25-March-2015: Centripetal Acceleration vs. Angular Speed

Lab 8: Centripetal Acceleration vs. Angular Speed

Purpose: To determine the relationship between centripetal acceleration and angular speed.

For this lab, the class used the same data. Due to budget restraints, Mr. Wolf demonstrated how the lab worked and ran the experiment. The data was then saved and all calculations were done individually by all the groups.

Materials: Rotating Disk, Accelerometer, Photogate, Scooter wheel

Apparatus: The accelerometer was taped down to the rotating disk and measured the disk's acceleration. The photogate would then help record the wheel's period (time for 1 rev).  The scooter wheel provided the force required for this lab. The scooter wheel was in contact with the disk; so as the wheel turned, the disk also turned. The speed of the scooter wheel was regulated by reducing or increasing the voltage allotted to it.

 

Procedure:
In order to determine the relationship between centripetal acceleration and angular speed, we conducted five runs with the scooter wheel at different speeds and calculated the following for each run:

• Centripetal Acceleration
• Period of Rotating Disk
• Angular Speed

The results where then plotted on an Acceleration vs Angular Speed graph to determine their relationship.

Data:

Measured Radius (Distance from Accelerometer to Center of Rotating Disk): 13.84 cm

Run 1:  In run 1, 4.4 volts was given to the scooter wheel. The rotating disk began to rotate and an acceleration was recorded.

Trial 1 Acceleration

As you can see above, the accelerometer recorded the disk's acceleration during the run and we calculated an average of the data.

• Calculated Acceleration: 1.557 m/s^2

Through use of the photogate, we were determined the rotating disk's period (time for 1 rotation).

The table above shows the times in which the rotating disk passed through the photogate. In order to calculate the period, we subtracted the time the disk last passed through the gate minus the first time it passed through the gate. We then divided the difference by the number of rotations that occurred during that time. Every two spots on the table is one rotation.

• Run 1: (16.461 sec - 1.6716 sec)/8 rotations = 1.85 sec
• T=1.85 sec for 1 rotation

We then found the angular speed using w = (2(pi))/T

• Angular Speed (w) = 3.4 rad/sec

This process was repeated for four other runs.




Run 1: 4.4 Volts

• T=1.85 sec
• w= 3.4 rad/sec
• a= 1.557 m/s^2

Run 2: 6.4 Volts

• T=1.035 sec
• w= 6.07 rad/sec
• a= 5.074 m/s^2

Run 3: 8.6 Volts

• T=.7192 sec
• w= 8.74 rad/sec
• a= 10.70 m/s^2

Run 4: 9.6 Volts

• T=.6731 sec
• w= 9.33 rad/sec
• a= 11.89 m/s^2

Run 5: 10.8 Volts

• T=.5488 sec
• w= 11.44 rad/sec
• a= 18.15 m/s^2

Plotting Results: 
In this part of the lab, we determined the relationship of centripetal acceleration and angular speed. We determined the relationship was proportional. In order to prove this assumption, we graphed        a vs w^2. The equation for the graph should look like this a = rw^2. In this case, r is the slope of the graph. If done correctly, r should match the radius measured from the accelerometer to the center of the disk. To recap, r = 13.84 cm.
We plotted the following data:

Graph Data



a=rw^2

• X Column: w^2 (angular speed)
• Y Column: a (centripetal acceleration)

Then, we took a proportional fit of the data to determine the slope.

• Slope of Graph: r= 0.1384 +/- 0.0006436 m

As you can see, the slope of the graph (r) is identical to the measured radius.
In both cases, r = 13.84 cm. This proves that the relationship between centripetal acceleration and angular speed is proportional.

Sources of Error:  The sources of error in this lab come from the precision of the instruments we used. The experiment depends upon the precision of the accelerometer and photogate. As you can see, the slope of the graph has a very small uncertainty. This uncertainty is the result of squaring the angular speed. In all, the sources of error were limited since our calculated and measured radii matched.

23-March-2015: Trajectories Lab

Lab 7: Trajectories

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

Materials: Aluminum "v-channel", steel ball, board, ring stand, clamp, paper, carbon paper

Apparatus:

For this lab, we used the apparatus shown above. The channels were set up as to form an incline; so the ball would gather speed and fall to the floor. As seen below, there is a piece of paper and carbon paper taped to the floor. This was used to record the ball's impact point.

Procedure:
In order to predict the impact point of a ball on an inclined board, we first determined the ball's initial velocity.

 

Finding Initial Velocity (Part 1)

First, we measured the distance from the floor up to the v-channel. Then, we launched the ball repeatedly from the same point on the incline. The ball then traveled along the channel and fell to the floor. We repeated this step five times.

