intercept form where a=x, F=y, and mass=the slope. The y-intercept in this case would be the amount of friction because theoretically the force should be zero when the acceleration is zero, but in our equation, the force = the y intercept when acceleration is zero.

Percent error: (|Actual-Theoretical|/Theoretical)*100 (|0.650-0.666|/0.666)*100 = 2.4%

2. What would friction do to your acceleration? Would you need a bigger or smaller force to create the same acceleration? Was your slope too big or too small? Can friction be a source of error in this experiment? Redo the calculation of acceleration WITH friction to show its effect.

The friction should theoretically decrease the acceleration because friction works opposing other forces like tension or normal force. In the equation to find acceleration, you would subtract the friction from the force in the numerator. To keep the acceleration the same but still account for friction, the force would need to be larger. Because friction reduces acceleration, and because we did our calculations ignoring friction, then our calculations are larger than the actual values. Friction could be a source of slight error in this experiment. However, since friction is consistent basically with each changing of the mass, it would only make our results less accurate, while they would retain precision. The slope would still be linear, but the fit should have a y-intercept of 0 and the slope would be more accurate. When we redid the calculations including friction, we still got a linear line of best fit with a r^2 value of 1, but the force was more accurate and the y-intercept is extremely close to 0. We found the friction by looking at the y-intercept of the graph.



3. Discuss the precision of your data. Because our R^2 value is one, our data is extremely precise. For this lab, our data should be precise because all the masses were known and the mass of the pulley was negligibly small. In addition, the data studio which calculated the acceleration is very precise. Therefore, all of our measurements are accurate.

4. The real pulley and mass arrangement is not as simple as we assumed. In fact the pulley is not massless and frictionless means that it does require a net “torque” (a turning force) to make it rotate – this is supplied by the tension in the string. The rotational inertia of the pulley then adds an equivalent mass to the total mass being accelerated, where the equivalent mass for the pulley is approximately equal to ½ of the mass of the pulley. If the mass of each pulley is 5.6 g, could the pulley mass account for a significant potion of your error in the experiment?

In order to figure out if the results are better, we took into account the mass of the pulley by adding it to the total mass, and observing the result. We look at the equation, and see if the graph data (theoretical) is close to the actual data. By including the mass of the pulley in the denominator when finding acceleration, the acceleration was smaller.

Percent error when mass of pulley is taken into account:

Conclusion:
In this lab, the relationship between force and acceleration was determined. The lab would determine if Newton's second law could be supported. Newton's second law states that Force = mass * acceleration. The hypothesis stated: According to Newton's Second Law, if the force is changed and the total mass is kept the same, then the acceleration should change proportionately according to the equation F=ma. The graph should be linear and as the difference in the masses is increased, the acceleration should increase because of their inverse relationship. Furthermore, force and acceleration should be directly proportional. Using an Atwood machine, we were able to show that Newton's second law is legitimate. The results from the experiment supported the hypothesis by showing that as force increased, and mass remained constant, acceleration also increased. The results from the graph of our experiment also supported the hypothesis. The line is a linear line, with a positive slope. This shows that as force (y) gets larger, acceleration (x) does as well, supporting Newton's second law. On the graph, there is an R^2 value of 1, which is perfect. The percent error between our graph and the collected data is 2.34%, which is very small. While experimenting, we ignored friction and the force exerted by the pulley. However, if we took these into account, our data would be even more accurate.

=Lab 2: Newton's Second Law= = =
 * Group Members:** Ariel Katz, Jillian Laub, Nicole Margulis, and Ani Papazian
 * Class:** Period 2
 * Date Completed:** November 29, 2010
 * Date Due:** November 20, 2010

Purpose:
The purpose of this lab is to determine if Newton's second law, F=ma, is a legitimate equation. By observing graphs made with DataStudio recordings of acceleration, given values of mass and calculated values of force, we can observe if the relationship between mass and acceleration and between force and acceleration is the same as Newton's law.

