15-Mar-2017: Trajectories

Since we have already learned how a free fall object moves, we can draw the trajectory of the object. However, in our reality, the object does not always land in the flat; sometimes, it will land on the incline. Therefore, in today's lab, we did some experiences on a free fall ball with horizontal velocity to see how the trajectories work when landing in an incline.

Purpose:
To use your understanding of projectile motion to predict the impact point of a ball on an inclined board.

Set up:
1. We set the apparatus as following.

2. Then, we put the ball at the top of track, then release it, and collected the data when hit the ground. (Repeat it to make sure at the same point). 

3. Next, we set up an incline to begin our experience. (The slope is 49º ± 1º)

4. Repeat step 2 to make sure to get a similar point. 

Analyze:

In the first part we get the X = 75.5 ± 0.1 cm, H = 96.3 ± 0.1 cm. 

By functions

We can get

then, we get v ≈ 1.703 m/s.

In the incline, we got the functions

Then, we got 

D ≈ 1.038 m , and the real D = 1.055 m 

Propagated uncertainty:

Part 1:

dv ≈ ±0.00331m/s

Part 2: 

dD ≈ ±2.096 m

since the uncertainty for angle is too large, the propagated uncertainty does not really work. 

Conclusion:

Since we have trajectories of free fall objects, we can always predict where will the objects land in any situation, and in a slope and horizontal velocity is known system, we can use the function

to find out the land point. In the meanwhile, if we know the distance and the angle of slope, we can the the initial velocity; if we know the distance and the initial velocity, we can calculate the angle of slope. 

(And the propagated uncertainty does not really work when one the uncertainty is large.)

13-Mar-2017: Modeling the fall of an object falling with air resistance

Every day, most of the people need to drive to school, work, or some other places. When we drive on the freeway, and we need to accelerate, then we will step deeper on the gas pedal. However, even though we did not slide out the gas pedal, the car will not keep accelerating forever. The main reason for that is there is a friction force increase while the car accelerated. The following lab is designed to figure out what relationship is between velocity and friction force.

Purpose:

Determining the relationship between air resistance force and speed.

Set up:

By the handout from http://spiff.rit.edu/classes/phys311/workshops/w6b/drag_expt.html, we know that the relationship between air resistance and velocity is

(The k term takes into account the shape and area of the object)

To collect the data of speed and air resistance, Professor Wolf brought us to building 13, where has a large interior space. Then, professor threw 7 times coffee filters with 6 different numbers from the top. After we collect by video, we took them back to the lab to detect the data.

Analyze:

From the data we collected, we can found the there is a relationship between velocity and mass (which can convert into force).

Then, we plugged them into the Logger Pro and modeled them by mg = kv^n.
We got k = 0.004877, n = 2.245. (from Logger Pro)
(Sadly, I forgot to take a screen shot for that, so the following is done on Excel, which is slightly different)
Through the fuctions

We know it can convert into a y = kx + b function. Then, graph by Excel, we got

Next, we put the data we collected to model the first experience.

The data came out that they are matched in the end, which means this model succeeded to model the situation of coffee filters.

Similarly, we got some close answers when we plugged into other experiences.

Conclusion:

From the lab, we could know that the relationship of coffee filters between velocity and friction force is

However, in our daily life, because the different shape and size, the coefficient k would change. But in this method, we have a new way to roughly detect the relationship between friction force and velocity.

8-Mar-2017: Non-Constant acceleration

During the time we just learned motion, we know that acceleration is the change of velocity, and therefore, we can find out the velocity by the function ∆v = a • ∆t. It's easy because we can get the velocity by multiple the acceleration and time, but what if acceleration is changing from time to time? It is no longer fit this function. So how can we get the velocity by acceleration? The following lab tells a simple example to calculate the velocity and gives a way to solve them.

Purpose:

Solve a problem that its acceleration is changing all the time, and think a way to solve other acceleration changing problems.

The problem:

A 5000-kg elephant on frictionless roller skates is going 25m/s when it gets to the bottom of a hill and arrives on level ground. At that point a 1500-kg rocket mounted on the elephant's direction of motion. The mass of the rocket changes with time (due to burning the fuel at a rate of 20 kg/s) so that

m(t) - 1500 kg - 20 kg/s • t

Find how far the elephant goes before coming to rest. 

Set up: 

Because during a very small time we could think acceleration's change is very small that we can ignore. Therefore the function ∆v = a • ∆t still works. Then in the same time, we can think the change velocity is also ignorable, which gives us the distance traveled. As a result, (By F = ma, ∆X = v∆t)

t              ∆a              a            ∆v              v            ∆X              X

0              0               a1             0              25             0                0

∆t           ∆a1          a1 +∆a1     a1• ∆t     25 - ∆v1     v1•∆t           ∆X1

2∆t         ∆a2          a2 +∆a2     a2• ∆t     25 - ∆v2     v2•∆t        X +∆X2

…            …             …          …              …           …               …

After we plug them into Excel, we get

Then, we found it's between 19s and 20s. But we can tell that the change within 1s can not be ignore, because it is pretty large. So we changed the ∆t to get some accurate value. 

