22-May-2017: Finding the moment of inertia of a uniform triangle about its center of mass

Lab 17: Finding the moment of inertia of a uniform triangle about its center of mass

22-May-2017
Purpose: Determine the moment of inertia of a right triangular plate around its center mass, upright, and on its side.

Summary:

Our set up here is identical to that of the angular acceleration lab's disk-hanging mass system, where we have a sensor at 200 counts per rotation. Thus we obtained graphs of angular position and angular velocity vs time, because with our angular velocity vs time graph, getting the upward and downward slopes of this graph  gives us our values of angular acceleration. The hanging mass exerts a torque on this system. Getting the aforementioned data would therefore let us find the inertia of the various systems.

1) Subsequently, we first record the upward, downward, and calculate the average of, the angular acceleration for the steel+aluminum disks+holder.

2) Then, we weigh, and measure the dimensions of our triangle plate, then place the triangle in an upright position with the triangle's center-mass placed directly into the holder. Then we record the upward, downward, and calculate the average of, the angular acceleration.

3) I repeat step 2, but rotate the triangle 90 degrees.

Knowing the angular acceleration of these systems lets us calculate for the inertia of each individual total systems (holder, upright triangle+holder, sideways triangle+holder), and because we can also derive the moment of inertia for the machine with just the holder, we can also derive the inertias of both triangle orientations by subtracting the inertia of the holder-system from our last two tests.

Analysis

The goal of this lab is to calculate the inertia of the triangles at upright and sideways orientations rotating about their center mass. To that effect, using the spinning disk-hanging mass system required us to find the inertia of the disks with the triangle-holder, the upright triangle+holder, and the sideways triangle+holder. Subtracting the inertia of the disks+holder from the last two cases lets us do this.

As aforementioned, our setup for finding the inertia is identical to our angular acceleration lab. In order to calculate the inertia, we measured the mass of the various parts of our apparatus, and used logger pro to find the angular acceleration, which was a matter of finding the slopes of logger pro's angular velocity vs time graphs as the hanging mass goes up and down. For our calculations, we obtain the average of these values. 

Afterwards, it's a matter of plugging in our values into our given equation of inertia, once again ignoring the friction that may be present in the disk, and in the pulley itself. This is so, for all three of my trials. These values represent the experimental part of the lab, because we want to be able to compare this to calculations for the same inertias, using derivations for the moment of inertia of a right triangle.

Angular accleration and dimensions of our triangle

Derivation of inertia for the three systems.

For our theoretical values for the moment of inertia, we use the equation for the moment of inertia of a right triangle that we calculated in the lecture videos. We first derived an expression for the center mass of the triangle, then the moment of inertia about the leftmost edge, and using the parallel axis theorem, solved for the moment of inertia about the triangle's center mass, and again, plug in our values from the experiment.

Doing that, we performed percent error calculations between the theoretical and experimental results for our triangles' center mass, and found them to be rather similar. Between our upright triangle, I got a percent error of 7%, and for the second, 4.7%, values, demonstrating a reasonable accuracy between my calculations and experiments. This is how we determined the moment of inertia of a right triangular plate around its center mass, upright, and on its side.

Calculating moment of inertia about the center mass of our triangle

Calculating our theoretical values for Icm using our derived expression.
+
Final percent error calculations

Measured Data:

Graphs:

Angular acceleration of the holder system

Angular acceleration of the holder+upright triangle system

Angular acceleration of the holder+sideways triangle system

Calculated Data:

Conclusion:

Because of the small calculated percent errors for inertia, I conclude that my values for the inertia of the right triangles were reasonably valid.
Regardless, there were uncertainties in this lab that explains the percent errors that I got.
As usual, uncertainties like the scale wouldn't affect our calculations that much -- our most important source of error carries over from our angular acceleration lab - the friction present in the disk and the pulley.

Though the air pressure reduces the friction between our disks significantly, it is still present, both between the disks, and the string-pulley. In our angular acceleration lab, we ignored this friction in calculating inertia -- and we do so again, for our experimental values of the triangle's inertia. This means a value for our angular accelerations that are inaccurate, which creates uncertainty for all of our experimental inertia calculations.