The ball landed around the 50 cm mark each time. In order to appropriately record the horizontal distance, we hung a plumb bob from the edge of the v-channel down to the floor and set that point as our 0 cm-mark. Next, we took an average of the distances between the impact points and our 0 mark.

• Measured Vertical Distance (height): 94.2 +/- .1 cm
• Average Horizontal Distance: 50.5 +/- .3 cm

Now, we determined the initial velocity of the ball with kinematics equations.

• Equations used: Height = V0y + (1/2)at^2 , Horiz. Dist. = V0x*t

First, we determined the ball's flight time. We used Height = V0y + (1/2)at^2, and solved for t.

• Height= 94.2 cm
• V0y= 0 cm/s
• a= 981 cm/s
• t= .438 sec

Then, we determined the ball's initial velocity. We used Horiz. Dist. = V0x*t, and solved for V0x.

• Horiz. Dist. = 50.5 cm
• t = .438 sec
• V0x = 115.235 cm/s or 1.15 m/s

Now, with the ball's initial velocity, we can finally predict the impact point of a ball on an inclined board.

 

Determining Impact Point (Part 2)

 

In this portion of the lab, we used the same apparatus; but we added an inclined board to the setup and measured the angle of incline. The impact point of the ball is now on the board and not on the floor. Our job is to determine this point (d).

• Measured Angle of Incline: 49 +/- 2 degrees



 

 

First, we solved the point of impact theoretically.

 

• Equations used: Height = V0y + (1/2)at^2 , Horiz. Dist. = V0x*t

We determined the ball's flight time by using Height = V0y + (1/2)at^2, and solved for t.

• Height= -d*sin(49)
• V0y= 0 cm/s
• a= -981 cm/s
• t = (d*sin(49)/490.5)^1/2

Next, we determined the point of impact using Horiz. Dist. = V0x*t and solved for d.

• Horiz. Dist.= d*cos(49)
• V0x= 115m/s or 1.15 m/s
• t = (d*sin(49)/490.5)^1/2

The calculated point of impact (d) is 47.47 cm

 

Calculated Uncertainty:

 

 

Measurements taken from previous step:

• x = 50.5 cm, dx = .30 cm
• y = 94.2 cm, dy = .10 cm
• alpha = 49 degrees, d(alpha) = 2 degrees = .035 radians



The point of impact is dependent of three variables (x ,y ,alpha). Therefore, we took the partial derivative of each variable and multiplied it by that variable's uncertainty. We, then added the products to determine the uncertainty.

• Calculated Value of Point of Impact: 47.47 +/- 5.88 cm 

 

Now, we ran an experiment to find the value of d. The procedure was the same, as in the previous part of the lab. The only difference is that point of impact is now on the board. We dropped the ball from the same point five different times and recorded the points of impact on the board. Just as we did in the previous step, we used paper and carbon paper to pin point the point of impact. We then calculated the average.

• Measured Average of Point of Impact: 48.02 +/- .52 cm

Measured vs. Calculated Point of Impact

•  Measured Average of Point of Impact: 48.02 +/- .52 cm
• Calculated Value of Point of Impact: 47.47 +/- 5.88 cm 

Our calculated point of impact falls within our measured point of impact. However, our calculated uncertainty seems a bit large for the experiment. This could be due to sources of error, such as: measured angle of incline, measured height of v-channel and measured horizontal distance ball traveled. There can also be error, if the ball was not launched from the same location on the inclined v-channel. This can also be the reason why our uncertainty is so large. Other than that, the measured and calculated point of impact are within reason.



18-March-2015: Modeling Friction Forces

Lab 6: Modeling Friction Forces Lab

Purpose: To understand the different types of friction and to develop models that predict the coefficients of friction.

Part 1: Static Friction

Static friction describes the friction force acting between two bodies when they are not moving relative to one another.

Static Friction Force (Fs) = Coefficient of Friction (Us)* Normal Force (N)

Fs is less than or equal to Us*N

 

 

For this portion of the lab, we had to determine the coefficient of static friction between a block and the table top.

 

 

Let's say the picture above is in equilibrium. This means that normal force provided by the cup is not  enough to make the block. From this perspective, you cannot find the coefficient of static friction.

 

In order to determine the coefficient of static friction, we slowly added water to cup to the point that it would make the block move. At this point, the weight of the cup was greater than the static force, which in turn made the block move. We repeated this process four times, increasing the mass of the block each time. In doing so, we recorded the mass of the blocks and the mass of the cup with the added water for each trial.

