03-June-2015: Physical Pendulum Lab

The purpose of this lab was to derive expression for the period of numerous physical pendulums, and then verify the predicted periods using LoggerPro.

Part 1: A Solid Ring
To find the period of a solid ring oscillating at small angles, we needed to find an equation for it's acceleration as a function of a negative constant and displacement. From there we would be able to find its period. Once we had this value, we would then use LoggerPro to find the actual period of the pendulum using a photogate. (see Fig. 1)

Fig. 1
Using LoggerPro and a photogate, we were able to find the period of the solid ring as it swung through small angles.

Since the ring was solid, we had to take its moment of inertia relative to both of its radii. When we measured the two radii, we got values of 0.05765 m and 0.06975 m respectively. The equation for the period of this pendulum would be given by 2*pi divided by it's angular frequency. After going through our calculations, we found the predicted period of the pendulum to be 0.718 seconds. When we ran the actual experiment, we found the period to be 0.720 seconds. Thus, we had a percent error of 0.376 %. (see Fig. 2 and Fig. 3)

Fig. 2
We were able to calculate the period of the solid ring be setting the angular acceleration into this form.

Fig. 3
Using LoggerPro, we were able to find the period of the solid and found that we were off by only 0.376%!

Part 2: An Isosceles Triangle about its Apex
To find the period of an isosceles triangle about its apex, we had to find two unknowns first. First, we had to find its moment of inertia. Next, we had to find its center of mass. Our calculations for finding the center of mass of the triangle can be seen in Fig. 4 and for the moment of inertia in Fig. 5

Fig. 4

Fig. 5
NOTE: There should be a "M" for mass next to the 12 in the final equation.

Once we had our equations we mad our prediction for the period, which turned out to be 0.703 seconds. (see Fig. 6) To test this, we cut out an isosceles triangle and measured its height and base to find a prediction for its period. (see Fig. 7)

Fig. 6
We calculated the period to be 0.703 seconds and we were off by only 0.92 %

Fig. 7
We cut out the isosceles triangle that we would use to test our theoretical value.

We found that we had a percent error of 0.92 %.

Part 3: An Isosceles Triangle about the Midpoint of its Base
To find the period of an isosceles triangle about the midpoint of its base, we again had to find an expression for its moment of inertia. Once we had this, we use Newton's Second Law to find an expression for the angular acceleration as a function of a negative constant and displacement. Once we had this, we were able to find its period. Our calculation for the moment of inertia can be found in Fig. 8.

Fig. 8

We plugged in the dimensions of the triangle as needed to find its period, which was 0.616 seconds. (see Fig. 9). Again, we used LoggerPro to find the actual period of the triangle. (see Fig. 10)

Fig. 9
Again, we used the formula for period being 2*pi/angular frequency to find our value for the period.

Fig. 10
We found that we had a mere 0.92 % error in our calculation from that of LoggerPro.

We found that we had a percent error of 0.92 %.

Part 4: A Semi-Circle about the Center of its Curved Edge AND the Center of its Base
To find the period of oscillation of a semi-circle about the center of its curved edge, we decided to first find its center of mass. This would be required to find the period of oscillation about the center of its curved edge and the center of its base. To do this, we decided to plot the semi circle and take its dm with respect to many little rectangles that would compose the semi circle. To see our full calculation, see Fig. 11. We found that the center of mass 4R/3pi above the center base of the semi circle.

Fig. 11
Finding the center of mass. 

Once we had the center of mass, we needed to consider its movement in small angle oscillations. This indicated that we would have to use Newton's Second Law, Torque= Moment of Inertia * Angular Acceleration. We would have to take the component of the disk's weight that was parallel to its motion (Perpendicular to the lever arm). However, to complete this equation, we needed the moment of inertia of the disks as they swung from different origins. Fig. 12 shows our calculations for both moments of inertia.

Fig. 12
In the red is our moment of inertia of the semicircle oscillating about the center edge of its curve. In the blue is
the moment of inertia of the semicircle oscillating about its center base.

