Kaiden,+Pontillo,+Cheng,+Schimmenti

Lab: Acceleration on an Incline
12/17/10 Due Date: 1/3/11


 * Objective:** Determine how the acceleration of an object down an incline depends on the angle of the incline.


 * Hypothesis:** As the angle of the incline increases so does the acceleration of the object. The measure of the angle and the acceleration of the object are directly proportional to one another. This is known from the equation derived in the pre-lab for the acceleration of an object down an incline:


 * Procedure:**

Materials: Wooden Block, Aluminum Track, ring stand, clamp, Meter stick, 2 Photogates, picket fence, tape, photogate stands, pulley, string, masses, mass hanger.

Part A Set-Up:

Part B Set-Up:

__Part A__
 * Data:**

EXMAPLE: 20 degree angle; runs 16,17,18 Run 17: By averaging the slopes of these three runs, we find a number to represent the acceleration of the object down the slope at a particular angle.

The Excel Table of Our Trials: The table shows how we derived all the information for our sin of angle vs acceleration graph.

Class Data:


 * = Name ||= Slope ||= y-internet ||= R2 ||= slope with y-int = 0 ||
 * = Rachel ||= 10.398 ||= -1.9849 ||= 1 ||= 3.99 ||
 * = Evan ||= 10.476 ||= -1.8539 ||= 0.9976 ||= 4.03 ||
 * = Nicole ||= 9.0393 ||= -1.9593 ||= 0.978 ||= 3.72 ||
 * = Justin ||= 8.7236 ||= -1.7436 ||= 0.9782 ||= 2.31 ||
 * = Andrew ||= 13.406 ||= -3.6733 ||= 0.9839 ||= 3.38 ||
 * = Ryan ||= 11.223 ||= -2.5289 ||= 0.9696 ||= 2.63 ||
 * = Emily ||= 10.089 ||= -2.0541 ||= 0.9987 ||= 2.23 ||
 * = Jessica ||= 9.7156 ||= -1.8744 ||= 0.982 ||= 3.68 ||
 * = Jae ||= 10.393 ||= -2.6813 ||= 0.9891 ||= 4.36 ||
 * = Deanna ||= 9.8699 ||= -2.264 ||= 0.9888 ||= 3.01 ||
 * = Chris H ||= 10.089 ||= -2.0515 ||= 0.989 ||= 4.3 ||
 * = Eric ||= 10.0376 ||= -2.3396 ||= 0.9951 ||= 4.21 ||
 * = Sam ||= 12.299 ||= -3.0632 ||= 0.9935 ||= 4.15 ||

**Graph:** Part A 

__Part A__
 * Calculations:**

Sample Calculation: θ = 20 º m = .363 kg µ = .2  a = 1.51 m/ s^2

Force Diagragm:

Equation From Graph: y = 12.299x - 3.0632 y = acceleration x = sin θ

Trial 1 (when θ = 20º): a = 1.14 1.14 = 12.299 (sin(20)) - 3.0632 1.14 = 12.299 (.3418) - 3.0632 1.14 = 1.143306

Percent Difference: Class Average Slope = 10.443 Experimental Slope =12.299

Percent Error: Theoretical g = 9.8 Experimental g = 12.299

__Part B__ Theoretical Acceleration: t = 1.5 s d = .5 m vi = 0

d = vit + 1/2 at^2 .5 = 0 + 1/2 (1.5)^2 a a = .22 m/s^2

Force Diagram:

θ = 20º m = .380 kg a = .22 m/s^2

hanging mass = mass b <span style="font-family: 'Times New Roman',serif;">block = mass a

<span style="font-family: 'Times New Roman',serif;">N - way<span style="font-family: 'Times New Roman',serif;"> = ma <span style="font-family: 'Times New Roman',serif;">N - way = 0 <span style="font-family: 'Times New Roman',serif;">N = way <span style="font-family: 'Times New Roman',serif;">N = mgcos20 <span style="font-family: 'Times New Roman',serif;">N = 9.8 (.380)cos 20 <span style="font-family: 'Times New Roman',serif;">N = 3.49 N

T - wb = ma <span style="font-family: 'Times New Roman',serif;">T - wb = .22 mb T - 9.8mb = .22 mb T = 10.02mb

T - f - wax = ma T - µN - 9.8(.380)sin20 = .380(.22) T - 3.49µ = 1.357 hanging mass # 1(based on coefficient of friction derived from graph from part a) µ = .3326