Hypothesis:
If the force acting on a system and the total mass of a system are compared, then the relationship between the two will be an direct relationship following the equation: F=ma. Also, if the acceleration of a mass and the total mass of the system are compared, then the relationship between the two will be an inverse relationship following Newton's second law equation: F=ma.

Materials:

 * 1) Dynamics cart
 * 2) Masses (5g, 10g, 20g)
 * 3) Track
 * 4) Photogate Timer
 * 5) Data studio program
 * 6) Super pully
 * 7) Clamp
 * 8) Mass hanger
 * 9) Metal stopping block
 * 10) Mass balance
 * 11) Level
 * 12) Base and support rod

Procedure:
Procedure 1. Set up the Dynamics Cart with Mass A. make sure the string is level with the wheel 2. Change the force A. change the masses to..., while keeping m1 + m2 constant (m1 refers to mass in cart (497.5 g) and mass of cart, m2 refers to hanging mass and mass hanger (5 g)) a. m1 = 522.5 g and m2 = 5 g b. m1 = 517.5 g and m2 = 10 g  c. m1 = 512.5 g and m2 = 15 g  d. m1 = 507.5 g and m2 = 20 g  e. m1 = 502.5 g and m2 = 25 g B. collect data in data studio a. for each change in mass, do 3 trials b. to calculate acceleration, use the slope of a v-t graph C. record data in excel, and populate graph 3. Alter the mass being accelerated A. change the masses to: a. m1 = 2497.5 g m2 = 20g b. m1 =2247.5 g m2 = 20 g c.m1 =1997.5 g m2 = 20 g  d.m1 = 1747.5 g m2 = 20 g  e. m1=1497.5 g m2 = 20 g  f. m1 = 1247.5 g m2 = 20 g  g. m1 = 997.5 g m2 = 20 g  B. collect data in data studio a. for each change, do 3 trials C. record data in excel, and populate graph

media type="file" key="newton 2nd law trial video 4.mov" width="300" height="300"

Data:








CHANGING FORCE:


 * m1= 522.5 g and m2 = 5g**


 * m1= 517.5 g and m2 = 10g**


 * m1= 512.5 g and m2 = 15g**




 * m1= 507.5 g and m2 = 20g**


 * m1= 497.5 g and m2 = 30g**

CHANGING MASS 1 (changing total mass):

 * m1= 2497.5 g and m2 = 20g**


 * m1= 2297.5 g and m2 = 20g**


 * m1= 1997.5 g and m2 = 20g**


 * m1= 1747.5 g and m2 = 20g**


 * m1= 1497.5 g and m2 = 20g**
 * m1= 1247.5 g and m2 = 20g**


 * m1= 997.5 g and m2 = 20g**

Analysis Questions:
a. If linear: What is the slope of the trendline? To what actual observed value does the slope correspond? How does it compare to this actual observed value (find the percent error between the two)? Show why the slope should be equal to this quantity. What is the meaning of the y-intercept value? The slope is 0.5066, the x-coefficient. This slope should correspond to mass (m), since the y value will represent force (F), and the x value will represent acceleration (a) and F = ma. The actual total mass equals the mass of the cart and the hanging object. That value is equal to 0.5275 kilograms. The percent error is 4.13%:

--- x 100 theoretical
 * theoretical-actual|

The y-intercept value represents friction.

b. If non-linear: What is the power on the x? What should it be? What is the coefficient in front of the x? To what actual observed value does the coefficient correspond? Find the percent error between the two. Show why this value should be equal to this quantity. The power of x is -1.0379. The power on the x should be -1. This is because a=F/m. The y value is acceleration (a). The x value is mass (m) and the coefficient in front of the x represents force (F).A negative exponent will cause the mass value to be the denominator. The coefficient represents the force causing acceleration, or the mass of the hanging object (.02 kg) times gravity (9.8). The coefficient on the graph is 0.1681 kg. This value should equal about .196 (the hanging object's mass x gravity) because y represents acceleration and x represents mass and a = F/m. The percent error is: --- x 100 theoretical
 * theoretical-actual|

2. What would friction do to your acceleration? Would you need a bigger or smaller force to create the same acceleration? Was your slope too big or too small? Can friction be a source of error in this experiment? Redo the calculation of acceleration WITH friction to show its effect.