Once we had the moments of inertia for the two differently orientated semicircles, we then proceeded to plug in our numbers and put angular acceleration into a SHM form. Once we did this, we took our omegas and calculated our periods. After running the experiments, we compared our expected values. (see Fig. 13 and Fig. 14)

Fig. 13-1
We found that our value for the period was only 0.6% of LoggerPro's value.

Fig. 13-2
Period LoggerPro found us for the semicircle oscillating about the center edge of its curve.

Fig. 14-1
We had to take into account the altered shape of the semi circle. It was not a completely circular curve, which may have correlated to our higher percent error.

Fig. 14-2
Period LoggerPro found for us for the semicircle oscillating about the center of its base.

We found that our first prediction for the semicircle oscillating about the center edge of its curve was good, in that we yielded only a 0.6% error. However, our second run for the semicircle oscillating about the center of its base had a larger 1.3% error. This may have been due to the fact that our semicircle was not completely round, thus throwing of its actual moment of inertia.

Error and Conclusion:
Although the lab represented little error, we must still account for some issues. In cutting the shapes, we were unable to make perfect and exact shapes, which in turn may have altered the actual period of oscillation of the system. Another issue was finding the perfect point to have the triangle oscillating. Using cable rings, we tried to get the edge of the triangle and semicircle as close as possible to being at the pivot.

In all, however, the lab was successful and allowed us to verify the period of multiple oscillating pendulums.

20-May-2015: Conservation of Energy and Angular Momentum Lab

The goal of this lab was to predict the height at which a simple pendulum would rise after colliding with a stationary piece of clay.

Setup:
To set up this lab, we used a rotary motion sensor as a pivot for a meter stick to act as a pendulum on. Once the meter stick was attached to this pivot, we setup a small piece of clay on the ground directly in path the meter stick would be falling. To ensure the collision would be inelastic, we wrapped the clay and meter stick with tape, sticky side out. Ideally, they would stick together after the collision (see Fig. 1). We then measure the mass of the meter stick as well as the clay.

Fig. 1Setting up the experiment.

Once we had the masses, we found the position of the pivot point for the simple pendulum. This was located roughly 0.0085 m away from the top end of the meter stick. Once we had all this basic information, we were ready to begin finding our prediction.

Calculating the Height:
To find the height at which the clay and meter stick system would rise, we split the collision into three components. First was the initial fall of the meter stick after being held horizontally. Second was the change in angular velocity after the collision. Third was the final height the system would reach after the collision.

PART 1: Initial Angular Velocity
To find the angular velocity of the meter stick directly before the collision, we used conservation of energy. Initially, as the meter stick was position in a completely horizontal manner, we only had gravitational potential energy. If we called the center of mass of the meter stick in the vertical direction our zero point for GPE, then our final energy would be rotational kinetic energy. (see Fig. 2)

Fig. 2
To do this calculation, we set our zero at the vertical center of mass of the meter stick.

Directly before the collision, we found our angular velocity to be 5.445 rad/s.

PART 2: Final Angular Velocity
To find the angular velocity of the system directly after the collision, we used conservation of angular momentum. However, to find the moment of inertia in the final stage of the collision, we had to include the moment of inertia of the clay as a point mass. (see Fig. 3)

Fig. 3
We used conservation of angular momentum in this section and included the moment of inertia of the clay.

Directly after the collision, we fond our angular velocity to be 3.858 rad/s.

PART 3: Finding the Height
To find the final height of the clay, we again used conservation of energy. To make the problem simpler, we called the pivot point of the meter stick our zero for GPE. This in turn would make our GPE values negative. In the final energy section of the equation, we multiplied our heights relative to the cosine of the angle at which the meter stick would make with the vertical. (see Fig. 4)

Fig. 4
To make this problem do-able, we tried to find the angle at which the pendulum swung rather than the actual height of the clay.

We found the theoretical angle of the pendulum to be 63.6 degrees. To find the height, we assumed that the clay was directly at the end of the meter stick. Thus, the height it rose would be the difference between the length of the clay to the pivot point, and the cosine of the length of the clay to the pivot point. In final, we predicted the height of the meter stick to be 0.5514 meters.