T - 3.49(.3326) = 1.357 T = 2.518 N

T = 10.02mb 2.518 = 10.02mb mb = .251 kg hanging mass # 2 (based on coefficient of kinetic friction derived from coefficient of friction lab) µ = .179

T - 3.49(.179) = 1.357 T = 1.982 N

T = 10.02mb 1.982 = 10.02mb mb = .197 kg

hanging mass # 3 (based on coefficient of kinetic friction derived from coefficient of friction lab) µ = .169

T - 3.49(.169) = 1.357 T = 1.947 N

T = 10.02mb 1.947 = 10.02mb mb = .194 kg

hanging mass # 4 (based on coefficient of friction from class average data) µ = .2693

T - 3.49(.2693) = 1.357 T = 2.297 N

T = 10.02mb 2.297 = 10.02mb mb = .229 kg

These were the possible masses we planned to test that we derived based on several different values for the coefficient of kinetic friction from our equation, the class data, and the last lab. We unfortunately did not have enough time to gather any data for this portion of the lab, but we were able to anticipate some of the results.

**Discussion Questions:**

This graph shows the direct relationship of the sine of the incline angle and the acceleration of the object. Our table shows that as the angle increases, as does its sine; therefore, increasing the angle causes greater acceleration, which intuitively makes sense. The line mapped should represent the equation . Therefore, the slope represents gravity. Ideally, the slope of our line should be 9.8. The y intercept is friction.
 * 1) //Discuss your graph. What does the slope mean? What is the meaning of the y-intercept?//

2. //If the mass of the cart were doubled, how would the results be affected?// If the mass were doubled, then the sine of the angle and gravity would be unaffected. However, the force of friction would change, causing the acceleration to decrease. Because the force of friction is directly proportional to the normal force, an increase in weight would cause an increase in friction. Even on a slant, the y-component of weight increases, which in turn, causes the normal force and friction to increase. Doubling the mass doubles the friction on an object.

3. //Consider the difference between your measured value of g and the true value of 9.80 m/s2. Could friction be the cause of the observed difference? Why or why not?// Friction is most certainly the cause of the observed difference in our measure value of g and the true value. First, as friction increases in the equation it must compensate by increasing the value of g as well. However, in our case we had friction along the side of the block as well for we allowed the block to slide down the track as it was in contact with the edge of the track. This would have added another friction component thus increasing friction in our equation even further and consequently increasing the value of g in our equation.

4. 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 do not have results for Part B. However, considering the objective of part B to have our block travel .5 m in 1.5 seconds we would want our margin of error to be small. If our margin of error was larger than 2% we would obviously not be achieving our objective, for either the time would be too long or short or the distance would be greater or shorter than half a meter. This type of error would defeat the purpose of this portion of the lab.


 * Conclusion:**

In this lab, we proved our hypothesis correct, and satisfied the purpose above. Our purpose was to understand the effect angle of incline has on the acceleration of an object. We hypothesized that angle of incline was directly proportional to the acceleration of the object down the slope. In Part A, our data showed a fairly consisted linear connection between the two variables. The slope of the line should represent the acceleration, and our slope is about 12.3**.** Because our R^2 value was very close to one, our results were precise, but not very accurate. The precision shows that there must have been a constant cause of error throughout all the trials.

Our high error could have risen from several causes. For example, our friction value (y-intercept) may not be accurate either. We accounted for the friction of the the slope on the bottom of the object, but we did not account for the friction metal on the side of the slope. The track had a risen metal wall on the side, which our object often slid against. In addition, movements of the masses on the block could have affected the acceleration. As we increased the angle, the masses had a tendency to move, even with tape holding the masses in place. Furthermore, our block often did not move in a straight line. The theoretical values require that the object be moving in the straight line with no derivations; however, this was not the case, as we so our block slide from one side to the other, and hit the metal wall as it moved down the track. Lastly, although we cleaned the track as best we could, we have no guarantee that the track is in fact, completely rid of all dust and dirt particles. To improve the accuracy of results from the lab, alterations should be made to the track. The track should eliminate the walls, because they are inconstant sources of friction in the lab (they are inconstant, because the objects tend to move from one side to another when sliding). To address the movement of weights, the lab should limit the angles of the weights or provide stronger methods of holding the additional masses on the wooden block, whether it be a cavity from the weights or stronger tapes.