Mass of Cup= 2.9 +/- .1g

• Trial 1: M(1 block)= 125.7 +/- .1 g , Mcup+water= 54.0 +/- .1 g
• Trial 2:  M(2 blocks)= 260.1 +/- .1 g , Mcup+water= 95.2 +/- .1 g
• Trial 3 : M(3 blocks)= 381.4 +/- .1 g , Mcup+water= 120.7 +/- .1 g
• Trial 4: M(4 blocks)= 516.4 +/- .1 g , Mcup+water= 172.8 +/- .1 g

 We then inputted these results into a graph on LoggerPro.



 

Normal Force of blocks(x) vs. Static Friction Force (y)

 

• Normal Force (x): Normal Force of Blocks = (Mass of block/1000)*9.81
• Static Friction Force (y): Weight of Cup = (Mass of cup+water/1000)*9.81

We divided the given masses by 1000 to convert grams to kilograms

 



This setup of x and y will give us the coefficient of static friction of the block. The slope of the graph equals the coefficient of static friction. In this case the slope of the graph is 0.2958.



 

Coefficient of static friction (Us) = 0.30

 

 

Part 2: Kinetic Friction

 

For this portion of the lab, we assume that kinetic friction is proportional to the normal force, and independent of the area or speed of the moving object.

Therefore: Force of kinetic friction (Ff)= Coefficient of kinetic friction (Uk)* Normal Force (N)

 

In order to find the coefficient of kinetic friction, we tied a force sensor to a block with a piece of string and dragged the block along the tabletop with constant velocity.

 







Before running the experiment, we calibrated the sensor by using a .5 kg hanging mass.

 

With the sensor calibrated, we then dragged it along the table with constant velocity and recorded the results. The force registered by the sensor is the friction force of the block. We did this step for 1,2,3, and 4 blocks.

• M1=124.9 +/- .1 g
• 
  
  M2=167.5 +/- .1 g
• 
  
  M3=149.2 +/- .1 g
• 
  
  M4=106.1 +/- .1 g

 

Measured Friction Forces



Each line represents a different trial. We then took an average of each run, to determine each block's friction force.

• Run 1 (M1): Friction Force (Ff) = .3447
• Run 2 (M1+M2): Friction Force (Ff) = .6857
• Run 3 (M1+M2+M3): Friction Force (Ff) = .9356
• Run 4 (M1+M2+M3+M4): Friction Force (Ff) = 1.198

Like in Part 1, we used the measured data to create a Friction Force vs. Normal Force graph. The slope of this graph will give the coefficient of kinetic friction for the block.

Normal Force of blocks(x) vs. Friction Force (y)

• Normal Force (x): Normal Force of Blocks = (Mass of block/1000)*9.81
• Friction Force (y): Measured Friction Force for each block (given in previous step)

Slope of graph is .2247+/- .006835. Therefore Uk = 0.2247+/- .006835

Part 3: Static Friction From A Sloped Surface

In this portion of the lab, we determined the coefficient of static friction for a block on a sloped surface. We placed a blocked on a horizontal surface and slowly raised one end of the surface, tilting it until the block started to slip. At this point, we recorded the angle of the surface at which the block began to slip. We then drew an FBD of the block on a sloped surface and solved for forces in the y and x-direction. From here, it was rather simple to solve for the coefficient of static friction.

 

First we weighed the block, m = 0.1249 kg. Then, we measured the angle at which the block began to slip, theta= 18 +/- 2 degrees. Finally, we drew an FBD of the mass and solved for forces in the y and x-direction. Here, we solved for the coefficient of static friction. Us = .33

Part 4: Kinetic Friction From Sliding A Block Down an Incline

In this part of the lab, we set up the block on an incline with a motion sensor. The block was released and the motion sensor then measured the block's acceleration. This helped us determine the coefficient of friction between the block and the incline.

 

 

First, we weighed the block. Then, we measured the angle of the incline.

• m= 168.9 +/- .1 g
• Theta= 21.3 +/- .1 degrees

Then we used the motion sensor to measure the block's acceleration.

• a=.8091 m/s^2

From here, we drew an FBD of the block and solved for the force in the y and x-direction.

First, we plugged in our known values and then simplified the equation. Finally, we solved for Uk.

• Coefficient of friction (Uk) = .30

Part 5: Predicting The Acceleration Of A Two-Mass System

In this portion of the lab, we used the calculated coefficient of friction from part 4 to derive an expression that would predict the acceleration of the system shown below.

As you can see the block is positioned on a flat, horizontal surface. The motion sensor is placed at one end of the track. In this scenario, the hanging mass will cause the block to accelerate and the motion sensor will record the acceleration. From here, we drew and FBD and solved for acceleration. If done right, the calculated acceleration and measured acceleration should match.

First, we weighed the block and the hanging mass.

• Mass of block = 168.9 +/- .1 g
• Hanging mass = 70 +/- .1 g

We then measured the acceleration of the system with the motion sensor.

