March 25, 2015; Lab 8: Centripetal acceleration vs. angular frequency

Purpose:

To use the theoretical relationship between centripetal acceleration vs. angular frequency (ω) to determine the radius of the rotating disk and prove that the theoretical relationship works.

Experiment: 

1.  To establish a relationship between "a" and "ω", we start from the displacement equation of an object moving in circle. After two derivatives, we obtain a centripetal acceleration equation as shown below,

-To find magnitude of a(t) from x and y coordinate, we use pythagorean theorem and get,

2.    A setup requires a disk, when an object (accelerometer) mounted on top such that the object is distance "r" away from the center of the disk. We place the object almost at the the edge of the disk (making r= the radius of the disk). By using different voltage to set the nearby wheel to rotate and in contact with the metal disk, the metal disk promptly begins to rotate afterward, at a constant acceleration. Piece of a tape hanging out of accelerometer goes through the photogate (the black c-shape). Then Loggerpro is used to record the acceleration and time interval for a certain number of rotations. 

3.   We should have the following to put into excel:

• voltage
• number or rotations
• Δt for those rotations
• recorded acceleration
• rotations converted to radians (theta)
• ω: angular speed (rad/sec)= theta/Δt

Total of 5 trials

Data Analysis:

graph of acceleration (y-axis) vs. ω^2 (x-axis)

Theoretically, a= rω^2. When we graph "a" and "ω^2", we basically get the slope (from proportional fit) as a value or disk radius "r". 

We also measure the radius of the disk and it's between 0.138m and 0.14m.

Experimental r= 0.1383m. Hence, the experimental value is in the range of the actual value.

Conclusion:

Possible error or uncertainty is the fact that acceleration can only stay relatively constant and that external force such as friction during rotation got in the way. Our proportional fit gives r^2 of 0.9997, which can mean that there's 0.03% error in our data. Nonetheless, the experimental "r" is in the range of the actual "r" and we conclude that experimental data agree with equation: a= rω^2.

March 23, 2013; Lab 5: Trajectories

Purpose:

To use projectile equations to derive an expression for point of impact distance on an incline board   along with calculating the propagated uncertainty.

Experiment:
Part 1
1. For the first part of the lab, we let a ball roll of from the top of an incline for it to have some value of velocity when leaving the table (as shown in the setup below).
2. Then we record the distance it travels by the mark it leaves on carbon paper. Several trials are used for purpose of precision.
2. We measure the height of the table and the average value of the distance the ball travels.
our data: h=.881m; Δx= 0.618m.
3. Using kinematics for projectile, we can use the above data to find initial velocity (the velocity right before the ball leaves the tip of the table) as shown below:

- Initial velocity in y coordinate is zero, hence h= (v-initial)t + (0.5)gt^2 = 0.5gt^2
- Δx=0.618m is an average from several trials.
Part 2:
1. An incline is attached to the table with angle θ with the floor such that when the ball rolls down, it hits the incline a distance "d" from the tip of the the table.
2. Since the setup on the table doesn't change, we can use initial velocity from part 1 and derive an equation that only uses θ and Vo to calculate "d".
3. Using right angle identity, as shown below:

At a measured 44.1 degree, actual "d" should be 0.587m 

3. After we find what "d" value is supposed to be, we proceed to the experiment. By attaching an incline to the end of table, keeping the 44.1 degree as we first measured, let the ball roll down and hit the board that will leave mark on the attached carbon paper, we collect several trials for "d". As shown in the picture above, our experimental value is 0.6032m.

apparatus of part 2 lab

4. Next, we calculate the propagated uncertainty.

5. Since the initial velocity is calculated, we have to replace it with the equation that has measured values in it. Hence Vo= Δx*[g/(2h)]^1/2

6. Through partial derivatives, (taking derivative of "d" with respect to x, h,θ) and these following equations are obtained,

- Notice that dθ has to be converted to radians.