The relationship between an object's acceleration and the angle of incline is relevant in real life, especially when driving. When drivers move up and down a hill, the angle of the hill affects the driver's speed and acceleration, causing the driver to apply more or apply less gas on the pedal. The equation we derived from this lab also help us understand the reason for this acceleration involves the effect of gravity and the friction between the object and road.

=**Lab: Coefficient of Friction**=
 * 12/10/10**
 * Objective:** To measure the coefficient of Static Friction between surfaces, to measure the coefficient of kinetic friction between surfaces, and to determine the relationship between the friction force and normal force.


 * Hypothesis:** The tension of the string pulling the wooden block is equal to the friction force on the block. The relationship between the friction force and normal force will be direct. The coefficient of static friction will be larger than the coefficient of kinetic friction.


 * Materials:** Force meter, USB link, wooden block, miscellaneous masses, string, aluminum track, and clamp.


 * Procedure:**
 * Part A:**

9. <span style="font-family: Times New Roman,serif;">Highlight the straight line part and click  <span style="font-family: Times New Roman,serif;">. Record the MEAN as the value for Tension at Constant Speed. <span style="font-family: Times New Roman,serif;"> 10. <span style="font-family: 'Times New Roman',serif;">Highlight the maximum point and record that value as the Maximum Tension. <span style="font-family: 'Times New Roman',serif;"> 11. Repeat twice more with the same mass.
 * 1) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Mass the wooden block.
 * 2) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Clamp the surface board to the table top.
 * 3) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Place the block on the surface and put 500-g on top of it.
 * 4) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Tie a short (15 cm) string to the block at one end, and to the force meter on the other.
 * 5) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Plug the force meter into your computer. Choose Data Studio, and “Create Experiment”. A force-time graph will automatically open.
 * 6) <span style="font-family: Times New Roman,serif; margin-bottom: 0in; page-break-after: avoid;"> Go to SETUP and check // Force – Pull Positive // and uncheck // Force – Push Positive // . Then on the graph display, click the y-axis label to change the name to // Force- Pull Positive //.
 * 7) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Leaving the string slack, press the button “ZERO” on the sensor.
 * 8) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Press START on Data Studio, and gently pull the block with the force sensor.
 * 9) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Be sure to pull with a very slow constant speed once it starts to move.
 * 10) <span style="font-family: Times New Roman,serif; font-weight: normal; margin-bottom: 0in; page-break-after: avoid;">Hold the string parallel to the board.

These are the trials performed for each mass on the wooden block. The letters represent who pulled in the string, because we performed three tests for each trial. Example of Calculation for Coefficient of Static Friction: Because there is no acceleration, the force of tension is equal to the force of friction in these trials. By measuring for the tension, we measured friction.
 * DATA:**
 * .5 kg**
 * 1 kg**
 * 1.5 kg**
 * 2 kg**
 * 2.5 kg**

For our graph's equation to correspond to the equation of friction, the x value became normal force. Normal force is equal to weight, when acceleration is zero; therefore, we solved for weight (mg), and used the magnitude for normal force, too. Then, we listed the average tension/friction forces during constant speed, and at the moment when static friction is overcome.


 * Part B:**


 * Procedure:**
 * 1) Attach a protractor to the track, using the nut-screw assembly that slides onto the track.


 * 1) Secure a 200-g mass to the block with masking tape, and place at the raised end of the track.

>
 * 1) Slowly lift the end of the track until the block just begins to slide down the aluminum surface. Record the angle at which this occurs. Have each member of your group conduct this step 2 times, recording each angle measurement, then take the average.
 * 1) Place the track on an incline by clamping it to a ring stand. Make the angle just slightly less than the angle measured in step 3. The block should NOT slide down on his own. When you nudge the block just slightly it should continue down the ramp at constant speed. Have each member of your group conduct this step 2 times, recording each angle measurement, then take the average.