• a = .2775 m/s^2

Next, we solved for acceleration with an FBD of the system.

First, we solved for the x and y forces of each block. We then set them equal to each other because the unknown Tensions (T) are the same. We also inputted Uk from the previous step. From here, we combined like terms and solved for acceleration.

• Calculated Acceleration = .79 m/s^2

To recall, our measured acceleration was 0.2775 m/s^2.

As you can see, our accelerations do not match. We suspect the error comes from our measured acceleration. The logic behind the calculated acceleration is sound. We calculated the acceleration three times and came up with the same result. The uncertainty in our measurements is small enough that we do not consider it as the source of error, since our accelerations are far apart.Therefore, we assume that the measured acceleration is incorrect.

11-March-2015: Air Resistance Lab

Lab 4: Modeling the fall of an object falling with air resistance

Part 1: Determining the relationship between air resistance force and speed

Purpose: To determine the relationship between air resistance force and speed

Assumption: Air resistance force on a particular object depends on the object's speed, its shape, and the material it is moving through: F= k*v^n

Data Collection: For this experiment, we used the video capture feature on LoggerPro.

Procedure:
The procedure for this lab was fairly simple. We had to capture video for 1,2,3,4, and 5 coffee filters falling from the balcony of building 13. First, we did a test run in the classroom to familiarize ourselves with capturing video and inputting the data into a position vs. time graph.  Then we headed over to the Design Technology building to complete the experiment. The indoor balcony gave us the height necessary so that the falling coffee filters would reach terminal velocity before landing on the ground. After taking video for the five trials, we headed back to the classroom to analyze the results.

Data:
Using LoggerPro, we tracked the coffee filter's position over time. The data was then inputted into a P v T graph. We then took a linear fit of the plot graph and calculated the terminal velocity. In this case the slope of the graph equals the coffee filter's terminal velocity. This step was done for 1,2,3,4, and 5 coffee filters.

 

 

P v T graph: 1 coffee filter



P v T graph: 1 coffee filter

Terminal Velocity:

• 1 filter: Slope= 0.816 m/s
• 2 filters: Slope= 1.216 m/s
• 3 filters: N/A
• 4 filters: Slope= 1.650 m/s
• 5 filters: Slope= 1.865 m/s

There was an error in the video capture of three coffee filters falling. So the data for three coffee filters will be forfeited.

Only applies at terminal velocity

Finding values for k and n:

As stated above F= k*v^n. In order to determine our missing values, we must plot F-air vs. V (terminal).

From the free-body diagram of a coffee filter, we see that F-air = mg, where g= 9.8 m/s and m= mass of coffee filter.
In this case 1 coffee filter= .000926 +/- .002 kg.

• We weighed 50 coffee filters.                                 For a total of  46.3 +/- .1 g
• Mass of 1coffee filter = 46.3g /50

Now, we plotted this data; where
x(column):Terminal velocity of coffee filters
y(column):Calculated weight of each filter= F-air

 As you can see in the graph above, we only plotted four points because the data for three coffee filters was not available.

Finally, we solved for k and n by taking a power fit of the graph.
From the graph: k= 0.01279 +/-0.0006846 , n= 2.042+/-0.09734

Part 2: Modeling the fall of an object including air resistance

Purpose: To apply the mathematical model developed in part 1 to predict the terminal velocity of the various coffee filters

From part 1, we found that F-air resistance= k*v^n, where F-air= mg.
Therefore Fnet=ma=mg-k*v^n
We then solved for acceleration, as seen below. In theory when a=0, the falling object is at terminal velocity.

From here, we used to Excel to model the fall of an object with air resistance.
First, we established six constants so that we could easily change the mass of the filters and not have to set up the columns again on a new spreadsheet.
The constants are:

• g= 9.81 m/s
• n= 2.042
• m= Mass of filter(subject to change)
• k=0.01279
• Delta t= 0.0001

Excel Setup:

• First Column (Set to 0): Time= Change in time (.0001)
• Second Column (Set to 0): Change in velocity= a1*change in time
• Third Column (Set to 0): Velocity= v1+ change in velocity
• Fourth Column (Set to 9.8): a= 9.8- (k/m)*v^n
• Fifth Column (Set to 0):  Change in x= (v2+v1)/2*Change in time
• Sixth Column (Set to 0): x= x1+change in x

If setup properly, the spreadsheet should look like this:

Numerical Data for 1 coffee filter



•  Notice how velocity is increasing and acceleration is decreasing.

Eventually, the acceleration reaches zero. At this point the filter has reached terminal velocity.

Numerical Data for 1 coffee filter

 As you can see from the spreadsheet, when a=0, the terminal velocity= 0.846 m/s. So the terminal velocity for 1 coffee filter is 0.846 m/s
This process was repeated for the various coffee filters.