7. Our assumed dθ, dx, and dy get us the above value, although the theoretical d= .587m is not in our range.

Conclusion:
 Since initial velocity is used in all parts of this lab, one of the keys to more accurate data is to keep the setup on the table unchanged. That way, we have the same initial velocity throughout the experiments. In addition, although our theoretical value is not in the range, our closest experimental "d"= .5978m, which is approximately 2% error from what "d" should be. The most reasonable argument here is that we pick our  dθ, dx, and dy to be too small. Nonetheless, we can conclude that the experimental data makes sense with the theoretical data we use projectile equation to find.

Mar 16, 2015; Lab 7: Modeling Friction Forces

Purpose:

We want to measure the experimental both the kinetic and static coefficient of frictions, make predictions on acceleration, and compare the experimental to the predicted acceleration value.

Experiment:

Part 1:Static Friction
To measure coefficient of static friction, we use several wooden blocks and a cup of water as a system. The continuation of adding water into the cup eventually accelerates the system and at that point we can measure μ-static.

The masses are also measured before each trial

The following is our data table. We treat the weight of water (m-water*g) as static friction since that is the maximum amount of force needed to outrun the coefficient of static friction. 

Since f-static is lesser than or equal to N*μ-static, graphing N and f-static gives μ-static as a slope:

μ-static=.2829; proportional fit

Part 2:Kinetic Friction

In this part, we assume a constant velocity while pulling block(s) with string attach to force sensor. The force sensor is initially calibrated to recognize the sensitivity in mass.The sensor, therefore, record the data of 4 distinct trials shown below. We are only interested in the mean of each pull. Thus, we use this value as f-kinetic to find μ-kinetic of the system.

Since f-kinetic= μ-kinetic* N, and we keep our N the same as part 1, μ-kinetic is the slope of f-kinetic and N graph.

4 means of forces are collected from here for the graph below

N values remain the same. Proportional fit.

Therefore, coefficient of static friction for the lab table= .2829, μ-kinetic= .2482

Part 3: Static friction form a sloped surface

By measuring just an angle of when the block starts to accelerate, we can find μ-static.

This block takes more than 13° incline to start sliding

We approach this problem by first drawing FBD and solve for μ-static. As it turns out, everything cancels except for the angle that's needed for calculation.

Therefore, μ-static of the metal track= 0.325

Part 4: Kinetic friction from sliding block down incline

We need a motion sensor for this part to measure the acceleration of a block. Knowing that it takes 18° to overcome static coefficient, we increase the incline to 22.3° so that the system moves. The graph below is a screenshot of our data: a= 0.3230m/sec^2 (as a slope of velocity graph).

With given mass-block, angle, and acceleration, we can calculate μ-kinetic as the following:

a= 0.3230 m/sec^2

μ-kinetic= .3745

Part 5: Predicting "a" of 2-mass system

With a heavy enough hanging mass, the system will accelerate. This part also requires a motion sensor to collect the data of acceleration. We also use μ-kinetic= .3745 from the last part.

The block moves away from the sensor as the system accelerates.

acceleration= slope of velocity graph= 1.444m/sec^2

We predict the minimum hanging mass needed to accelerate the system and it's .047kg. So we decided to use a hanging mass of .074kg. On the right side, we derive an expression for acceleration and plug in previously measured values to find acceleration.

Error= -8.67%

Conclusion:

From this lab, we've noticed that μ-static > μ-kinetic, in most cases. The coefficient of friction (static and kinetic) depends only on the surface texture, not the area of contact nor mass of the block. Possible errors such as the one in part 5 can be from the fact that μ-kinetic of the metal track is larger compared to its μ-static, which means there are propagated errors in μ-kinetic.

Mar 11th, 2015: Lab 4; Modeling Fall of an Object with Air Resistance

Purpose: 
We want to measure the terminal velocity of a falling object, and to propose a model to show a relationship between air resistance and weight of an object.