 * DATA**

Calculations Coefficient of Friction a = 0 <span style="font-family: 'Times New Roman',serif; margin: 0px 0px 0in; padding: 0px;">wx = mg X sin θ <span style="font-family: 'Times New Roman',serif; margin: 0px 0px 0in; padding: 0px;">wy = mg X cos θ <span style="font-family: 'Times New Roman',serif; margin: 0px 0px 0in; padding: 0px;">N - wy = ma <span style="font-family: 'Times New Roman',serif; margin: 0px 0px 0in; padding: 0px;">f - wx = ma  µ <span style="font-family: Times New Roman,serif;"> =

<span style="font-family: 'Times New Roman',serif; margin-bottom: 0in;">f = mg X sin <span style="font-family: 'Times New Roman',serif;">θ <span style="font-family: 'Times New Roman',serif;">N = mg X cos θ µ <span style="font-family: 'Times New Roman',serif;"> = µ <span style="font-family: 'Times New Roman',serif;"> = tan θ

Static a = 0 <span style="font-family: 'Times New Roman',serif;">θ = 10.75 <span style="font-family: arial,helvetica,sans-serif;"> µ <span style="font-family: 'Times New Roman',serif;">s = tan (10.75) µ <span style="font-family: 'Times New Roman',serif;">s = .189

Sliding a = 0 <span style="font-family: 'Times New Roman',serif;">θ = 9.59 <span style="font-family: arial,helvetica,sans-serif;"> µk <span style="font-family: 'Times New Roman',serif;"> = tan (9.59) <span style="font-family: arial,helvetica,sans-serif;"> µk <span style="font-family: 'Times New Roman',serif;"> = .169


 * Discussion Questions:**

The coefficient of static friction for wood and aluminum is between .2 and .6 (this is the range for the coefficient of friction for wood and any clean metal). Our measured results do fall within that range, we found the coefficient of static friction in our experiment to be .207. [|SOURCE]
 * 1) Why does the slope of the line equal the coefficient of friction? Show this derivation. The slope of the lines in the graph are in the form y=mx+b. Because we set x the x axis to measure normal force and y axis to measure tension, the x value of the slope is normal force, and the y value is tension. The equation for friction is very similar. The equation for friction is, [[image:friceq1.gif width="41" height="22"]]. The form of the equation is very similar, with friction in place of y, and normal force in place of x. Therefore, the m value of the graph's equation must be the coefficient of friction.
 * 2) Look up the coefficient of friction between wood and aluminum. Discuss if your measured results fall within the range of theoretical values. Be sure to cite your source!
 * 1) What variables affected the magnitude of the force of friction? What variables affected the magnitude of the coefficient of friction?

Because in this lab the tension force on the rope is equal to the friction force, the amount of tension placed on the system will affect the magnitude of the friction force as well as the surface that the system is sliding against. The normal force and the friction force both affect the coefficient of friction. Because the coefficient of friction, normal force, and weight force on a system are all directly related, there will be a large normal force on the system causing the coefficient of friction to be larger as well. The force of friction is indirectly related to the coefficient of friction therefore, the more friction the system has, the smaller the coefficient of friction will be.
 * 1) How does the value of coefficient of kinetic friction compare to the value for the same material’s coefficient of static friction?

The coefficient of static friction is larger than the coefficient of kinetic friction. Static friction is the force that is attempting to accelerate or simply move the system whereas kinetic friction is the force that keeps a system in motion. It takes more force to accelerate a mass from rest than keeping a mass in motion. Because of inertia when a system is at rest, it remains at rest if no other force is added to it. The applied force has to then overcome both static friction and the inertia. When a system starts to move, it tends to remain moving at a constant speed. The applied force does not need to overcome the mass's inertia, just the friction.


 * 1) Did putting the track on an incline significantly change the coefficient of friction? Why or why not?

The coefficient of friction should dramatically change when placed on an incline. Because the friction force is now calculated by just using one of the axises, it is just one part of the entire system's weight, not the entire weight. Only using the x-axis eliminates all of the forces on the y-axis, therefore there is no tension and normal force. This would significantly change the coefficient of friction because both the friction force and the normal force would greatly affect the coefficient of friction.

The goal of the lab was to find the coefficient of friction between wood and the metal tracks. We also tried to determine the relationship between friction force and the normal force. Our hypothesis that "the tension of the string pulling the wooden block is equal to the friction force on the block" was proven correct. We are able to prove this because with the wood being pulled at constant speed acceleration equals 0. Since acceleration equals 0, the net force must equal 0. The two forces on the x axis are friction moving in the negative direction and tension moving in the positive direction. Therefore, friction must equal tension.
 * Conclusion:**

Our hypothesis that there will be a direct relationship between friction and normal force. We were able to prove this correct with a graph of tension/ friction compared to normal force. The graph thus, shows a direct relationship between friction and normal force. We were then able to prove that the slope was equal to the coefficient of friction because of this equation. Which is the equation which represents our graph. Our data also helps prove that the coefficient of static friction will be larger than the coefficient of kinetic friction. In part a, the coefficient of static was larger than the coefficient of kinetic friction also in part a .207 to .179. The same went for part b .189 to .169. This is explained by, that static friction is how much force is needed to keep an object at rest, while kinetic friction is the friction force on the moving object.