PART 4: Comparing to the Actual Height:
Using video analysis, we recorded the collision of the meter stick and clay. Setting up a horizontal and vertical axis, we placed a point directly at the maximum height of the clay. (see Fig. 5). We found this to be 0.4842 m.

Fig. 5
Using LoggerPro, we were able to find the actual height the clay rose.

We found our percent error to be 13.87 % at the conclusion of the lab. (see Fig. 6)

Fig. 6
Percent Error Calculation.

Error and Conclusion:
There were a few sources of error in this lab. For one, we assumed that the meter stick was completely uniform and it's center of mass was located directly at the 50 cm mark. Having degraded over time, this may not have been entirely true. Also, we chose to ignore the fact that energy was lost in other forms during the collision, such as in sound or friction. Thus, it makes sense that our predicted value for height is a little bit over the actual value.

Overall, the lab was a success and we were able to prove that energy and angular moment was conserved in this inelastic rotational collision.

13-May-2015: Moment of Inertia of a Uniform Triangle

The purpose of this lab was to find the moment of inertia of a triangle about its center of mass using physical methods.

Part 1: Deriving our Theoretical Value:
To begin the lab, we first wanted to find an expression for the moment of inertia of a uniform triangle by derivation. To do this, we took two approaches.

In the first approach, we considered the triangle to rotate about the edge of its base. We found an expression for this moment of inertia and then used the parallel axis theorem(I_cm=I_{around one edge} + M(d_{parallel axis displacement})²) to find its moment of inertia about its center of mass. (see Fig. 1).

Fig. 1
Although the formula uses "b" as its term for the side perpendicular to the axis of rotation, this is actually the "height" of the triangle.

In the second approach, we considered the triangle to rotate about the edge of its height. We used the same method as we did in the first approach. (see Fig. 2)

Fig. 2
Using the same method as before, we altered the axis at which the triangle rotated.

Once we had derived from both approaches, we found that our expression for the moment of inertia about the center of mass of the triangle was going to be I= (1/18)Mb^2, where "b" is the base perpendicular to the axis of rotation.

^{Part 2: Physically Finding the Moment of Inertia:Once we had our expression, it was now time to actually find the moment of inertia using the apparatus from our previous angular acceleration lab}. Using the formula I=(mgr/a)-mr^2. (a=angular acceleration), we could find the moment of inertia of the apparatus in three separate trials. We used a hanging mass of 25.0 g, and a torque pulley of 2.51 cm radius. The only information we needed was the mass that was being dropped and the radius of the torque pulley, given that LoggerPro would give us the angular acceleration of the system. 

The idea is that if we find the moment of inertia of the system by itself, and then the moment inertia of the system and the triangle (orientated either way), we could find the moment of inertia of the triangle by taking the difference of the two. 

We ran three trials: One without the triangle (moment of inertia of just the system), one with the triangle long-ways up (moment of inertia of the system and vertical triangle- Fig. 3), and one with the triangle long-ways down (moment of inertia of the system and horizontal triangle- Fig. 4). In each setup, we wrapped a string around the torque pulley and allowed the system to free fall. Using LoggerPro, we were able to find the angular acceleration of the system by taking the slope of the angular velocity vs time graph. However, since there is some friction in the system, we had to take the average value of angular acceleration (going up and down) as our usable value  (Fig. 5)

Fig. 3
We placed the triangle vertically up, giving us a specific value for the moment of inertia.

Fig. 4
The moment of inertia of the triangle would be the difference of its combined inertia with the system and the inertia of the system by itself.

Fig. 5-1
We needed to find the average angular acceleration in order to attempt to cancel out the torque due to friction.

Fig. 5-2

Once we had all of our angular accelerations, it was time to begin calculating the moments of inertia of each setup. (see Fig. 6) 

Fig. 6

We were left with a final moment of inertia of the center of mass of a triangle rotating about its vertical height being .00023902 kg*m^2. The moment of inertia of the center f mass of a triangle rotating about its horizontal base was .0005586 kg*m^2.

