Activity: Trajectories

Activity: Trajectories
Purpose: Using the understanding of projectile motion to predict the impact point of the ball on an inclined board.
Materials: We need aluminum "v-channel", steel ball, board, ring stand, and camp to set up the following apparatus.

 We also need paper and carbon paper so we can clearly record the collision point of the steel ball.

Procedure:
To make sure the angle of the apparatus doesn't change, and to make sure our ball has exact the same horizontal initial velocity when it take off from the table, we have to carefully tape every joint of the apparatus, and make sure we launch the steel ball at the exact same point on the track.

First of all, we need to remove the inclined board and let the ball hit the ground, so we can find out the initial horizontal velocity of the ball when it take off from the table.

Then, we will set up the inclined board with a certain angle, and use the initial horizontal velocity we found and the inclined angle of the board to predict the point the ball collide with the inclined board.

Finally, we will let the ball go with the inclined board installed on the apparatus, and compare the experimental distance it travel along the board to the one we predicted.

Measure and Prediction:
We first put the carbon paper on a sheet of paper at the point the ball will most likely hit. We launched the steel ball at the exact same point on the track, and measured and record the horizontal distance it traveled from the table in 5 trials. We took the average of the 5 trials and we got that the horizontal distance it traveled after taking off from table is 66.9cm, and we also measured the height of the table and found it was 96.5cm.

We then did the following calculation to get the initial horizontal velocity when the ball take off from the table:

The initial horizontal velocity v0 = 1.51 m/s
The angle of between the inclined board we set up and the floor is 51.6 degree.
With these two information we did the following prediction:

Our theoretical distance the ball traveled along the inclined board d = 0.95 m

Finally, we tape the carbon paper on a sheet of paper and put them on the board where the ball would most likely hit, and we again launched the ball at the exact same point on the track as we earlier did, and record and take the average of the  distance the ball traveled along the board in 5 trials. The experimental distance the ball traveled along the inclined board d = 0.91 m ± 0.1 m.
%error= (0.91-0.95)/0.95*100% = -4.2%
Conclusion:
In this activity, the final lab result has a -4.2% error compare to the theoretical prediction we made. The difference might be caused by the imprecision measurement of the height of the table and the distance the ball traveled. We also found that our table was a little inclined, which would cause the initial took off horizontal velocity to be a little faster. Finally, in the prediction we made, we didn't take air resistance into account.

Find out the relationship between air resistance and speed of a falling object

Find out the relationship between air resistance and speed of a falling object.

Introduction: We see certain relationship between air resistance of a falling object and its speed, and in this lab, we will do an experiment to try to define this relationship.

Apparatus:

 We use this camera that can connect to a computer to film the falling motion of coffee filters, and use logger pro to create v vs. t graphs of each trial.

In this lab, we use coffee filters as the falling objects because it has a small weight and large surface, which can help us see the change air resistance make to the falling motion. and we can easily change the weight of the falling object by stacking more  than one coffee filters together.




Theosis:
We expect the magnitude of the air resistance force on a particular object depends on the object's speed. We expect the equation to look like: F = kv^n
Procedure:
1. We first filmed the free fall motion of different numbers of coffee filters, and mark the position of coffee filters in each split second. Logger pro then produced a position vs. time graph with the points we marked.
2. We will get total of 5 graphs since we did 5 trial with different weight, and we will find the slope of the part of the graph that is straight because the straight part of the graph indicate the coffee filter has reached its terminal speed. The slope will indicate the terminal speed of coffee filters.
3. We will then create a resistance force vs. terminal speed graph, where the resistance force will be the same as the weight of the coffee filters.
4. We will then use the graph to find n and k in the equation: F=kv^n.

Data and Analysis:
Following is the v vs. t graph of our 5 trials

First trial with one coffee filter, which has a total weight 0.010143N, the terminal velocity where resistance force =total weight is 1.489 m/s

Second trial with two coffee filters, which has a total weight 0.020286N, the terminal velocity where                             resistance force =total weight is 1.651 m/s 

Third trial with three coffee filters, which has a total weight 0.030429N, the terminal velocity where                                     resistance force =total weight is  1.844 m/s

Fourth trial with four coffee filters, which has a total weight 0.040572N, the terminal velocity where                         resistance force =total weight is 2.146 m/s

Fifth trial with five coffee filters, which has a total weight 0.05071N, the terminal velocity where                               resistance force =total weight is 2.236 m/s

We can get this following chart from the above graphs:

Coffee Filters	Terminal Velocity(m/s)	Force(N)
1	-1.489	0.01014
2	-1.651	0.02028
3	-1.844	0.03042
4	-2.146	0.04057
5	-2.236	0.05071

The example of calculate:(20 coffee filters=20.7 grams)
1 coffee filters: F=mg=(20.7/20)/1000 *9.8=0.01014

Then we make Resistance Force vs. Velocity graph:

Then we use the fit feature in logger pro and find out that A = k = 0.003913 and B= n =3.154.