Error Analysis Percent Difference (experimental from part a and calculated from b) % difference =

Static part a = .207 part b = .189 % difference = % difference = 4.55 %

Kinetic part a = .179 part b = .169 % difference = % difference = 2.87 %

Percent Difference (experimental from graph and class average) Static experimental = .207 class average = .2114 % difference = % difference =1.05 %

Kinetic experimental = .179 class average = .1847 % difference = % difference =2.72 %

As shown above our experimental results were not far off from the class average nor were they far off from the calculated values. This shows we were certainly in the correct range given the limits that all groups experienced in the lab based on the procedure and equipment we used. This also shows we were consistent, considering our experimental and calculated values were very close. The only possible sources of error in this lab that did not include the obvious human error could have been encountered during part b. It is quite possible the block was not moving at constant speed or was forcibly nudge so it may have appeared to be moving at constant speed when it was not really. A way we tried to prevent this, was running 8 trials, two per person. If we were to get more technical with this we could use data studio to try to make sure the piece of wood was traveling at constant speed instead of using our eyes. Overall, our experiments produced good results that we can feel confident supported our hypothesis.

**Newton's Second Law Lab**

 * Objective**: To prove Newton's Second law about the relationship between force and acceleration, and acceleration and mass.

__**Materials**: cart, smart-pulley, ramp, string, 5/10 pound weights (total 30grams), data studio USB plug and wire connection to pulley__ __**Hypothesis**: Because the total mass does not change, increasing the acceleration of the cart increases the force. When we keep the force constant, the acceleration and mass increase and decrease inversely.__

__**Procedure**:__ __1. Place the ramp close to the edge of a table, with the correct (can stop the cart) end on the edge.__ __2. Tie one end of the string to the edge of the cart.__ __3. Tie the other end to the mass hanger.__ __Part A:__ __1. Place 5grams on the hanging mass and 25 grams on the cart.__ __2. With each trial, move five grams from the cart to the hanging mass until the cart has 5 grams and the hanging mass is 25 grams.__ __3. Graph v-t of each trial on data studio.__ __Part B:__ __1. Place 300 grams on the cart and 50 grams on the hanging mass.__ __2. With each trial, remove 50 grams from the cart.__ __3. Graph v-t of each trial on data studio.__ media type="file" key="SPKCprocedure.mov" width="300" height="300" media type="file" key="SPKCprocedure2.mov" width="300" height="300" __**Data**__



__**Graphs**__ __Data Studio (v vs. t):__ __Excel Graphs:__ __Constant Mass (F vs. a)__

__Constant Force (a vs. m)__

__**Analysis** **Questions**__ 1. (a) The slope of the trendline represents mass of the hanging object, .03 kg as used in our experiment. While total mass stays the same, decreasing the mass in the cart in proportion to the increasing mass in the hanging object will increase the net force of the hanging object. The y-intercept value of the trendline refers to friction. In this experiment, we performed the experiments as if they occurred on frictionless surfaces. For Run # 1 we calculated our force to be .026 N. When we plug this number into the equation of our line it should give us the acceleration of the cart we got from data studio. The equation of a line: y = mx + b y = force m = total system mass being accelerated x = acceleration b= friction

y = .03x - 4 E-16 .026 = .03x - 4E-16 x = .867 m/ s^2 Since the mass in the equation matches the total mass of the system being accelerated in the lab we know our graph and data are correct. And as shown above clearly it is because it produces the correct values for acceleration as we determined. The only difference in the equation and that affected our results was friction represented in the y-intercept, however, as shown above since the value for friction was so small its affect can be considered negligible. Our error would not technically be zero but would be very insignificant due to this.