Once we had our theoretical values, it was time to find the percent error from the actual values. With our equations from before, we measured the height of the triangle to be 14.936 cm, its base to be 9.844 cm, and its mass to be 456 g. (see Fig. 7)

Fig. 7
The low % error showed that our expression for the moment of inertia was legitimate. 

We found that our percent error was less than 3% for both cases, which indicated that the experiment in all was a success and our expression for the moment of inertia of a uniform triangle was good in both cases.

Conclusion:
By deriving an expression for the moment of inertia of a uniform triangle around some axis and then using the parallel axis to find it about its center of mass, we were able to find a good prediction of a real physical attribute of this rotating triangle. Considering that the main source of uncertainty would be in the case of friction altering our angular acceleration, I believe that our % error being less than 3% is very reasonable.

11-May-2015: Moment of Inertia and Frictional Torque Lab

The purpose of this lab was to find the moment of inertia of a disk and use that determine the frictional torque that acted on it. We would then use this information to determine to solve an inclined cart problem.

Part 1: Finding the Frictional Torque:
The apparatus consisted of a large metal disk attached on a central shaft. The disk would spin on this shaft (see Fig. 1).

Fig. 1

To begin, we needed to understand what we were exactly looking for. Considering that the net torque is equal to the moment of inertia multiplied by the angular acceleration, we needed to find the moment of inertia of the entire rotating object first. To do this we needed to split the apparatus into three components: two shaft cylinders, and the main center disk. With the equation for the moment of inertia of a disk being I=(1/2)MR^2, we needed to find the radius of all three of the disks. To do this, we used calipers to precisely measure their diameters (see Fig. 2) and then divided this value by 2. Once we had the radii, we needed to find the mass of each individual piece. To do this, we used percent composition. That is, we found the volume of the disk and two cylinders, totaled it up, and found the percentage that each piece made of the whole. To find the volume, we simply found the thickness of each piece (see Fig. 3) and used the equation V=pi*r^2*h.If we had the percent composition, we could then find the percent composition of mass for each piece. On the disk was labeled the entire mass of all three components together. We broke this mass into the percentages we calculated and got our mass for each piece (see Fig. 4 for calculations).

Fig. 2
Measure the diameter of the disks, we used calipers to get a precise value, and then took a picture of the calipers reading to analyze it more thoroughly.

Fig. 3
Using the end of the calipers, we were able to measure the lengths of the smaller disks.

Fig. 4-1
Here we found the percent composition of the volume of the rotating apparatus. 

Fig. 4-2
We used the percent composition to determine the mass of each individual piece of the rotating apparatus.

Once we had our masses, we found the moment of inertia of the two cylinders and center disk, and then summed it all up. Our final moment of inertia of the entire apparatus was calculated to be 0.019986 kg * m^2 (see Fig. 5 for calculations).

Fig. 5

Once we had our moment of inertia for the apparatus, we then proceeded to find the deceleration of the disk due to its frictional torque. To do this, we set up a camera at one end of  disk, directly lined up with the central shaft (see Fig. 6). Once we opened LoggerPro and ensured that we were getting live video feed, we gave the disk a gentle spin and let it slow down to a stop.

Fig. 6
We lined up the camera to get an accurate visual of the disk.

Once we had our video capture, we went back to LoggerPro to analyzed the data. To do this, we placed a coordinate system into the video, placing our origin directly over the central shaft. From there, we marked one specific point on the disk and followed it frame by frame as the disk slowed down (see Fig. 7) NOTE: We placed a black piece of tape on the edge of the disk to clearly show the point which we were marking in LoggerPro.

Fig. 7
By placing dots at the black edge, we were able to obtain x- and y-coordinates for the position of the edge.

Once we had our points marked, we found that we had x- and y-components of velocity throughout the circular motion. Thus, in order to find our tangential velocity, we had to take the square root of the velocity in the x-direction squared and the velocity in the y-direction squared. We defined a column in LoggerPro to do this for us at every point (see Fig. 8).

Fig. 8
Using LoggerPro, we were able to find the tangential velocity at every point.