Our base assumption that we can ignore friction is what most likely lead to the modest percent errors - as 7% is not insignificant - therefore, this assumption was clearly flawed, and if we had taken it into account, no doubt our final result for the theoretical would've been much smaller, and much closer to that of our experimental results.

22-May-2017: Moment of Inertia and Frictional Torque

Lab 18: Moment of Inertia and Frictional Torque

22-May-2017
Purpose: Demonstrate and calculate the moment of inertia, and frictional torque, of a large metal disk on a central shaft.

Summary:

1) We first took measurements of the various parts of the apparatus above. We took the radius of the metal disk and the metal shaft, and recorded the mass printed on the disk, and from this we calculated the mass of the individual parts of the apparatus, which allows us to calculate the moment of inertia of the disk.

2) Using a stopwatch (when the video capture failed), we timed how long the blue tape on our disk would take to come to rest once I gave it an initial spin, and then recorded where it stopped. This would allow us to find the angular velocity, and subsequently the angular acceleration, which would let us solve for the frictional torque exerted by the shaft. We did this multiple times until we got values for angular velocity that was reasonable.

Down the slope

3) Then, we propped up a wooden board in front of the disk at an angle, and connected a dynamics cart to the shaft of our apparatus with a string.

4) Using our calculated value of frictional torque in step 2, we set up equations for the cart and the disks' torque, which lets us solve for the acceleration of the cart, and subsequently for the time it might take for the cart to travel down the ramp 1 meter. This calculation of time would be our theoretical value.

5) To get the experimental value of time for comparison, we simply let the cart roll down the ramp, and our group recorded the time it took for it to travel 1 meter down the ramp. Our experimental value of time was the average of all our measurements.

Analysis:

In order to calculate the moment of inertia for the disk, and subsequently the frictional torque, following my measurements of the disks' radius and mass, solving for its inertia was a matter of using the form of inertia for a circular disk of mass M and radius R. We took the radius of the metal disk and the metal shaft, and recorded the mass printed on the disk, and from this we calculated the mass of the individual parts of the apparatus - the large disk, and the shaft to calculate the moment of inertia of the disk.

Calculating m_1/m_3, mass of the left//right hand shaft, and m_2, mass of the disk using the total mass M printed on the disk.

Calculating the inertia of the disk using m_2.

Then, I had to somehow find the frictional torque of the disk and shaft. To do this, I needed the moment of inertia of the disk, and its angular acceleration, which would be equal to whatever torque the friction was causing. Because I just calculated inertia, I had to find an angular acceleration. To do this, I first thought about using video capture to plot the position of the blue tape over time, which in theory should give me an angular acceleration, presumably less than 1 rad/s^2, because I was not spinning the disk nowhere near that fast. However, the video capture gave me values for alpha that were ridiculously wrong (about 5.1 rad/s^2) - so I turned to using a stopwatch and calculating omega and alpha by hand.

To calculate this, I first found how long it took the disk to stop after its initial spin, how many times the blue tap crossed the top of the disk, and lastly, where the blue tape was stopped relative to the vertical. Determining how many radians the disk spun divided by our average time^2 would give us the angular acceleration, which came out to a much more reasonable 0.5874 rad/s^2.

Calculating for the frictional torque, I stated that the product of the disks' inertia and its angular acceleration is the frictional torque, because once I gave the disk a constant spin, no other force but friction would have been acting perpendicular to the disk. Using my calculation for Omega, solving for the frictional torque gave me a value of .012278 kg*m^2/s^2.

frictional torque

In the next part of the lab, in order to verify just how accurate our calculations for inertia and frictional torque have been so far, we needed to set up an equation that would allow us to predict how long it would take for the dynamic cart to roll 1 meter down the ramp.

It is important to solve for the cart's linear acceleration (which is also equal to the disks'), because we can then use kinematics to solve for time t:

I started by writing my axis in the x and y directions, and using newton's second law I came up with an expression for the dynamic cart and the tension in the string as they both moved down the ramp:

mgsin(0) - T = ma

Then writing out an expression for I*Alpha using torque:

T*r - τ = Iα

Because I want to solve for linear acceleration, 'a,' I convert Alpha to 'a' in my torque expression, and solve for 'a' by first solving for T in my second law expression, and substituting this into my torque expression gives me a value of the cart's linear acceleration to be 0.03233 m/s^2, and a short kinematics calculation gives me an expected time of 7.865 seconds on the ramp as our theoretical value.