Thus, our equation for relationship between air resistance force and traveling velocity is:

F = 0.003913v^3.154

In the Resistance Force vs. Velocity graph above, we found that our second trial and fifth trial is closer to the fit curve than other points, so we decided to use Excel to model our 2nd and 5th trial, and find out how much is the % error.

To model falling motion of each trials, we make the time interval to be 1/30 s. ΔV=a*Δt, V=V previous+ΔV,a=g-(kv^n)/m => a=9.8-0.003913*v^3.154/0.02028, Δx=(V previous+V)/2 *Δt, X=X previous+Δx.

2nd trial

The chart we modeled for the falling motion of our second trial indicate that the terminal speed of the falling object should be 1.685m/s, and our experimental value is 1.651 m/s^2.

% error = (1.651-1.685)/1.685 *100%= 2.0%

The chart we modeled for the falling motion of our fifth trial indicate that the terminal speed of the falling object should be 2.253m/s, and our experimental value is 2.236 m/s^2.

% error = (2.236-2.253)/2.253 *100%=0.75%

Conclusion:

The relationship between air resistance and speed of a falling object:              Fresistance= 0.003913v^3.154

The relationship equation we found for the air resistance depend on velocity ends up to work very well. The percent error is very low when plug in theoretical number compare to the experimental value. The possible reason for the 2% and 0.75% error is that every single coffee filter has a slightly different weight, but we count every coffee filter's weight with the average of 20 of them; the way we marked the position of coffee filter in each split second of the falling motion in the video is not very precise because of the low resolution of the video and our eye can not sharply determine the real position of the coffee filter in each split second.

Non-Constant acceleration problem/ Activity

Non-Constant acceleration problem/ Activity:

Introduction:
In this activity, we are trying to solve a problem that involves an object moving with a non-constant acceleration:

Using Newton's second law, we can set up an equation for the acceleration of the elephant:

a(t)= Fnet / m(t)= -8000 N/(6500 kg-20 kg/s*t)= -400/(325-t) (m/s^2)

Although we can integrate a(t) to get functions of v(t) and x(t), the result of integration would be very complicated and very hard to solve. 

In this activity, we learned to use the software spreadsheet to solve complex problem like the one above.

We first set up the first few rows of Excel to simulate the movement of the elephant in tiny time intervals, and shows the acceleration, average acceleration, difference in velocity, velocity, and distance traveled in different columns. Initially we set the time interval to be 0.1 second, and fill the whole rows down to obtain data for about 220 rows. After doing so, the result looks like the following picture:

The velocity of the elephant when it comes to stop will be 0, and to find the distance traveled by the elephant before it stopped, we simply have look at the row at which v=0:

According to the chart we made, the elephant stopped between time interval 19.6 second and 19.7 second after the motion begins, and thus, we can conclude that the elephant traveled 248.695 meters before came to stop.

Questions:

1. Compare the results you get from doing the problem analytically and do it numerically.

   To do it numerically, we derived the function a(t) to get function of velocity v(t) and function of distance traveled x(t):

Then, we look for the value of t when v(t) = 0, and it turned out that our answer falls at t = 19.69 second, which is the same as the time we acquired from Excel.

2. How do you know when the time interval you chose for doing the integration is "small enough"? How would you tell if you didn't have the analytical result to which you could compare to you numerical result?

If the time interval is not small enough, velocity will never be very close to zero, but change from a positive number to a negative number suddenly. If the time interval is not small enough, the answer we found will not be precise enough to tell how much distance the elephant really traveled before coming to stop.

9/2/2014 Free Fall Lab: determination of the gravity on Earth.

9/2/2014 Free Fall Lab: determination of the gravity on Earth.

Introduction: Gravity is everywhere, and physics has so much to do with it. We have always been told that gravity accelerate objects at 9.8 m/s^2 when they are falling since high school. We have done an experiment in class to examine this statement.

Apparatus:

We used the free fall apparatus in class to get precise records of motion of the falling object. The falling motion was precisely recorded by a spark generator on a spark- sensitive tape. The spark was generated at 60 Hz, which left a dot on the tape for every 1/60 second of the falling motion. we eventually got series of dots that corresponded the position of the falling object for the each entire falling motion. To be more precise in this lab, we dropped the mass for many times, and each team in class analyzed one of the different tapes.









Thesis:
We assumed the value of the acceleration gravity gives an object while falling down is 9.8 m/s^2 toward the center of the earth (toward the floor in this experiment.)
We measured and record the distance from each dot on the tape to another, and put the set of value under a column named X into Microsoft Excel. In Excel, we apply the equation:  ∆X= X-Xprevious to another column ∆X. We then got the another series of value, velocity, by applying equation: V = ∆X/∆t, where ∆t was 1/60 second. Finally, we got the series of value, acceleration, by applying a=∆V/∆t in Excel, and the average of the series of acceleration would be our experimental gravitational acceleration. Hopefully, the experimental gravitational acceleration should be close to our theoretical gravitational acceleration, 9.8 m/s^2.