Numerical Terminal Velocity:

• 1 filter: v= 0.816 m/s
• 2 filters: v= 1.19 m/s
• 3 filters: N/A
• 4 filters: v= 1.67 m/s
• 5 filters: v= 1.860 m/s

 Terminal Velocity from Part 1:

• 1 filter: v= 0.816 m/s
• 2 filters: v= 1.21 m/s
• 3 filters: N/A
• 4 filters: v= 1.650 m/s
• 5 filters: v= 1.865 m/s

 As you can see the terminal velocities from Part 1 and Part 2 are fairly similar. This proves that our model works.  The velocities are not exactly the same due to the uncertainty in the k and n values. Nonetheless, if we tinkered with the numbers long enough; the velocities would match. However, there may be some limitations to our model because our data sample is simply not big enough. There can also be a few sources of error. For example, the scaling in the videos for the falling coffee filters could be off. The video capture could also show error, if the timing was off between the person capturing video and the person dropping the coffee filter. In all there will always be room for error. For the time being, we proved that our model works and accurately predicts the terminal velocity of various coffee filters.

9-Mar-2015: Unknown Mass and Propagated Uncertainty Lab

Lab 4: Unknown Mass and Propagated Uncertainty

Purpose: To calculate the propagated error of measurements and to determine the mass of an unknown mass.

This lab had two parts. In Part 1, we calculated the propagated error in density measurements. In Part 2, we determined the mass of an unknown mass.

Part 1: Propagated Uncertainty of Density

Pre-Lab: For this portion of the lab, we learned how to use a caliper.  If you are new to them, please head on over to: http://sites.laverne.edu/physics/courses/general-physics-labs/measurement-lab/
It is important to learn this skill, in order to accurately record data for this lab.

 

Materials: -Scale      - Vernier Caliper       -Three Metal Cylinders (Steel, Aluminium, Copper)

In this portion of the lab, we calculated the density of each cylinder and the propagated error in each of our measurements and ultimately found the uncertainty in our calculated value of density.

Procedure: First, we used the caliper to measure each cylinder's diameter and height. Then, we weighed each cylinder to find its mass. In order to account for uncertainty, we also included a small +/- value to each of our measurements. Our results are the following:
D=Diameter, H=Height, M=Mass

• Steel: H= 5.00 +/- .01 cm, D=1.41+/- .01 cm, M=61.1 +/- 0.1 g
• Copper: H=5.12 +/- .01 cm, D=1.28 +/- .01 cm, M=57.5 +/- 0.1 g
• Aluminum: H=4.87 +/- .01 cm, D=1.45 +/- .01 cm, M=21 +/- 0.1 g

     The next step was to calculate the density of each cylinder.

For this step, we used Density= Mass/Volume, V=pi*H*(D/2)^2 (Volume of a cylinder)

We simplify the equation and get Density= 4M/(pi*H*D^2)
Now, we use this equation and calculate the density for each cylinder.

Density:

• Steel= 7.83 g/cm^3
• Copper= 8.73 g/cm^3
• Aluminum= 2.61 g/cm^3

As stated above, Density= 4M/(pi*H*D^2)
Here, we notice that density is a function of three variables (M,D,H). In order to find the uncertainty in the calculated value for density, we must add the products of the partial derivative of each variable and that variable's uncertainty.  When taking a partial derivative, you derive only one variable and treat the rest of the function as a constant.

Note: Take the absolute value of the partial derivatives.

Uncertainty in Calculate Value of Density:

• Steel: dp= .14
• Copper: dp= .17
• Aluminum: dp= .05

Final Calculated Values of Density:

• Steel: 7.83 +/- .14 g/cm^3
• Copper: 8.73 +/- .17 g/cm^3
• Aluminum: 2.61 +/- .05 g/cm^3

Calculated Values vs. Accepted Values

• Steel: 7.83 +/- .14 g/cm^3  vs.  7.75 to 8.05 g/cm^3
• Copper: 8.73 +/- .17 g/cm^3  vs.  8.94 g/cm^3
• Aluminum: 2.61 +/- .05 g/cm^3  vs.  2.7 g/cm^3

We find that the calculated values of the three cylinders are within range of the accepted values.

Part 2: Finding the mass of an unknown mass

In this portion of the lab, we will find the mass of two different unknown masses.

Materials: - Rods  -  C- clamps -    String  -     Spring Scales  -       Unknown Masses

The setups can be seen below:

Mass 8

Mass 7

 

As you can see, the masses are suspended in equilibrium. The unknown mass is held in equilibrium by two strings, each with its own spring scale. The strings are then tied to rods at a distance, resulting in a hanging mass.