Calculating the theoretical time

Lastly, our experimental value for time was determined with three of us timing how long it takes for the dynamics cart to roll down the ramp inclined at 47°, and finding the average of our three values (provided they weren't too different from each other, which they weren't). Our average recorded time came out to be a remarkably close 7.68 seconds, with a percent error of only 2% compared to our  theoretical value. Clearly, our calculations had been very accurate, and demonstrated that the calculations of inertia and torque simply worked.

Experimental results for time

Measured Data

Time measurements

Calculated Data:

Conclusions:

Based on the accuracy of my calculated result of time compared with the experimental result of the dynamics cart rolling down the slope, I conclude that the moment of inertia, and frictional torque, of a large metal disk on a central shaft can be calculated with a high degree of accuracy, and this does indeed work.
Regardless, there were a lot of uncertainties in this lab.
Apart from the usual, trivial uncertainties (like the scales having an uncertainty of a tenth of a gram, or the calipers a tenth of a millimeter) that would hardly affect all of our calculations, the first one to note was the mass printed on the metal disk. While we assumed and trusted what it said, the fact that we were not able to verify its accuracy leads me to assume that our subsequent calculations for each of the individual components of the apparatus and our frictional torque - also have an uncertainty as a consequence.
Next, when calculating omega, because we did not use logger pro's video capture -- we timed it instead -- our measurement of the period and the radians it travels in that time have significant uncertainties of their own. Obviously, this means that my results for the angular acceleration, and subsequently the inertia of my disk, also have an uncertainty.
Lastly, when rolling the dynamics cart down the ramp, we recognize that some energy is lost whilst the wheels make contact with the ramp, and that this would mean our recorded values of time would be inaccurate as well, because we treated the cart as if it were just a sliding mass at some angle in our calculations.
Ultimately, however, our percent error demonstrates that these were not significant enough to detract from our conclusion.

15-May-2017: Angular Acceleration

Lab 16: Angular Acceleration

15-May-2017

Idea:

1. Use a torque-hanging mass system to demonstrate the effect of changing the hanging mass, the radius of the hanging mass' torque, and changing the rotating mass. 
2. Use this data to derive an expression for the moment of inertia for each of the disks, and solve.

Summary:

The torque-hanging mass system

Part 1

1. The first thing that we did was take measurements of all of our equipment. We took the mass and diameter of the disks, the small and large torque pulley, and the mass of the hanging mass.
2. In LoggerPro, we set the sensor to record at 200 counts per rotation. Then, we obtained graphs of angular position and angular velocity vs time.
3. Afterwards, we look at the effect various changes in the system have on the angular acceleration recorded by our sensor. We do this by coiling the string around the disk, and releasing the gas valve so that the hanging mass will rise, descend, multiple times.
1. To do this, we first record the α_down, α_up, and calculate their average with a single hanging mass, with a small torque pulley, with a steel topmost disk.
2. Then we do the same with a twice the hanging mass, with a small torque pulley, with a steel topmost disk.
3. Next, we have thrice the hanging mass, with a small torque pulley, with a steel topmost disk.
4. Now, we again have a single hanging mass, but with a large torque pulley, with a steel topmost disk.
5. Then a single hanging mass, with a large torque pulley, but with an aluminum topmost disk.
6. Lastly, we have a single hanging mass, with a large torque pulley, but with an aluminum topmost disk AND a bottom steel disk.

Part 2

1. In order to find the moment of inertia of each of our disks, we utilize our data from Part 1 to derive an expression for the moment of inertia, and solve.

Analysis:

Part 1

The purpose of the first part of this lab was to study the effects of various changes in the system, and how they ultimately affected the acceleration of the system. To that end, I discovered various things.

As shown in my data table for the various experiments two images below, changing the mass, changing the radius of the pulley, changing the mass of the disks, all had a direct impact on the angular acceleration.

• Doubling the mass doubled the angular acceleration. Likewise, tripling the mass tripled the angular acceleration, by about 1.97 times and 2.92 times, respectively. The "initial" condition of the system was: 1 hanging mass, a small torque pulley, and a top steel disk.
• Increasing the radius of the torque pulley by approximately 50% (from .0131 m to .01812m) doubled the angular acceleration of the system by by approximately 1.93 times..
• Changing the top steel disk to an aluminum steel disk that was 34% as heavy nearly doubled the angular acceleration by 1.921 times.