Friction would make the acceleration less. It would slow down the cart because it is pushing in an opposite direction. To create the same acceleration, you would need a bigger force which could be done by adding more weight to the mass that is hanging. Our slope was too small (.5066) compared to the actual value (.5275). Friction could be a source of error in this experiment. They y-intercept represents friction. With friction, the y intercept will be more negative. Without friction, the y intercept will be zero.

Conclusion:
The purpose of this lab was to find the relationship between mass, force, and acceleration. We came up with our hypotheses about these relationships based off of Newton’s second law which states that F=ma. From this equation, we hypothesized that force vs. acceleration and force vs. mass would have direct relationships, while mass and acceleration would be inversely related. The results from our experiment supported our hypotheses, showing the acceleration increase as the force increased and that the acceleration decreased as the mass increased. Our graphs illustrate these results and give a clear visual comparison of the relationships. Although our values weren’t accurate (acceleration and force didn’t exactly fit the equation F=ma) our results were precise. The trends in the graphs imitate the trend of actually acceleration vs. mass and force vs. acceleration graphs. Each of our graphs had extremely high r^2 values (both over 0.99), indicated that the hypothesized relationship was indeed the best fit for our data, supported Newton’s second law. Although we had percent error between theoretical values (according to the equation) and our actual data, based on the trend lines, our data was very precise with an extremely low percent error. Causes of discrepancy between our actual and the theoretical values could be simple human error due to the set up of the lab, lack of a large sample size of trials, or very slight friction. For future experiments, we would try even harder to make sure the string attached to the cart was parallel to the track and do many more trials with many different mass totals. Although the track was basically frictionless, the numbers we were dealing with were very small, so that might account for some constant difference in our results.

percent error (actual- theoretical)/(theoretical) * 100

acceleration vs. mass

picking a random mass and plugging it into our equation y=.1681x^-1.0379

total mass= 1.4 (random point)

y=0.1681(1.4)^-1.0379 y= .1185

acceleration = .1185 m/s

plugging acceleration into our second equation we can find force...

y=.5066(.1185) + .037 y= .097

force=.097

(actual)

F=ma F=(1.4)(.1185) F=.1659

(theoretical)


 * actual-theoretical|/ Theoretical *100


 * .097-.1659| / .1659 * 100

41.5% error

=Lab 1: Inertial Mass= **Group Members:** Ariel Katz, Jillian Laub, Nicole Margulis, and Ani Papazian  **Class:** Period 2  **Date Completed:** November 22, 2010  **Date Due:** November 23, 2010

Purpose:
The purpose of this lab is to create a graph representing the relationship between mass and inertial period. With this graph, we can interpolate and extrapolate mass values for unknown masses given their inertial period.

Hypothesis:
If the mass is greater, and it's period is timed using a inertial balance and stopwatch, then the period will be greater. Therefore, a mass and it's inertial period have a direct relationship.

Materials:
1. Inertial balance for measuring periods of masses 2. Known masses a. 10 g b. 50 g  c. 100 g  d. 300 g  e. 400 g  3. Stopwatch 4. Clamp 5. Rubik's cube

Procedure:
1. Set up inertial balance 2. Place 10 g object on the inertial balance 3. Measure the duration period (10 vibrations) using the stop watch a. 3 trials 4. Record in excel 5. Place 50 g object on the inertial balance 6. Measure the duration period (10 vibrations) using the stop watch a. 3 trials 7. Record in excel 8. Place 100 g object on the inertial balance 9. Measure the duration period (10 vibrations) using the stop watch a. 3 trials 10. Record in excel 11. Place 200 g object on the inertial balance 12. Measure the duration period (10 vibrations) using the stop watch a. 3 trials 13. Record in excel 14. Place 300 g object on the inertial balance 15. Measure the duration period (10 vibrations) using the stop watch a. 3 trials