In this two ∆t, four beginning digit number remain the same. Then we can say the ∆t is small enough, and the answer is 248.7 m.After this, we did a similar problem that is changed the value of mass (7000 kg in total), speed of burn (40 kg/s), and force (13000 N). We can still use the same way to solve it. 

 

Then, we found out the answer is 164.0 m.

Analyze:

This process is actually the same as integral, when we change ∆t into infinitely approaching to 0, we can get the function:

In first problem, because F = ma, m(t) = 1500 - 20t. a(t) = 8000/(6500 - 20t)
After we solved the integral, we could find out the time that makes v(t) = 0.
Then plug the number into X(t), we got Xf = 248.7 m, which is the same as what we got.
Then in similar way, we can get the answer of second problem.

Conclusion:
Today, we learned a way to solve the problems of changing acceleration, which is by dividing the change into many small part, and then find the sum of them. This idea is the same as integral. It means that the relationships among X, v, a can be written into  
                   
Moreover, the idea of integral can be used in many other value changing problems.

6-Mar-2017: Calculate Propagated Uncertainty

When we measure a thing, it is hard to avoid making an error so we can put ± to show the uncertainty. However, how would the uncertainty changed after we calculate the uncertainty into some new numbers? The following lab shows the way of calculating propagated uncertainty. 

Purpose: Use the original uncertainty to calculate the new uncertainty. 


From the video we watched before the class, it gives us functions to calculate propagated uncertainty.

     dA = |∂A/∂L| • dL + |∂A/∂W| • dW          OR         df = √[(∂f • dx / ∂x)^2 + (∂f • dy / ∂y)^2]

By those functions, we can find out the propagated uncertainty.

Following is a problem on using functions to solve propagated uncertainty.

Measure the density of metal cylinders. Calculate the propagated error in each of your density measurement. 

Set up:

We used some measurement to measure weights, heights, and diameters of two metal cylinders.

Cylinder 1: m = 49.7 ± 0.1 g, h = 5.07 ± 0.01 cm, D = 1.25 ± 0.01 cm
Cylinder 2: m = 28.9 ± 0.1 g, h = 3.26 ± 0.01 cm, D = 1.28 ± 0.01 cm

Analyze:

Density = Mass / Volumn , Volumn = π • h • (D / 2)^2
∂V = √[(∂V • dh / ∂h)^2 + (∂V • dD / ∂D)^2] = √[(π • D^2 • dh / 4)^2 + (2π • h • D • dD / 4)^2]
∂ρ = √[(∂ρ • dm / ∂m)^2 + (∂ρ • dV / ∂V)^2] = √[(dm / v)^2 + (m • dV / V^2)^2]

Therefore, 
Cylinder 1: V = 6.22 ± 0.10 cm^3, ρ = 7.99 ± 0.13 g/cm^3
Cylinder 2: V = 4.19 ± 0.067 cm^3, ρ = 6.90 ± 0.11 g/cm^3

Conclusion:

Today, we learned two functions to calculate propagated uncertainty. 
     dA = |∂A/∂L| • dL + |∂A/∂W| • dW          OR         df = √[(∂f • dx / ∂x)^2 + (∂f • dy / ∂y)^2]
It allows us to calculate some new uncertainty with our calculation. When we measure things in real life, it helps us to describe objects more accurately.

1-Mar-2017: determination of g and some analyzing data

When we solve problems about motion, we always regard gravity to be a constant (9.8 m/s^2, or 9.81 m/s^2 in some case) in every situation. However, the real earth is not plat or a good-looking sphere. The real earth is like following.

(From GFZ, @GFZ_Potsdam)

From the image, we know that the hight is not the same in every place, which cause different gravity (from law of universal gravitation F = G • M1 • M2 / R^2). So, what is the gravity in Walnut? 

Moreover, after we grab all datas from different groups. We found we get different values about gravity. How is that happen, and which value should we use? 

The following lab will show the answer of all those questions. 

Part I:

Purpose: determine g.

At the beginning, Professor Wolf showed how to use the sparker and get a spark paper. 

Procedure:

1. Pull a piece of paper tape between the vertical wire and the vertical post of the device. Clip it with a weight to keep the paper "tight".
2. Turn the dial hooked up to the electromagnet up a bit. 
3. Hang the wooden cylinder with the metal ring around it on the electromagnet. 
4. Turn on the power on the sparker thing. 
5. Hold down the spark button on the spark box. 
6. Turn the electromagnet off so that the thing falls. 
7. Turn off power to the sparker thing. 
8. Tear off the paper strip.