Equation: Fair resistance = kV^n

Experiment:
Part 1
_ The actual experiment took place in Bld. 13. Coffee filters are dropped from the balcony, each time, adding one more coffee filter to the total mass. LoggerPro records the video in 30 frames/sec. We had to manually change the height in the video and pinpoints as dots the position the coffee filter falls into. By doing this, LoggerPro can gather the data and plot them into a position vs. time graph, as shown:

example of x vs. time for 1 coffee filter

_ mass coffee filter= 0.000926 kg
_ The terminal velocity is obtained from the slope (m) from doing linear fit.
_ After 5 terminal velocities are collected, we calculate the Fair resistance. The assumptions here are:

• If coffee filter moves twice as fast, air resistance is twice as much: Fair resistance ~ V
• If the velocity moves as it's multiply by itself and air resistance stays the same: Fair resistance ~ V^2
• If the first 2 are to fail, we must add variable to the equation: Fair resistance = kV^n. As "k" accounts for the sensitive value in surface area and volume of filter when falling.

_ Assuming that at some point during the fall, resistance force= mg, we can label the y-axis below as so. The x-axis is composed of 5 terminal velocities previously collected.

Power law fit used here to find "k" and "n"

Part 2
_ To assure our video capture's accuracy, we also use numerical analysis on Excel.
_ The usual setups are required here as shown below:

Cells setup for 5 coffee filters (only the mass change for each setup); time increment= 1/50

_ From the third assumption (from part 1) and Newton's second law: Fnet: mg-Fair resistance = ma; a=g-[(kV^n)/m].

_ This is why "k" and "n" are needed from part 1. (as shown in the cells setup above)

_ We start with a time increment of 1/30 since that's how much frames the video captures each second. Then we change it to 1/50 just for sake of accuracy.

These two screenshots are from the same trial (mass of 5 filters)

_ Notice how acceleration goes to zero and the velocity stays at 1.88666m/sec. For part 1's value, for the same amount of filters, terminal velocity= 1.8680.  Percentage error ~1%.

Conclusion:

 The above comparison turns out to be the best among the five, and not to mention the propagated uncertainty in mass measurement: dm= +/- 0.1g, error in LoggerPro manual input, and video capture. However, our data gives values of "n" and "k" that are very close to our second assumption. All in all, the quality of lab equipments still prevent us from getting a more suitable data for our air resistance and velocity relationship.

Mar 4, 2015: Lab 6; Propagated Uncertainty in Measurements

1. Measuring the Density of Metal Cylinder

Purpose:

We want to measure the density and its uncertainty for 3 metal cylinders, given the uncertainty value of the measuring equipments.

Procedure:

 _Use caliper to measure the height and diameter of the cylinders
 _ Use scale to measure the mass

Red: copper; White: Aluminum; Rusty green: steel

 Data and Analysis:
_ with measured uncertainty of +/- 0.1g and +/- 0.1mm, we obtain the following:
                           Height                        Mass                   Width
Aluminum:        50.9mm                    17.2g                  12.8mm
Copper:             50.9mm                     56.6g                  12.8mm
Steel:                 50.9mm                     53.8g                  12.8mm  

_ ρ, density is calculated by mass/volume, where volume of cylinder is πh(D/2)^2:

Density: Aluminum= 2.626g/cm^3

              Copper= 8.641g/cm^3

              Steel= 8.214g/cm^3

      Where: dρ- uncertainty in density (the calculated value)

                 dm- uncertainty in mass measurement

                 dd- uncertainty in diameter measurement

                 dh- uncertainty in height measurement

_ By using partial differentiation, we obtain:

        dh=dd; since we use the same equipment for 2 measurements;
_ Calculations,

One example of a detailed calculation of propagated uncertainty

Aluminum: dρ= 0.06136 g/cm^3
Copper: dρ= 0.1672 g/cm^3
Steel:  dρ= 0.1593 g/cm^3

• Aluminum: 2.626+/- 0.0614 g/cm^3

• Copper: 8.641+/- 0.1672 g/cm^3

• Steel: 8.214+/- 0.1593 g/cm^3

_ Comparison to theoretical density: Aluminum: 2.70g/cm^3;  Copper: 8.96g/cm^3;
       Steel: 7.75-8.05g/cm^3.