1. (b) The power of x is -.24. It really should be -1. The equation represents a= f/m, so acceleration equals net force divided by mass. The power of x should equal -1 because acceleration and mass are inversely related. Our coefficient of x is equal to .771N which represents the force causing the acceleration. The coefficient should be equal to the hanging object's mass (.05 kg) times gravity(9.8 m/s^2). The coefficient value should be equal __.49N.__

2. For this experiment, friction would cause the acceleration to be less than it would be without friction. Friction goes against motion and causes the cart to slow down more than it would without friction. To create the same acceleration, there would need to be a greater force, because force and acceleration are directly related. If you need your force to increase, the acceleration must increase as well. In our first graph, for when mass is constant, friction was negligible clearly because our results and the equation of our line were so exact. Had friction played a role our equation would have changed, for mass in the equation would no longer equal the exact mass we used in the lab. In the first lab, we increased the mass of the hanging object, and decreased the mass of the cart, which causes friction to be more and more negligible. In the second lab, we added more and more weight onto the cart, which increases the effect of friction on the motion of the cart as the cart becomes heavier in comparison to the hanging object.

3. Our data was very precise in the first lab except for the influence of friction. In the first lab, the data collected showed almost the same different slope between each point on the force-acceleration graph. The slopes between every point were either 0.03 (the m value on the trendline) or 0.299. The second lab was not as precise. The R^2 value was lower than that in the graph of force vs. acceleration. However, with an R^2 value of 0.99, our data was still fairly consistent in maintaining the same quantitative relationship between the variables.

4. Yes, the pulley mass probably did affect our data. The pulley added 2.6 grams of total weight. The string had to turn the pulley. If the pulley was heavier than originally expected, the tension of the string becomes less effective in turning the pulley, thus causing quantitative errors in our results. However, the relationship between the variables should have remained the same.

The data collected from our experiment supported our hypothesis. Our hypothesis had the idea that because the total mass stays the same, as acceleration increases so does the force. Thus, acceleration directly relates to the force. Our data collected supports our hypothesis because our acceleration increased, so did the force. This can be seen in our acceleration vs. force graph above. For the data, we have great accuracy, which is measured by the R^2 value which is 1. The other part to our hypothesis was if we keep force constant, the acceleration and mass would increase and decrease inversely. The equation obtained from the graph of this data shows the inverse relationship. It is shown with, x to a negative number. In real applications, this data helps calculate what is needed to lift objects while using a pully system. It can be used by calculating what acceleration is needed to lift the objects. Our hypotheses prove Newton's Second Law of Motion, a rule recognizable to most people, but many times not fully understood. The law provides the equation F=ma, which when manipulated, show the relationships between the three variables that we proved in the lab. It is important for us to understand the effect motion has on all three of these variables.
 * Conclusion:**

We found a lot of error in our calculations in both experiments due to friction. Friction between the cart and the track caused the cart's acceleration to be lower. We also did not account for friction between the string and pulley. As a result, our calculated acceleration was less than expected due to the fact that we essentially ignored friction. We did not add it into our net force. Another source of error was that we were not able to keep the track and the pulley at a steady height.The track often shifted from the normal force of the stopping device pushing against the cart, and consequently the track (the track and cart gripped each other with friction). In addition we did not account for the mass of the pulley or the string. These masses add additional forces that may have interfered with the results. In other terms, we assumed a frictionless, idealized environment when we did not, in fact, perform our experiment in these conditions.
 * Error Analysis:**

=Rubiks Cube Lab=


 * Objective:** Find the mass of an object only using its inertia.
 * Materials:** Inertia balance, known masses, stopwatch, and clamp
 * Hypothesis:** The greater the mass, the less vibrations will occur in a period of time therefore the smaller the mass, the more vibrations.

1. Clamp the inertia balance to desk or table 2. To cause the weight from not moving in the balance, place a paper towel inside the balance. 3. Place a weight in balance and begin vibrations of the balance. 4. With a stopwatch, calculate the amount of vibrations occur in 10 seconds. 5. Repeat steps above with various different masses 6. After many data points have been calculated and graph has given a function equation, repeat steps 1-4 with rubiks cube 7. When the amount of vibrations has been found, plug that number into the y of the equation given. Solve for x, (that is the mass).
 * Procedure:**



> Gravitation did play a part. The more mass there was, the more gravity was affecting the object. The larger pull would make the object heavier. With more of a pull, the balance would have shorter periods, relating to the weight of the object. > An increase of mass lengthened the amount of time in one period of motion. The heavier the mass, the longer it took to make one period. > Yes, stiffer side arms would shorten the periods. They will be shortened because the arms will not bend as much as they do now because they are stiffer. > Yes, the more gravitational mass an object has, the more inertia it will have. As the mass of an object increases, its resistance to changes in state in motion (inertia) also increases.
 * Follow-Up Questions**
 * 1) Did gravitation play any part in this operation? Was this measurement process completely unrelated to the “weight” of the object?
 * 1) Did an increase in mass lengthen or shorten the period of motion?
 * 1) The greater the mass, the less vibrations occurs, causing it to have a slower acceleration and the opposite occurred for a weight that weighed less, causing it to have a faster acceleration.
 * 2) 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?
 * 1) Is there any relationship between inertial and gravitational mass of the object?