Data and Analysis:
Here is the table that we have analyzed in Excel in class:

After trying to graph t vs. v with all the data we have, we found that our data is off by too much from cell F19 to F23, so we decided to only use the part of data that seems to be more reasonable, which is the data selected in the above picture, to graph:
 v(cm/s) vs. t(sec) (reasonable data)

In our data, the average value of acceleration is the average value of all the data include those we decided not to use, so we used the linear fit feature to get the slope of the best fit of the graph, which will still equal the average acceleration of our falling object.

The slope of  v(cm/s) vs. t(sec) is the average acceleration because we know that derivative of velocity function is acceleration:
v'(t)=a(t)

After acquired the gravitational acceleration of our sample falling object, we compare our value with the value other groups in the class got:
Comparison of all the gravitational acceleration for standard deviation from 10 groups in the class, and ours is number 2. 

The standard deviation for 10 different values our class got is 10.74

Conclusion:
Discussing assumptions you made in doing the lab
From the actual data we found, we ignored the data from cells F19 to F23 because we assume the acceleration is a constant value 9.8 m/s^2 such that the pattern in our data is that value of v was increasing as time pasts, and data from cells F19 to F23 was jumping up and down. 
Discussing the effect of experimental uncertainty
Since the ruler we used to measure the distance traveled by the falling object has precision up to 10th decimal place of a centimeter, all the length we got have uncertainty of                + or - 0.05 cm.
Differences between expected and experimental values
All the data and calculation was collected and calculated in centimeter, so the average gravitational acceleration according to our class was actually  9.51 m/s^2. According to standard deviation theory, 68% of 10 groups got the gravitational acceleration, g = 9.51 m/s^2 + or - 0.11 m/s^2. That is, 7 groups out of 10 groups have their value falls in one standard deviation from the mean value.
According to the statement: "In the absence of all other external forces except gravity, a falling body will accelerate at 9.8 m/s^2," our experimental value has % error equals to         (9.51-9.8)/9.8*100% = -3.0 %.

8/31/2014 deriving formula for Inertial balance's SHM

Inertial balance lab

Introduction: Mass is the inertia of an object, and is an independent value that will not be affected by the strength of gravity. Inertia is the ability of an object to resist the change in an object's state of motion, so we can expect the motion of the inertial balance to be slower when the mass in the inertial balance is large. In this lab, we are going to find out how mass in the inertial balance affect the period of the simple harmonic motion of the inertial balance, and generate an function to find out the relationship between mass and period of an inertial balance.
Apparatus:

Inertial balance: which will do simple harmonic motion once we give it a start. We will put different weight on it to make the inertia of the entire system different, and observe the relationship between its total mass and period of its motion.

Photogate: This device will help us record the period of each one of complete harmonic motion of inertial balance. We will place inertial balance in between its C-clamp.

Thesis: Because mass is actually inertia of an object, and inertia is the quantitative measure of an object's ability to resist change in motion, we know there must be some kind of relationship between the mass of the object on the inertial balance and the period the inertial balance finish each one of complete simple harmonic motion.
               We assume the relationship between the mass and the period can be showed in power-law type equation:                                                 T= A(m+Mtray)^n
               In this experiment, we will have three unknown number: A, Mtray, and n. To make it easier to get those three value, we will take the natural logarithm of each side :
                                                              lnT=n*ln(m+Mtray)+lnA
               We will then be able to find out n and lnA through graphing lnT vs. ln(m+Mtray) using Logger Pro, and to get an accurate value of Mtray, we will need to plug in different parameter numbers until we get correlation coefficient that is very close to 1. Since the we can not find the exact number of Mtray, all of our unknown number can only be defined to a very small interval instead of exact number.
Data and calculation:
Here is the data we collected in class with a photogate:

If we let lnT=y; ln(m+Mtray)=x; n=a, and lnA=b,
we can look at the equation in the form of y=ax+b, which means:
the slope of the graph m=a=n;
y intercept of the graph b=lnA,
and now we can try to different number to be parameter Mtray until we find the numbers that will make correlation coefficient to be as close to 1 as possible. Once we make the correlation coefficient to be close enough to 0, we can narrow lnA and n to a very tiny interval.

In these two graphs, the correlation coefficients are both 0.9947, which is the best correlation coefficient my team and I could found.

lnT vs. ln(m+Mtray), Mtray=0.397

Parameter Mtray=0.397, and we found that b=lnA=-0.5109, and m=n=0.7138
which indicate that A=e^(lnA)= 0.6000

lnT vs. ln(m+Mtray), Mtray=0.397

Parameter Mtray=0.399, and we found that b=lnA=-0.5123, and m=n=0.7157
which indicate that A=e^(lnA)= 0.5991

Conclusion:

Conclude the number we got from both graphs, we get ab estimate equation that tells us the relationship between the mass and the period of the inertial balance: 

T=(0.5995±0.005)(m+Mtray±0.001)^(0.7148± 0.010)

With the apparatus that is available to us for this lab, this is the best equation that tells the relationship of mass and period of the inertial balance our team get. It obviously shows that when the total mass is greater, the period is also larger.