4 March 2015. Non-Constant Acceleration (how to integrate easier)

Purpose
to find an easier and faster way using numerical integration (excel) instead of doing an analytic integration in a problem with a non-constant acceleration symbolized as a function of t (time).


The Question:

A 5000 kg elephant on frictionless roller skates with V=25m/s is going to the bottom of a hill. When it reaches the bottom, a rocket in his back generates F=8000 N opposite direction of the motion.
The mass of rocket changes with time, because the fuel inside it burns at a rate of 20 kg/s. Therefore, we have: m(t) = 1500 kg - 20 kg/s x (t).
Find the distance until the elephant reach Vt=0m/s.

Using analytic Integration:

• shown in the picture below.

• Now, we just insert the number. For the elephant to come at rest, the final velocity = 0 m/s.

• We have t = 19.69075 s and X = 248.7 m.

Using numerical Integration:

• First, we have to understand the concept written in the picture below.

• Next, we compute this to excel and vary the ∆t.
• With time interval of 1 s, we got:

• With time interval of 0.1 s, we got:

• With time interval of 0.05 s, we got:

• With time interval of 0.02 s, we got:

Conclusions:

• Using numerical integration (excel) is much easier because the program itself do not care if we want to smaller the ∆t into per milliseconds. It is going to generate the result in just a fraction of a second. Meanwhile, using analytic integration is really risky. At first, we do not even know whether the equation itself is able to be integrated or not. Moreover, it takes time and generates many opportunities to make mistakes along the way.
• But, when we need to stop "smaller-ing" the ∆t? We know when to stop if the smaller value of ∆t does not generate much of a difference in the result (in this case x, the distance).
• Take a look when ∆t = 1 s. It generates t = 20 s and x = 248.628 m, when v=0 (elephant at rest).
• Then when we change ∆t to 0.1 s, we got  t = 19.7 s and x = 248.68 m, when v=0.
• Then we change the ∆t it again to 0.05 s, we got t = 19.7 s and x = 248.69 m, when v=0. At this point, the value of the result does not change much. This is the right time to stop.
• But, I did change the value of ∆t again to 0.02 s and got the exact same answer to six significant figures for the distance. This is to show that it is a little bit pointless to do the procedure one more time.
• The result that we get from doing both numerical and analytic way is more or less the same, to an extent when ∆t is minimize, with the differences (∆t=0.02 s) for the distance = 0.002 m.

25 February 2015. Determining Gravity from a Free Fall Experiment (using excel)

Purpose
to examine the statement that states: "In the absence of all other external forces except gravity, a falling body will accelerate at 9.8 m/s²", by analyzing the motion of free-fall body.

Apparatus:

1. A set of a heavy tripod base with leveling screws, a fee-fall body, a weighted clip to anchor the spark paper, and an electromagnet with power supply as shown in the picture. This device is purposely used to observe the motion of a free-fall body.
   
   
   
   
2. A spark generator: a generator of electric oscillations that utilizes the discharge of a condenser through a spark gap as the source of its alternating-current power. In this experiment, the frequency of the spark is 60 Hz. That means it will produces a dot every 1/60 s.
3. A tape and a ruler (metric units).

Procedure:

The Professor will do the step 1-7.

1. Turn the dial hooked up to the electromagnet up a bit.
2. Hang the wooden cylinder with the metal ring around it on the electromagnet.
3. Turn on the power on of the spark generator.
4. Hold down the spark button which will generate 60 Hz frequency and leave dots on the paper.
5. Turn the electromagnet off, so that the spark generator fall.
6. Turn off the power of the spark.
7. Tear off the paper strip to examine the series of dots found.

Data Analysis:

• Notice that we have series of dots on the paper corresponding to the position of the falling mass every 1/60 s.

• Then we measure the length of each dots using the metric ruler.
• In Excel, we compute the time, which is added by 1/60 s every time we counter the next dot, in comparison with the length from one dot to another. (Shown in the picture below)

• Next, we compute difference in distance (∆x), which is X-two - X-one; the mid-interval time which is t-two - t-one divided by 2; and the mid interval speed (V) which is the ∆x/(1/60).

• Then, we graph "Distance versus Time" and "Velocity versus Mid-Interval Time".

• Now, from logical thinking, we can see that from the "Velocity vs mid-interval time" that they have a linear relationship, which means that the slope (the acceleration, since accelerations is the derivative of velocity) is constant. Therefore, the instantaneous velocity in the middle of any time interval is the same as the average velocity because the average value of a straight line is always at its middle.
• We can also prove this by calculation as shown below.

• Next, from the graph "Distance vs Time" we can find our value of acceleration by differentiating twice.

• From the graph "Velocity vs mid-interval time" we can find our value of acceleration by differentiating once.

"Errors" in Data Analysis:

• As we can see from the calculation above, we can see some "errors" in comparison with the "true" value of gravitation (as shown below).

• Now, we are going to analyze the class data for the value "g", using excel to calculate the standard deviation of the mean.

• The standard deviation of mean that we got is 0.201. Now, if our data are properly distributed according to the graph below, we can conclude that:
• If we say g= 9.56 +/- 0.201, we can state that we are 68.26% sure that it is true,
• If we say g= 9.56 +/- 0.402, we can state that we are 95.44% sure that it is true, and
• If we say g= 9.56 +/- 0.603, we can state that we are 99.74% sure that it is true.