6. Why do we almost always use gravitation instead of inertia as a means of measuring mass of an object? Inertia is similar to mass, because mass is a measure of how much inertia an object has- how much it resists change in state of motion. Because all these measurements occur on earth, the effect of gravity is important to account for. Weight measures gravity and the object’s resulting attraction to the earth. Inertia does not exist purely, because gravity (and sometimes other forces such as friction and tension) acts on objects.

7. How would the results of this experiment be changed if you did this experiment o<span style="font-family: Arial,Helvetica,sans-serif;">n the moon? <span style="font-family: Arial,Helvetica,sans-serif;">The relationship between mass and inertia would remain the same. As mass increased, inertia would increase, and vice versa. The change in gravity would not affect the lateral movement of the vibrations. Therefore, the results would be the same.


 * Data**


 * Mass (g) || # of vibrations ||
 * 200 || 19 ||
 * 100 || 22 ||
 * 50 || 28 ||
 * 20 || 31 ||
 * 400 || 13 ||
 * 108.15 || 23 ||


 * Graph**

orange point = rubik's cube

find mass of rubik's cube: y = .0001 x^2 - .0965x + 32.267 23 = .0001 x^2 - .0965x + 32.267 0 = .0001 x^2 - .0965x + 9.267 x = 108.15 g
 * Calculations**

Actual Mass: 101.41 g % Error = (|experimental - theoretical| / theoretical) X 100 % Error = 108.15- 101.41/ 101.41 X 100 % Error = 6.64 %
 * Percent Error- how far off from real answer caused by limitations of equipment of techniques**

Class Average: 107.63 g % Difference = (|average - experimental| / average + experimental) X 100 % Difference = (|107.63 - 108.15| / 107.63 + 108.15) X 100 % Difference =. 24 %
 * Percent Difference- tells you if it was the lab itself that produced the error**

Upon considering our percent error an the precent difference of our result to the class average it is pretty clear that the error in this experiment lay in the limitations caused by the equipment and techniques used. Other groups would most likely have produced similar percent error because the entire class used similar equipment and techniques.


 * Conclusion**

Our hypothesis proved correct. We predicted that an increase in mass would cause less vibrations in a given amount of time (a decrease in frequency) and more time per vibration (longer periods). When we weighed the known objects, we found (and graphed) an overall trend of decrease of vibrations as the masses became larger. After identifying the equation of the line, we could predict what the weight of the rubiks cube was, based on the inertia it showed through its vibrations. A greater mass produces less vibrations. This is useful to know, because there is often much confusion between inertia and gravity, like mass and weight. Objects with greater mass have greater inertia. The heavier objects had more inertia, because they more effectively resisted a change in state; they were at rest on the tray prior to the initial push of of the tray to cause the vibrations. The tray was not able to move as rapidly as it would have with lighter object that is less able to resist the change from rest to vibratory motion.

Potential sources of error might have occurred from several reasons. Firstly, the weights may have shifted. The larger weights especially, did not fit as easily on the tray. Because the rubiks cube was so large, it often slid when the tray moved and vibrated. This change in state of motion could have contributed to inaccuracies in the pattern of vibration for the tray. If the object was shifting, it would have weighed unevenly in the tray and put inconsistent force on the tray. We did try to avoid this by adding paper towels underneath the cube to avoid moving. The problem was that Rubik's cube's width was too large to fit into the tray of the inertia balance. Furthermore, human reaction time is not perfect. When we timed the number of vibrations per ten seconds, the time would often vary by milliseconds. In addition, there is time lost during signals between the person counting the vibrations and starting the vibrations, and the person that watches the clock for the moment to stop the timer. To decrease sources of experimental error, the lab could use a different instrument that does not depend on time. This eliminates the issue of human reaction time. In addition, the tray could have been larger and offered more friction to prevent movement of the objects.