Part 2

The purpose of the second part of this lab was to first examine how friction could complicate our deriving of an expression for the moment of inertia (although we ignore it in our actual calculations), and more importantly solve for the moment of inertia for each of our disks. Using the measurements we obtained in part 1 will allow us to solve for the moment of inertia for each of our disks--the aluminum, top steel, and bottom steel. While our measurements let us directly solve for the inertia of 

• Both top+bottom steel, 
• Top steel by itself
• The top aluminum disk 

We cannot directly solve for the inertia of the bottom disk. However, because we can find the inertia of both top and steel disks directly, we can subtract from that the inertia of the top steel in order to solve for the bottom steel disk.

Because the bottom steel disk and the top steel disk have similar masses and identical dimensions, I expected their moment of inertia to be relatively identical as well--and they seemed to be, with a final percent error of approximately 18.1%.

Measured Data:

Mass and radii of all masses in the system

Data recorded from Logger Pro

Graphs:

Expt #1

Expt #2

Expt #3

Expt #4

Expt #5

Expt #6

Calculated Data:

Calculated Data table

Deriving expression for moment of inertia

Calculating inertia of the aluminum disk, and the top steel disk.

Calculating the inertia of the total steel-disk system, and then solving for the inertia of the bottom steel disk by subtracting the inertia of the top steel disk.

Conclusion:

Based on my analysis of my data in part 1, I determined that angular acceleration does indeed change directly proportional to mass and radius, and furthermore, was able to see that such changes were reflected in the changing of various components of the disk-machine. Lastly, I was able to calculate the various inertia of the disks in the system using the data I found in part 1.

Apart from the usual minor sources of error - with our measuring scale, the imperfect string, and a slightly inconsistent power source, the biggest source of error was friction - its impact could be seen on our graphs, as our angular velocity continued to decrease over time, indicating a gradual loss of energy. This means that our values of angular acceleration read by LoggerPro is inaccurate. In addition, friction would complicate our calculations for the moment of inertia. All in all, ignoring friction in our calculation for inertia, and while recording our data, lead to a final percent error of approximately 18.1% 

03-May-2017: Ballistic Pendulum

Lab: Ballistic Pendulum

03-May-2017
Idea: Demonstrate an inelastic collision and determine the launch velocity of a ball fired from a spring loaded launcher.

Summary:

1) Having a ballistic pendulum, as shown below, we first make sure that:

1. The device is level. We make sure that the ball does not roll around when placed flat.
2. The block connected to the strings is also directly facing the gun and that when fired, the ball will enter the block.

2) After measuring the length of the string attached to the block, mass of the ball, mass of the block, we proceed to fire the gun five times and record the angle that the block-ball mass goes to, as indicated by the metal arm, and then I took the average of these angles. This will allow me to calculate the launch velocity.

3) Then, we proceeded to launch the ball horizontally, and see how far the ball would land, which would allow us to again calculate the same launch velocity, using a black carbon paper. This is so that we could verify our calculations done in step 2, and perform propagated uncertainty calculations.

Analysis:

Part 1:

Our goal in the first part of the lab was to use conservation of momentum in order to find the firing speed of the gun. The ball undergoes an inelastic collision with the block once fired, and the two swing to a certain height with a certain angle, theta. In order to calculate Vo, I had to use the equations for conservation of momentum and energy, and furthermore, know exactly how high the block-ball swings up to.

To do that, I simply said that the starting level of the gun-block was zero, and because the block was connected to a string and rose to a certain angle theta, that would allow me to find the Y component of the string when tilted at that particular final angle. It turned out to be L-Lcos(theta).

Therefore, I was able to calculate the launch velocity using my momentum and energy equations by setting up two different equations, and solving for Vo.

Part 2:

Next, the task was to verify this calculated result by putting the block to the side, and launching the gun off a tabletop. By finding the horizontal distance that the ball travels, we could use kinematics in order to determine the launch speed of the ball based on the trajectory measurements.