Conclusions:

• Our pattern of the values of the calculated g are basically randomly distributed around the average value of g.
• Our average value of g in comparison with the true value of g as had been calculated above is:
• 0.1879 in average absolute differences, and
• -1.914% in average relative differences (also called percent error).
• Random error by definition is caused by any factors that randomly affect measurement of the variable across the sample. It will not affect the average (since the amount of positive errors and negative errors are the same, they will cancel each other out), it just adds variability. In this experiment, the random errors could be the "guessing" part of the last decimal places of the length between dots is not really accurate (it is different between each students). 
• While systematic error is caused by any factors that systematically affect measurement of the variable across the sample. Since all equipment  used in this experiment are adequately precise and accurate, there is likely no systematic error that occurred.
• While we are trying as hard as possible to gain a perfect result with no errors, in reality there will always be errors caused by humans directly or indirectly. Nevertheless, with good apparatus, we managed to gain a fairly small deviation of 0.201 which is quite satisfying and enough. Just imagine if the sparker is less precise, with 0.1 s delay every time it sparks a dot. It would really affect the whole calculation and we will be way off from the true value of gravitation at the end. Therefore, good precise apparatus, which is also expensive, really help us in avoiding unwanted errors.

23 February 2015. Deriving a power law for an inertial pendulum.

Purpose
to find a relationship between mass and period for an inertial balance. After the relationship is modeled by an equation, individuals can find the mass of the unknown objects by measuring their period by the inertial balance device.


Apparatus:

1. C- Clamp: c-clamp is a device that is used to secure object between the flat end of the screw and the flat end of the frame.
2. Photogate: A photogate sensor is a timing device used for very precise measurements of high-speed or short-duration events. It detects an object which will move through and blocks the beam of light between the source and the detector.
3. Logger pro: A set of devices and software that can analysis data taken from an experiment.
4. Inertial balance: a device that acts and helps objects to oscillate inertially.
   

Procedure:

1. Set up the photogate, c-clamp, and the inertial balance as shown in the picture, such that when the balance is oscillating, the tape completely passes through the beam of the photogate.

     2.  Connect the photogate to the logger pro (open application pendulum timer).

     3.  Try some oscillations then record the time with a stopwatch to make sure that the application works.

     4.  Record the period without any mass in the tray, and vary it from 100g-800g (adding 100g each time).

Data analysis:

• We know from the experiment that the graph with mass in comparison to period is as follow:

• From the graph we guess that the period is related to mass by some power-law time of equation which can be denoted by:

T=A(M_Tray+M_Added)ⁿ

• From this equation, we have three "unknowns"; A, M_Tray, and n. We have to determine the values and make a perfect guess for the mass of the tray, by taking natural logarithm of each side >>     

lnT = n ln(M_Tray+M_Added) + lnA, which looks like Y = mx + c

• Therefore, we need to plot lnT vs ln(M_Tray+M_Added) which will gives us a straight line (if we guess the mass of the tray perfectly), with the slope n and y-intercept lnA. (We can adjust the value of the mass of the tray by going to the userparameter menu in LoggerPro).
• In our experiment, we find the nearest-to-1 correlation coefficient we can get is 0.9998, with the mass of the tray equals from the range 0.253 kg until 0.295 kg.

Calculation

• Here is the graph with the higher mass of the tray = 0.295 kg, with slope = 0.6650 and y-intercept = -0.3971. (correlation = 0.9998)

• Here is the graph with the lower mass of the tray = 0.253 kg, with slope = 0.6213 and y-intercept = -0.3718. (correlation = 0.9998)

• Calculation done in the picture as shown:

EXTENSION

• From the calculation above, we got our two models of equation. Therefore we can use the equation to determine the mass of the unknown objects by measuring their periods of oscillation. In our group, we use calculator and tape as the objects.
• Remember, because of the uncertainties in the mass of the tray before, we found two models of equation, meaning that the mass of the calculator and tape will not be exact, but they are going to be presented in range.

• We measured and found out that the period of oscillation of the calculator is: T=0.4742 s, and the tape T=0.382198 s by using the inertial balance device and logger pro.
• From the equation that we modeled, now we can calculate the mass, as shown in the picture below.

• As a comparison, we compare the masses found from the equation with the direct measurement of the calculator and object on the digital balance.

• We can see that the mass of calculator is off by: 0.308kg-0.297kg = 0.011 kg. The error may occur due to some external factors such as:
• the inconsistency of the force of oscillation. The differences of magnitude of force to make the inertial balance oscillate (with or without mass) will produce different results.
• rounding figures while calculating. Due to some complicated calculations, the amount of significant figures are reduced to simplify equation. This would result into a slight differences in the outcome of the final result.
• other non-calculated factors.
• While the mass of the tape is off by: 0.134kg-0.134kg = 0 kg, which means that the model of equation is accurate this time.

Conclusion

Therefore, the relationship between mass and period for an inertial balance is modeled by the range of equations of:

• For higher Mtray, T = 0.672268 (M_{Added +} .295) ^0.6650
• For lower Mtray, T = 0.6894921 (M_{Added +} .253) ^0.6213