 

Procedure:

In order to determine the unknown mass, we must record the angles of each string from the horizontal axis and the forces found on the spring scales. From here, we draw a FBD for each mass and only solve for forces in the y direction because in this case forces in the x direction are irrelevant, since the mass is in equilibrium. 

 

When we solve for forces in the y direction: m= [(F1*sin(theta1))+(F2*sin(theta2))]/ g

 

Mass 8:  Unknown Mass = 1.05 kg

(F1*sin(theta1)) is the y-component of F1.

(F2*sin(theta2)) is the y-component of F2.

 

Same process is done for Mass 7

Mass 7: Unknown Mass= .84 kg

(F1*sin(theta1)) is the y-component of F1.

(F2*sin(theta2)) is the y-component of F2.

Now that we have the calculated value for the unknown masses, we must find its uncertainty.

 m= [(F1*sin(theta1))+(F2*sin(theta2))]/ g

Here, we can see that m is dependent of four variables (F1,F2,theta1,theta2).
m=m(F1,F2,theta1,theta2).

Therefore, in order to find the uncertainty in the calculated value for mass, we must add the products of the partial derivative of each variable and that variable's uncertainty. 

We are essentially doing the same steps as in Part 1 of the lab.

Propagated Uncertainty

Mass 8: dm= .104
Note: d(theta1) & d(theta2) are in radians, ( 2 degrees = .035 radians)

Mass 7: dm= .099
Note: d(theta1) & d(theta2) are in radians, ( 2 degrees = .035 radians)

 Final Values:

• Mass 8: 1.05 +/- .104 kg, 10% uncertainty
• Mass 7: 0.84 +/- .099 kg, 12% uncertainty

In this lab, we learned how to use a caliper and calculated the propagated error of measurements and determined the mass of an unknown mass. In Part 1 of the lab, we measured three metal cylinders and calculated their density. Then , we found the uncertainty for the calculated values of density and compared those values to the accepted densities of steel, aluminum, and copper. In Part 2 of the lab, we determined the mass for two unknown masses by drawing a FBD and solving for m. Finally, we calculated the propagated uncertainty of the masses.

2-March-2015: Lab 3 Non-Constant Acceleration

Lab 3: Non-Constant Acceleration Problem

Purpose: To solve a non-constant acceleration problem numerically and to learn essential skills on Excel along the way.

Problem: A 5000-kg elephant on frictionless roller skates is going 25 m/s when it gets to the bottom of a hill and arrives on a level ground. At that point a rocket mounted on the elephant's back generates a constant 8000 N thrust opposite 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)= 1500kg - 20 kg/s*t.
Find how far the elephant goes before coming to rest.

Procedure : We solved this problem in two different ways.

In the first method, we gathered the given quantities in the problem and created an a(t) function. We then took the integral of a(t) to find the equation v(t). From here, we took the integral of v(t) to find an x(t) equation. In order to find how far the elephant goes before coming to rest, we set our v(t) = 0 an solved for t. Finally, we inputted the calculated value of the t into our x(t) equation and solved for x.

First Method: From the problem stated above, we gathered the following information:

• Elephant Mass = 5000kg
• Rocket Mass = m(t) = 1500kg - 20kg/s*t
• Mass of System = 5000+1500-20t = 6500kg - 20kg/s*t
• Initial Velocity (Vo) = 25 m/s

From here we create the acceleration function a(t):
a(t) = Fnet/m(t) = -8000N/ 6500kg - 20kg/s*t = -400/325 - t (m/s^2)

We then integrate the acceleration from 0 to t and then derive an equation for v(t):
v(t) = [25 - 400ln(325)] + 400ln(325 - t)

We then integrate the velocity from 0 to t and then derive an equation for x(t):
x(t) = [25-400ln(325)]t + 400[(t-325)*ln(325-t) - t + 325ln(325)]

To find how far the elephant travels before coming to rest, we set v(t)=0 and find that t = 19.69075(s)
We plug t into our x(t) equation and find that x = 248.7m
This means that the elephant traveled 248.7m before coming to rest.

The steps shown above are simply a summary and does not show each step required in order to solve the problem analytically.
For further explanation, please refer to the Lab 3 handout.

In the second method, we simply solved the problem numerically by using Excel. This method is easier and so much faster because solving the problem analytically is time consuming and creates a lot of room for error.

Second Method: We opened up a new Excel spreadsheet and set up 8 columns as seen below.