To do this, we recorded the distance from the gun to the table edge, and then the edge to where the ball lands, as the X distance the ball travels. We do so similarly for the Y distance, we measure the height of the table, and then the height from the tabletop to the gun. These allow us to do our kinematic equations.

With such measurements, we also went ahead and calculated our propagated uncertainty for our result, based off the measurements we recorded with our meter sticks, and the mass scale. Doing so yielded a value of Vo very close to those of part 1, which was to be expected.

Measured Data:

Data Table of Measured Data

Calculated Data:

Data Table of Calculated Data for Part 1 and 2

Calculations of Vo for Part 1

Calculations of Vo for Part 2

Propagated Uncertainty Calculations for Part 2

Conclusion:

Comparing the calculations for velocity from part 1 and part 2, I notice that the two are remarkably similar, especially once I take my propagated uncertainty into account. Doing percent error calculations gives me a difference of little more than 1%, meaning that whether I find the speed using kinematics or momentum/energy, because both methods are equally valid, I should end up with final values that are very similar--if my measurements are accurate.

In terms of uncertainty,

• In measuring the height and length the ball travels in part 2, there were so many measurements to take (distance between the table and gun, distance between gun and black carbon paper, height of table, etc) that my measurements, if they are, are most likely off due to human error. I would not be surprised if I mistakenly missed or added a centimeter here or there, which would contribute to inaccurate lengths for X and Y, and subsequently an inaccurate value for V in part 2.
• The collision with the ball and the block is not perfect, the position of the gun required some meddling so that it would at least go into the block, which of course would lead to an inaccurate angle, on top of the fact that the angle indicator itself has an uncertainty of +- 0.5 degrees. An inaccurate angle leads to a value of height too tall or too short, subsequently messing up our solving for the initial speed with the conservation of momentum and conservation of energy.

24-April-2017: Collisions in Two DImensions

Lab 15: Magnetic Potential Energy Lab

24-April-2017
Idea: Create two-dimensional collisions and verify whether momentum and energy are conserved.

Summary:

1) Using a leveled glass table, I set up my phone camera that will capture the collision of

1. Steel ball with steel ball, one is initially stationary
2. Steel ball with smaller aluminum ball, also initially stationary, at 240 fps.

2) Thus the setup looked like such:

3) I record two videos for the two cases, and export the videos to Logger Pro, where I begin to plot the position of the two balls. Then, I scale the image based on the width of the glass table, and position an xy-axis over the video in order to create my graphs, like the following.

4) Therefore this gives me four sets of data for each of the two test cases: positionof x before and after the collision for both balls, position of y before and after the collision for both balls. Because I am trying to verify whether momentum & energy are conserved, what is important here is only the components of velocity in the x-y direction before & after the collisions: these will be used to calculate the momentum and energy, and as such the slopes of the position-time graphs gives me these velocities.

5) Lastly, I produce graphs of the position x-y center mass for both systems throughout the collisions, and also the velocity x-y center mass for both systems.

Data Analysis/Discussion

1)  In order to demonstrate that momentum is conserved, I obtain the slopes of the position-time graphs, which would allow me to find my values of velocity for the balls. After collecting this data, I calculated for momentum by writing out momentum (M=mv) before and after the collision, and the results I got did indeed verify this, getting results for Momentum initial versus Momentum final that were relatively close to each other:

Steel vs Steel ball

Mx: 0 = -.0508 N*S

My: .6548 = .5367 N*S

Steel vs Aluminum Ball:

Mx: .00556 = .0059 N*S

My: -.0163 = -.01459 N*S

To verify this further, I then performed percent error calculations, which confirmed this (12-27%), exception being for the momentum in the x direction for the balls of equal mass--since the way I positioned my axis meant that the initial x-component of velocity was zero, I was left trying to compare 0 to 0.0508, which was impossible to do in the percent error calculations, so I decided to leave that one to a percent error of 51%, definitely a source of uncertainty. But overall, my calculated data pointed towards demonstrating conservation of momentum for both cases as a result.

In proving conservation of energy I had a similar process of writing out the energy of the system before and after the collision with the changing kinetic energies, but unlike momentum, energy is a scalar--thus I used the following equation to obtain the actual speed of the two masses, NOT velocities in the x and y directions.

Sample calculation of how I calculated these velocities for my energy equations.