Conclusion

Including uncertainty in measurements and final calculations is important since it guarantees, (in a worst case scenario), the true value of density is in the range of that uncertainty.

2. Determination of an unknown mass:

Purpose:
We want to use a given angle, force, and our knowledge of free-body diagram to find the unknown hanging mass and its propagated uncertainty.

Experiment:
We want to find 2 unknown masses and their uncertainty by using these lab setups:

One of the lab setups

• The red spring measures the force in unit of "N"
  
• The yellow compass measures angle in degree, where we have to convert to radians for calculation. (to covert: degree*(π/180) )

yellow compass; unknown mass of the bottle

Data and Analysis:

   Object#7: 1st Force= 6N; 2nd Force= 8N; first θ= 21°; second θ= 46°

   Object#8: 1st Force= 7.5N; second Force= 10.5N; first θ= 11°; second θ= 48°

From free-body diagram: ΣFy: F1Sinθ1 + F2Sinθ2  = mg 

Hence,  m=(F1Sinθ1 + F2Sinθ2) /g ; all angles are converted to radians

Free-body diagram, calculation setups, and data collected

From this equation we get object 7's mass= 0.807 kg; object 8's mass= 0.94 kg

To find the uncertainty, we follow the same procedure as the first half of this lab, partial differentiation:

partial differentiation

dF= +/- 0.5N from the red spring scale; dθ= +/- 2° from the yellow compass

equation for partial differentiation for object 7 is similar to ob.8

Answer: obj.8's mass= 0.94 +/- 0.0989 kg

              obj. 7's mass= 0.807 +/- 0.0947 kg 

The propagated uncertainty of mass~ 1%

Conclusions:

In this second half of lab, instead of a density as a calculated error, we have mass as a calculated error. Our dF and dθ assure the measured uncertainty that turn out to be in the range of 1% error (or an error that only affect from the second decimal place of the mass).

Mar 2, 2015: Lab 3: Non-constant Acceleration

Purpose:

To use excel in a given non-constant acceleration problem to ultimately find the values of time and position; then compare it to the results done by integration by hand (analytical).

Data:
_ Analytical calculation (using integration):

• acceleration is a function of time and can be calculated through net force/mass of system, which are both given. Through Newton's second law, a(t)= F/m= (-8000N)/(5000+1500-20t)
• We have to integrate a(t) twice in order to get position as a function of time, x(t).
• We know that area under the curve of acceleration graph is the change in velocity,

Graphs of acceleration, velocity, position as functions of time. Vo=25m/sec

•  Solution from the lab handout gives the value, t= 19.69075sec, and x= 248.7m when the elephant comes to complete stop. (v-final = 0 m/sec)

__Numerical calculation (using excel)

        

       1.  Create the following cells in excel:

• a: function of time evaluated by intervals set by "t"
• t: manipulated time 
• a_average: average of points of "a" giving interval of time
• change in V: change in "a_ave" with given time interval
• V: function of time, an integral of a(t).
• V_ave: average of points of "v" given a time interval
• delta X: change in "V_ave" with given time interval
• X: adding changes in position delta X giving a running total.

         2.  Use time increment of 1second, note the highlighted row:

time increment= 1sec

Vfinal= -0.41~ 0m/sec; x=248.63m; t= 20sec

         3.    V-final= 0.12 ~0 m/sec; x= 248.69m; t=19.6sec

time increment= 0.1sec

         4.    V-final= -0.012m/sec; x= 248.698m; t= 19.70sec

0.05 increment

Conclusions:

1. Through 3 different numerical integrations, the results came out to be very close to answers from analytical solution.

2. With analytical solution to compare to, we can either take the same time and compare the difference between position in numerical and analytical or take the same position and compare the time. Without the analytical result, we are still able to find the position and time by looking at cell E (velocity). The value should be very close to zero since the elephant stops completely and that same row should have values for the position and time.

3. From this lab, we've learnt to use excel to do a shortcut to a rather long (by hand) integration problem. When comparing the answers from both ways, excel gives very reasonable values of position and time.