• First Column: Time
• Second Column: Acceleration
• Third Column: Average Acceleration over one time interval
• Fourth Column: Change in velocity for one time interval
• Fifth Column: Velocity
• Sixth Column: Average velocity over one time interval
• Seventh Column: Distance covered in one time interval
• Eight Column: Total distance covered at that time interval

Equations Used:

• First Column:  Time interval= .1 sec; small time interval = accuracy
• Second Column: a(t)= -400/325-t
• Third Column: a_av= a final- a initial/2
• Fourth Column: change in v= Change in a*t
• Fifth Column: V= Vo+change in v
• Sixth Column: Vav= Vf+Vo/2
• Seventh Column: Change in x = Vav*t
• Eight Column: Add change in x over each interval

In order to complete this process, the correct cell must be chosen and given an appropriate action.
For further explanation, please refer to the Lab 3 handout.

Note: All averages and change in velocity/distance occur over one time interval.

After we set up the columns with the appropriate equations, we then filled 220 rows by using Excel's fill down feature. It is essential that the equations in each column were filled properly, otherwise the data would not be accurate. If done properly the spreadsheet should look like this:

Now if you scroll down to the point where V = 0 m/s, t value and x value should be close if not the same to the values calculated in the first method.

To recap, we found that the elephant traveled 248.7m before coming rest and t=19.69075 sec.        (see first method)
In our spreadsheet, velocity is closest to zero between 19.6-19.8 sec. At these intervals, the distance traveled is 248.7m. As you can see, we found the same answer by solving the problem numerically.

Final Thoughts:

In this lab, we solved the problem in two different ways; analytically and numerically. Our results came out to be the same. We found that by solving the problem either analytically or numerically, the elephant still traveled 248.7m before coming to rest.

Choosing an appropriate time interval is essential in this lab. A large time interval can create a gap in the data and something too small would make the data difficult to analyze. In this case, we found that t=.1sec was appropriate through trial and error and also based on our analytic answer. If an analytical result is not available, you can use the numeric data and solve for the given information in the problem. In this case, I would use the numeric t-value and x-value and solve for Vo. The problem states that Vo=25m/s. So if my calculated value of Vo equaled the given Vo, I can conclude that the numeric data is accurate.

02-March-2015: Free Fall Lab-Determination of g

Lab 2 - Determining (g)

Purpose: To examine the validity of the statement:
   In the absence of all other external forces except gravity, a falling body will accelerate at 9.8 m/s^2.

Apparatus:  
The apparatus shown to the right consists of a heavy tripod
 with leveling screws, a free-fall body, spark paper, and an electromagnet. In addition, we also used a spark generator (not pictured). The electromagnet is seen at the top of the apparatus and it is used to hold the free-fall body. The spark paper is setup along the spine of the apparatus and it used to record the distance traveled by the free-fall body. The spark generator is essential because it will mark the location of the free-fall body on the paper. It will appear as a dot.
In order to fully understand how the apparatus work, please select the following link: https://www.youtube.com/watch?v=J6lqgJsA6Xk
Skip to the 4:40 mark to see the apparatus in action.




Procedure: Due to time constraints, our group did not use the apparatus. Instead a demo of the apparatus was given to the entire class and we were handed a spark paper from a previous experiment. Our job was to establish an origin on the spark paper and to record the distance from that point to the origin. The setup is seen below. The direction of fall is noticeable because the points begin to grow in distance.

The picture above shows the position of the falling mass at every 1/60th of a second.
We were able to record the distance of  20 points and then inputted the data into a spreadsheet.

 

The first column shows our time intervals, which increase at 1/60th of a second.

The second column shows the distances we measured for the falling mass.

The third column shows the change in distance over a time interval.

The fourth column shows the mid-interval time and increases at a rate of 1/120th of a second.

The fifth column shows the average velocity of the falling mass.

 

In constant, acceleration, the velocity in the middle of a time interval is the same as the average velocity for that time interval.

For example:

• Mid-Interval Speed =66

Average Velocity = Change in x/ Change in t

• V_av = 2.1-1/0.03333333-.01666667 = 66

We then used our data and created the following graphs:



 

Position vs. Time

 







Equation of line: y=477.96x^2 +41.348x+.1541
This equation strongly resembles that of  X=Xo+Vot+(1/2)at^2
Therefore, in order to find the acceleration you would simply multiply 477.96 by 2, which equals 956. The acceleration for the P vs T graph is 9.56 m/s^2 This differs from the accepted value of 9.81m/s^2

Velocity vs. Time

Equation of line: y=954x+42.529
In order to find the acceleration of our falling mass, we simply take the slope of our V vs T graph.
 From here, we can see that our V vs T graph resembles a straight line. This is appropriate because acceleration is constant during free fall. We recognize g as 9.81m/s^2 and based on our graph our value of g is 9.54m/s^2.