In doing so, and calculating for the energy of the system, I observed that notable amounts of energy were lost:

1. I calculated the final and initial energies to not be close enough to one another for them to be considered equal. When the masses were equal, I got a percent error of approximately 53% for that first case. The initial energy, .42J, was much greater than the final energy, .20J, showing that a significant amount of energy was lost during the collision. 
2. This was less of a case on the second trial where the final energy, .00709J, was much more closer to the initial energy, .0095J with a percent error of 25%, showing that the loss of energy was not as significant in this case. However, it is nevertheless a relatively high degree of error.

Therefore, I could see that the total kinetic energy before the collision was not the same as total kinetic energy after the collision, and could therefore conclude that unlike momentum, in this collision, energy unfortunately was not conserved.

I obtained numerous graphs that show:

• Initial and final position of x for each of the balls for both cases
• Initial and final position of y for each of the balls for both cases
• Initial and final position of x AND y for each of the balls for both cases

2) Afterwards, I had to create graphs of the x&y centers of mass vs time for both position and velocity of the balls. Reason being, I concluded, that these graphs would show that the center mass of position and velocity would demonstrate the conservation of momentum and energy further; that the system behaves as if all of its mass were at the center mass.

To demonstrate this, I used the following equations to solve for the various center masses:

I entered these equations for xcm and ycm for both position and velocity in new calculated columns, and the resulting graphs demonstrate the original assumption: the center mass moves linearly, and its velocity of this point appears to be constant for BOTH cases.

Overall, I obtained numerous graphs that show:

• Xcm + Ycm for case 1
• Vxcm and Vycm for case 1
• Xcm + Ycm for case 2
• Vxcm and Vycm for case 2

Measured Data/Table:

Measured data of:
steel ball, aluminum ball, x-y components of velocity for case Equal Masses (table 1) and Different Masses (table 2)

Calculated Data:

Demonstrating the Conservation of Momentum in the X and Y directions for both cases.

Calculation of the Conservation of Momentum in the X and Y directions for both cases

Percent error of Conservation of Momentum

Calculating the Conservation of Energy for both cases
showing that energy is not conserved except for case 2 , with a percent error of 25%.

Sample of how I calculated center mass, in this case, for position in the Xcm. The same was done for Ycm, Vxcm, Vycm, and so forth.

Graphs:

Steel balls of equal Mass:

X vs Time for steel ball with initial velocity

Y vs Time for steel ball with initial velocity

X vs Time for steel ball with zero initial velocity

Y vs Time for steel ball with zero initial velocity

All of the above graphs overlayed into one.

Different Masses of steel and aluminum ball:

X vs Time for steel ball AND  aluminum ball

Yvs Time for steel ball AND  aluminum ball

X-Y Center Mass & V Center Mass for Equal Steel Masses

Xcm vs t and Ycm vs t overlayed

Vx-cm vs t & Vy-cm vs t overlayed

X-Y Center Mass & V Center Mass for Steel and Aluminum Balls

Xcm vs t and Ycm vs t overlayed

Vx-cm vs t & Vy-cm vs t overlayed

Conclusion:

Based upon my calculated and measured data, I can reasonably conclude that momentum was conserved due to how numerically close their initial and final values were, even if the percent error on one of them was 51% because of zero. Besides that, the percent error was very reasonable:

Percent Error for momentum

 However, the same could not be said for energy. Because of just how large the numerical difference, not just the percent error, of the initial and final energies of the cases were from each other. For us, energy cannot, at least for the two balls of equal mass, be said to have been conserved. There was a clear and significant loss of energy:

Percent Error for energy: Far right column

Uncertainty/Error

• We can determine a level of uncertainty in the measurement of the ball mass. The scales had an uncertainty of .01 grams, which would affect the second case calculations for momentum and energy more so than the first, as the variable mass simply cancels out when the two balls are of equal mass.
• Next, human inaccuracies (compounded by video quality) in positioning the balls in the position vs time graph would lead to an inaccurate slope; velocity, and therefore introduce a degree of uncertainty in our calculations for momentum AND energy.
• Most importantly, we also see that real-world collisions see losses of energy from a variety of sources: friction, sound, vibration, etc, all of which clearly affected our data--most significantly, our calculation of energy of the equal masses.