Feb 25, 2015: Spark Free Fall and Experimental Uncertainty

Purpose:

 To test the theoretical statement that g= 9.80m/s^2; where the experiment has negligible air resistance and friction.

Procedure:

We sue a wooden cylinder as a free-fall body. It's hung at the top of the apparatus. Once secured, the sparker is turned on followed by turning off the electromagnet to release the freefall body. The sparker works at 60Hz, meaning it will leave a dot for every 1/60th of a second. The electromagnet is turned off so that the body could freely fall, without any magnetic interference. While free-fall happens, the sparker leaves dots on the white tape. Turn off the sparker and remove the tape from the apparatus and measure the displacement for each dot.

displacement is measured here

A white tape is placed behind  the metal wires

Data and Explanation:

In excel, the these rows are created:

• time: previous column + 1/60. Example, c2+1/60
• distance from origin: displacement
• delta x: change between each displacement
• Mid-interval time: midpoint of each time interval
• speed: midpoint of each speed interval; (delta x)/ (1/60)
• delta V: change between two speed midpoints.
• acceleration: (delta V)/(1/60)

final data

Two graphs are required for this lab:

• Position vs. Time: use polynomial fit,
• Velocity vs. Time: (or mid-interval speed vs. time) use Linear fit. The linear regression (R^2), is closed to 1.

• x(t) takes the form of 0.5mx^2 + bx+ c
• V(t)= mx+b
• a(t)=m
• Our average acceleration is the slope of velocity graph: a= 9.62m/sec^2

Questions 

1. The result of a straight line from V vs. t graph, shows that with constant acceleration, V-ave in a time interval is the same as the mid-interval speed in that same time interval.
2. From the velocity graph, we can look at the slope and that would be the average acceleration. The calculation of average of sum of (delta V)/(1/60) would give almost the same answer. We are (9.62-9.80)*100%/9.80= -1.94% off from the accepted value of g.
3. If v(t) takes the form of mx+b, then x(t)= 0.5mx^2 + bx+ c. Our value for "0.5m" = 4.815; making m= 9.63. Hence, we are (9.63-9.80)*100%/9.80= -1.94% off from the accepted value of g.

Conclusion:

_ We've made an assumption that there's no friction and the sparks are exactly 1/60th sec apart from one another.

_ From these assumptions, we see the pattern in our data as the following:

• a curve for position graph, and a straight line for velocity graph, an spread out data for acceleration graph.

_ Should we count in the factors of friction and other interference,

• acceleration graph would have had less deviation (variation) from one another.
• In our case, we choose the first 26 dots. Although this is a good sample size, more would have been better for a better standard deviation.

_Nonetheless, our uncertainty is quite small: Absolute difference= 9.63-9.80= +/- 0.17

Relative difference: (9.60-9.80)*100%/9.80= -2.04%

Errors and Uncertainty

Purpose: To find the uncertainty in value of g from the whole class using the techniques of mean and standard deviation.
Data:

mean= 9.56m/sec^2; std. dev= 0.20m/sec^2

_ We are 95% confident that the actual g-value is in (+/- 2std. deviation) range of 9.16 and 9.96m/sec^2.

_ Our collected values of g seem almost to be under the theoretical value. The standard deviation of each data from the average seem to vary every time.
_ Our average class value of g is also under the accepted value: (9.56-9.80)*100%/9.80 = -2.44%
_ Systematic error (bias) that account for difference between our value and other groups',

• small sample size to compare to (n=9)
• The friction during free fall vary each time resulting in different displacements
• Number of dots that each group decide to measure vary, hence the mean and deviation for each group are unique.

_One possible random error (noise) that account for difference is the uncertainty of the displacement measuring device.

_ Conceptually the less spread out the data, the more accurate and precise the experiment is. From this part of lab, we use the calculated means and deviations to statistically state our percent confidence in our experimental g-value compared to the actual g-value. Ideally, the data should give a 95% confidence that the actual g-value is in the range of 9.80+/- 0.02m/sec^2.