 

 Errors:
There are a couple errors in the lab because our calculated acceleration does not match 9.81m/s/s.
In order to calculate the percentage error we used the following formula:
[(Experimented Value-Accepted Value)/Accepted Value]x100%
Calculated Percentage Error: -2.75%

Furthermore, there was error in two recorded measurements, as seen in the v vs t graph. The ruler we used to measure the distances on the spark tape had a piece of tape that prevented us from seeing the correct measurement.

Class Data: After conducting the experiment on our own, we collected and analyzed the other groups' values of g.

The first column shows the different values of g and the calculated class average.

The second column shows the deviation of a group's value of g from the class average.

In the third column, the deviation was squared and an average was calculated.

The fourth column shows the standard deviation of the mean, which is the square root of the average from the third column.

 

The standard deviation of the mean for the class data is (20.12). This means that we are:

• 68% confident that the real value of g falls between 956 +/- 1 Dev.    Range:936 to 976
• 95% confident that the real value of g falls between 956 +/- 2 Dev.    Range:916 to 996

    An ideal value would be 980 +/- 02 with 95% confidence

The percentages were taken from Pg.7 of the Lab Packet.

We can be certain this data is correct unless our experiment had a systematic error; such as not accounting for the friction in the apparatus as the free-fall object dropped. If there were friction, it would make our calculated values smaller.

 

Final Thoughts:

 

1. As far as I can tell, I do not see any major patterns in our values of g. However, there are three values in the 960 range.
2. Our average value of 956 differs from the accepted value of 981.
3. The class' values of g range from 926 to 992, with the majority being close to 950.
4. There might be a small difference between the average value of our measurements and those of the class because the ruler we used to measure the distances on the spark tape had a piece of tape that prevented us from seeing the correct measurement. This can be classified as a random error. The apparatus can cause a systematic error because it may not have the precision required for this lab. A systematic error can also be caused by inputting the wrong data into the spreadsheet.
5.  At the end of every lab, we amass data and analyze the results. The point of this part of the lab was to learn how to analyze data properly. In order to analyze the data properly, we increased our sample size and compared our data to the accepted value of g. In doing so, we learned some essential skills on Excel. Ultimately, we learned that not all experiments will be a success but when given data; our job is to know how to interpret it properly and find its accuracy. 

 

 

 

 

 

23- Feb-2015: Deriving a power law for an inertial pendulum

The purpose for this Inertial Pendulum lab was to find a relationship between a mass and its period. In doing so, we created a model that will give an appropriate estimate of the mass of an object when its period is given or vice versa.

The picture above shows the setup that was used for this lab.



Materials: Inertial Balance, Photogate, LabPro, Masses 100g-800g



The procedure was fairly simple. As the balance would oscillate, the photogate recorded the period and the results would then show up on the computer.  After setting everything up,as seen in the picture above, our job was to record the period,T (sec) for a set of given masses. First, we recorded the oscillation of the inertial balance without any additional mass on it. Afterwards we conducted 8 trials, increasing the mass by 100 grams after each trial. In total we recorded the periods for 9 different masses. This is shown in the data table below.

In addition to the masses seen in the data table. We recorded the periods for two distinct objects. An eraser weighing 19 grams, period (.298 sec) and a calculator weighing 148 grams, period (.375 sec).
We then inputted all of this information into the computer.

Up to this point our model looked like this: T=A(m+Mtray)^n. Next we took the natural log of both sides resulting in: lnT=n*ln(m+Mtray)+lnA. From here our model can turn into y=mx+b.



 As seen below

This graph shows our data in a linear fit.

 

It will ultimately help us determine all of the unknowns from our model. Our job was to tinker with the mass of the tray until the correlation was as close as possible to 1. We succeeded in narrowing down the mass of the tray between a low of 290 grams and a high of 310 grams. The correlation in this range was .9999. Using the masses of 290g and 310g, we came up with the following two models, which will give us a range of the mass of an object.

 

Finally, we used our new models to appropriately estimate the mass of the calculator and eraser as mentioned above. To recall, the eraser has a mass of 19 grams and a period .298 sec, and the calculator has a mass of 148 grams and a period of .375 sec.

Unfortunately, we were not able to match the weight of the objects with our model. By inputting a period of .298 sec the model gave a range of 35-44 grams. With a period of .375 sec, the model gave a range of 89-99 grams. The calculations are below.

 

 

We suspect this is due to an error in the periods we measured for our masses. Due to either human or computer error, the periods for the eraser and calculator are not accurate.

 

Systematic Error: There was a systematic error in this lab. The measured periods for the different masses are not reliable because we did not verify the results with a stopwatch. Unfortunately, that part was overlooked in the lab handout. This explains why an object with a mass of 148 grams gives a result that is lower than the 100 gram used in the curve fit.

 

In all, we found a relationship between a mass and its period. In doing so, we also created a mathematical model that can estimate the mass of an object based on its period, under the right circumstances.