17. Measure the duration period (10 vibrations) using the stop watch a. 3 trials 18. Record in excel 19. Make graph of all of the data in excel 20. Using the previous data collected and the patterns and relationships observed, figure out the mass of an unknown object based on the duration of its period a. Figure it out by plugging in the length of the period into the equation acquired from the graph

Data: [[file:rubiks physiks.xls]]
Table:

Graph:

Determining mass of Rubik's Cube:
y = 0.001x + 0.2945 0.4040 = 0.001(mass) + 0.2945 mass = 109.50 g

Follow Up Questions:
1. Did gravitation play any part in this operation? Was the measurement process completely unrelated to the "weight" of the object? No. Gravitation is a downward force on the object and affects only the weight. The inertial balances moves side to side and only measures inertia of the object.

2. Did an increase in mass lengthen or shorten the period of motion? According to our graph, an increase in mass lengthened the period of motion.

3. How do the accelerations of different masses compare when the platform is pulled aside and released? The bigger masses went slower then the smaller masses. The acceleration was closer to zero as the masses increase.

4. Do you imagine that the period would have been different if the side arms were stiffer? Would this change lengthen or shorten the period of the motion? Yes, the period of motion would be shortened if the side arms were stiffer because the inertial balance arms wouldn't have as much eccentricity and flexibility. While the periods would be shorter, they would all be proportionately shorter - the graph would still be linear and the relationship between period and mass would still be direct.

5. Is there any relationship between inertial and gravitational mass of the object? Inertial mass is the mass of an object based on the equation F= ma. Gravitational mass is the mass of an object based on gravity. Inertial and gravitational mass of an object are in the same ratio.

6. Why do we almost always use gravitation instead of inertia as a means of measuring mass of an object? We use gravitation instead of inertia to measure the mass of an object because instead of counting the time it takes for one period on an inertial balance, it is much easier to use a scale. Also, it is more precise.

7. How would the results of this experiment be changed if you did this experiment on the moon? The results would not change, because gravity and weight don't affect an objects inertia. The proportions of the mass and inertia would still be the same.

Conclusion:
Our graph illustrated a direct linear relationship between the mass of the object and its inertial period. The larger the mass, the longer the period was. We used the equation of the line of best fit, y = 0.001x + 0.2945, to estimate the mass of a Rubik's Cube, which was unknown. After observing the period of the Rubik's cube to be 0.4040 seconds, we plugged the value into the y of the line of best fit. The r^2 value is 0.9979, which is very close to 1. when this value is multiplied by 100, it is very close to 100% m. Since the r^2 value is high, most of the data points are close to, or matching the line of best fit. Solving this equation, we solved for x, getting 109.5 g for the mass of the Rubik's cube. This is very close to the actual mass, which was given to be 101.41 g, with a percent error of only 7.39%. The reason for this much percent error is probably due to human error. Because the periods of the small masses are so short individually, we had to calculate the length of each individual period based on the length of ten periods. It was difficult to count exactly for the periods and to stop the stopwatch at the exact moment. Since all of the periods dealt with fractions of seconds, there is a large possibility that the time wasn't completely accurate. Also, our class average came to be 103.40, with a percent error of only 1.92%, showing very accurate results. From this, we can conclude that there wasn't a systematic error with any of the equipment, errors were simply human. The percent difference between our value and the class value was 5.57%. This value is close to our percent error, emphasizing the accuracy of the class average. Also, it displays the fact that our value was closer to the classes than to the actual. Therefore, error in the class value was that the class value was greater than the actual, similar to our error, because both our value and the class value were above the actual value.