When we plotted the tangential velocity versus time, we found that the velocity had a downward linear like trend. Thus, in order to find the tangential acceleration, we found a linear fit for the graph. (see Fig. 9). This value for acceleration could then be divided by the radius of large center disk and found our value for angular acceleration (see Fig. 10).

Fig. 9
By finding the slope of the tangential velocity graph, we are able to obtain the tangential acceleration.

Fig. 10

With all of our calculation completed, it was now time to plug into our net torque equation to find the frictional torque of the system. When we multiplied our moment of inertia and angular acceleration, we found the net frictional torque to be -0.01808 (kg*m^3)/s^2. 

Part 2: Testing Our Value:

To test our value for the frictional torque, we set up an inclined cart system. This cart system would consist of a cart that would be attached by string to one of the small cylinders on the rotating apparatus. The cart would be placed on an inclined track, where it would be release from the top and allowed to roll down. The goal was to make a prediction for the amount of time it would take the cart to go down the track and then test that value (see Fig. 11)

Fig. 11
The cart would slide down the track while we recorded the amount of time it takes to reach the bottom.

First, we had to find an expression for the amount of time it would take the cart to go down the track. To do this, we combined our net torque and net force equations. We found the mass of the cart using a scale and the angle of inclination using our phones. (see Fig. 12 for full calculations) Our prediction was 7.95 sec.

Fig. 12

When we ran the actual experiment, we obtained a consistent value of 8.64 sec. This yielded us a percent error of 7.99%. The fact that the actual value of time was larger indicated that there were some forces that we did not account for, which is reasonable.

Overall, we were able to justify our value for frictional torque with our inclined cart test. However, some possible sources of error include our measurements, our ability to properly place precise dots on our video capture, and our ability to press our stopwatches in a timely fashion.

4-May-2015: Angular Acceleration Lab

The purpose of this lab was to find the relationship between the angular acceleration of a system, the force causing it to rotate, and the mass which is rotating. Along with this, we were to find the moment of inertia of our spinning disks.

Part 1: Relationships
To perform this lab, we set up rotating system that would comprise of two disks, a pulley, and a hanging mass. The idea was to attach the mass to the system by a string, have it attached some radius from the center of the disks, and allow it to fall. This would in turn cause the system to accelerate. We would then measure this in LoggerPro and record several trials with different hanging masses, disk masses, and rotating radii. (see Fig. 1).

Fig. 1
The apparatus consisted of a hanging mass, rotating disks, an air hose connected to an air valve and a rotational motion sensor.

To begin, we first had to measure all of the equipment we were planning on using. The materials included:

• Two steel disks (one for the top and another for the bottom)
• An aluminum disk (used to change the mass of the rotating disk)
• A small and large torque pulley (used to find the effect of changing the radius at which the force acts on the object)
• A hanging mass (used to cause the system to begin rotating)

Using calipers, we measured the diameter of the two steel disks, aluminum disk, and torque pulleys. (see Fig. 2). We then found the mass of the two steel disks, aluminum disk, torque pulley and hanging mass. (see Fig. 3) We recorded all this data on our white board.

Fig. 2
Using calipers, we were able to gain precise measurements of each disk. 

Fig. 3
Data Table

As seen in Fig. 1, the apparatus is hooked up to our laptop, allowing us access to data that LoggerPro could extract for us. The idea is that the black and white dashes around the disk could be counted while moving by some external sensor on the rotational apparatus. This would allow us to find angular velocity and acceleration. To properly set up the sensor, we had to input how many marks there were on the top disk (200). Once we had this, we were ready to begin collecting data. 

We tied the first mass (24.7 g) by string to the small torque pulley (r = 1.26 cm) atop the steel disk (1348 g). We began collecting data with LoggerrPro as we released the mass and allowed it to accelerate down. We collected data as the mass dropped and as it rose again, giving us an angular acceleration down (cw) and up (ccw). (see Fig. 4) However, we had to note that there was some friction within the system. The apparatus allowed for a virtually "frictionless" surface for the disk(s) to rotate, however in reality this was not true. Thus, in order to counter the effects of friction, we would simply take the average angular acceleration. This average acceleration is what we would use when analyzing and evaluating our data.