After we get the spark paper, we picked a part of dots and measured the length between them. 

Then, we entered the datas into a Excel to make a graph on it. 

Then, we found out the function of distance is y = 4.7697x2 + 0.6901x + 0.0103

                         and the function of velocity is  y = 9.629x + 0.5247

Analyzing:

For any acceleration is constant velocity, there is function v = a*t + v0

Thus, the average velocity of any time interval will be (a*t1+ v0 + a*t2 + v0)/2 = a*(t1 + t2)/2 + v0

the middle of time interval is a*(t1 + t2)/2 + v0, which is the same as the average velocity. 

For the distance function: X = (a*t^2)/2 + v0*t + X0, 4.7697 = a/2, a = 9.54 m/s^2

It is only 0.12 m/s^2 away from average value, so it is accepted. 

For the velocity function: v = a*t + v0, a = 9.629 m/s^2

There is nearly no error from the average. It is a good data. 

Therefore, we found out our acceleration of gravity is 9.629 m/s^2

Part II:

Purpose: Analyzing the datas

Analyzing:

After all the groups done measurements, professor Wolf grabbed all the datas of gravity. And we found out that none of us get the exactly same result. 

There are various reasons (including professor gave us some fake value). One of the most important reasons is we will make some uncertainties, but as long as the error is acceptable, it will be fine. 

One way to value the error is using the function ∑i|xi-xavg|/N or √[∑i(xi-xavg)^2/N], which is called the standard deviation of the mean. 

As a result, we got 10.01 (cm) for the standard deviation of the mean, and it is acceptable. 

Analyzing:

Our value of g is the closest one to the average, and I think it is pretty good. For class, most datas are close to 9.6285 m/s^2, but there are two datas which are pretty far from the average value. 

I think most datas that are close to 9.6 m/s^2 are random errors, but those two far from average are systematic errors. 

In this lab, we got the real value of g in our classroom, but more importantly, we learned that there are some random errors that are hard to avoid. The best way to solve this problem is to do the lab over and over again, which shows the importance of comparing the value with other groups. What's more, we can check whether we make some mistake in the lab by checking the value acceptable or not. 

27-Feb-2017: Verifying a power law for an inertial pendulum

When we want to detect the weight of mass, we can use a spring or a machine to measure it. However, is there any other way to find out the weight without using those "old" items? The following gives us a new way to measure the mass.

Purpose: Finding a relationship between mass and period.

In the beginning, Professor Wolf gave us a function T = A(m + Mtray)^n  ( m is the mass we added, Mtray is the weight is the tray, T is the period, A and n are unknown constant. It turned our purpose into verifying the power law.)

Set up:

1. Use a C-clamp to secure the inertial balance to the tabletop. Put a thin piece of masking tape on the end of the inertial balance.
2. Set up a photogate so the when the balance is oscillating the tape completely passes through the beam of the photogate. 
   
   
   
3. Set up the LabPro with a power adapter, USB cable, and plug adapter plugged into the DIG/SONIC1 input. 
4. Open the Logger Pro application. 
   
5. Then, start to collect the data.
   

Analyze:

1. We can easily simplify the function T = A(m + Mtray)^n by using nature log. Then, we get lnT = n•ln(m + Mtray) + lnA. 
2. So the function turns into a linear function, which is y = kx + b. 
3. Because we do not know the mass of tray, we can suppose it into different value to make the function seem to be a straight line. 

First, we tried Mtray = 220g

We get correlation = 0.9986

Second we tried Mtray = 260g

We get correlation = 0.9994

Last we tried Mtray = 320g

We get correlation = 0.9994

However, this result is a little bit different from what we get in Logger Pro.
In Logger Pro, when we use Mtray = 260g, we get correlation = 0.9999

At that time, we assumed Mtray is around 260g.
In that situation, we know lnT = 0.6246•ln(m + 260) - 4.726

Extension:

• After that, we found the smallest value and the largest value for Mtray when correlation still equal to 0.9999 by changing its value. We get 253g and 280g.
• Before the end, we used the function lnT = n•ln(m + Mtray) + lnA to measure some object's mass.
• First, we tried an iPhone, it comes out the period is 0.387s.
• By using the different values of Mtray, we get is weight between 162g and 164g.
• Then, we tried a tape. The period of it is 0.663s.
• Different values of Mtray show the mass is between 666g and 671g.

Conclusion: Today, we learned a new way to find the weight by measuring the period of shaking on the clamp. However, there are many sources may made uncertainty. Firstly, the detector may be not accurate enough. Secondly, calculation has some approximation. Thirdly, we do not really know the real weight of tray.