Fig. 4
The spike at the top of this graph indicates the turn around point of the hanging mass. To find the angular acceleration, we had to take the slope of this graph.

We ran the same experiment six times, with some slight changes each time. To find the effect of different tension pulling on the system, we ran the first three trials with the same small torque pulley, same top steel disk, but instead added additional mass to the hanging mass each time. To find the effect on using a different rotating disk on the system, we ran the next two trials with a large torque pulley, but with the top steel disk in one trial and the top aluminum disk in another. Finally, for the final trial, we used the large torque pulley and allowed the top and bottom steel disks to rotate. We received data through LoggerPro (see Fig. 5) and recorded it (see Fig. 6).

Fig. 5
Here are the graphs for all six of our trials. We took the slope of one only as an example.

Fig. 6

NOTE: During the data collection, we made an error when calibrating the machine to 200 marks per rotation. Our graphs represent angular accelerations that are slightly off. However, to correct the issue, we multiplied our values for angular acceleration by 360/200.

With all this data, we drew a couple of conclusions. First, we found that when we almost doubled the hanging mass, we almost doubled the angular acceleration. Same went for tripling the mass. Second, we found that when we used the large torque pulley rather than the small torque pulley (doubling the radius at which the tension acted on the disk) we also doubled the angular acceleration. Third, we found that when we decreased the mass of the rotating disk three-fold (using the aluminum disk rather than the steel disk) we INCREASED the angular acceleration roughly three times. Fourth, we found that when we doubled the rotating mass (allowed two steel disks to rotate), we cut the angular acceleration in half.

The final step in part one of this lab was to find the relationship between the linear velocity of the hanging mass and the rotational velocity of the system. To do this, we simply ran one run the exact same way as we did the other six trials. We used a large torque pulley for the tension to act on. The mass of the disks spinning and mass falling become irrelevant, as we are simply analyzing velocity in the angular and linear sense. Using a motion sensor and note card, we began collecting data with logger pro, in both rad/s and m/s. When we graphed linear velocity in the y-axis and rotational velocity in the x-axis, we obtain the graph in Fig. 7. When we took the slope of this graph (positive or negative), we obtained a value of 0.02513 m which translates into 2.513 cm, which is practically exactly the radius of the large torque pulley.

Fig. 7
We found that the slope of the linear vs angular velocity was exactly the radius of the large torque pulley.

Conclusions for Part One:

For part one of this lab, we were able to verify the magnitude of the hanging mass and the radius at which a force was acting on a disk were directly related to the angular acceleration of the system. This meant that if we increased the hanging mass, we increased the angular acceleration. (same for radius). We were also able to verify that if we increased the mass that was rotating, we decreased the angular acceleration. All of these relationships were numerically proportional (ie. if you double the hanging mass, you double the angular acceleration and if you tripled the mass that was rotating, you cut the angular acceleration into 1/3 of what it used to be). As for linear and angular velocity, we found that they were related by the radius at which the force acting on them was being applied. Through this, we were able to verify the relationship v=rω.

Part 2: Moment of Inertia of Each Disk:

For the second part of the lab, we had to use our tools to find the moment of inertia of the single top steel rotating disk, the two steel rotating disks combined, and the top aluminum disk. To do this we had to derive an expression for the moment of inertia of this system using the force and torque equations. We are finally left with the equation I=(mgr/α)-mr^2 To verify if this is true, we used the given equation for the moment of inertia of rotating disk as our "actual" value. I=(1/2)MR^2 Our calculations can be seen in Fig. 9

Fig. 9

In all, we found that we had a percent error of less than 5% for the single steel disk, both steel disks, and aluminum disk. This indicated that our data and formula were good. 


Conclusion:In all, the lab was a success and allowed us to find the multiple relationships between angular acceleration and the factors that change it. We were also able to verify our equation for the moment of inertia of a rotating solid disk with our data. However, we had to account for a few areas of uncertainty. This included all our areas of measurements as well as the frictional torque we did not account for. We assumed that averaging the angular acceleration would completely rid the friction, which is not entirely true. However, in all we were able to prove our moment of inertia and relationships.