Lab 18 moment of inertia and frictional torque

Purpose: In this laboratory we are trying to determine the moment of inertia of the apparatus, and figure out a way to use video capture to determine the angular deceleration as it slows down. And calculate the frictional torque as it slows down.

The purpose of the apparatus is to use it to find the angular deceleration by taking a video of the apparatus spinning, and eventually use the ramp to find the time it takes the cart to travel one meter.

Procedure:
In this lab, the first thing we did was measure the diameters and lengths of the different cylinders that made up the rotating object on the apparatus. Then we found the volume of the two smaller cylinders and of the big disk to find the percent of volume the big disk was compared to the three put together. We then used that information to find the mass the large disk and the smaller cylinders because the mass would be the percent of the volume of each. We found the moment of inertia of the three parts together after finding the mass of each cylinder and the disk. Then we took a video of the apparatus rotating to a stop after giving it an initial push. Using the video analyzer we to marked an initial point on the apparatus and added another point every time the apparatus made a revolution so all our points were on the same spot and marked one revolution of the apparatus. Using the marked points we were able to make a position versus time graph of the points. Then we added a power fit line to get an equation of the position with respect to time. To get the deceleration we took the derivative twice of the position vs. time equation. Using newton's second law and angular acceleration we found the linear acceleration of the cart going down the ramp at the measured angle of the ramp. Plugging the acceleration in a kinematics equation we solved for the time it would take the cart to go a distance of one meter. Then we compared that value to the value we got by measuring the time it took the cart to go down the ramp a distance of one meter.

We measured the diameters and heights of the cylinders and the disks and used the measurements to find the volumes of each. Then we calculated the percent of volume the disk was so that we could find the mass because the mass percent is the same as the percent in volume. So we multiplied the percent to the mass of the whole that read on the three objects together and got the mass of the disk. And we got the mass of the smaller cylinders by subtracting the mass of the disk to the entire mass and divided it by two. The calculations for this are as follows:

To find the moment of inertia of the whole object we had to add the moment of inertia of the individual parts(the two cylinders and the disk). Individually the moment of inertia of a solid disk is 1/2mr^2 and since all three parts were solid disks we added this formula three times with the mass of the individual parts being m and r being the radius of each part. Our calculations were as follows:

Next we recorded a video of the apparatus spinning and coming to a stop, and plotted the points every time it made a revolution and put it on a distance versus time graph on logger pro. We power fitted the line to get an equation.

Using the equation of the line we took the derivative twice so that we could get the angular acceleration of the apparatus.

Using this calculated angular acceleration and converting it to linear acceleration using the equation alpha = a/r where a is the linear acceleration and r is the radius of the large disk. We used newton's second law to calculate the linear acceleration of the cart and the frictional torque of the disk.

Then we calculated the time it would take the cart would take to travel 1 meter using a kinematics equation:

This was our theoretical value so then we found the experimental value by letting the cart go down the ramp 1 meter and timing how long It would take. We did this three times and took the average time of the three.

Conclusion:
According to the data the experimental value for the time it would take the cart to travel 1 meter down the ramp was 7.98 seconds and the theoretical value that was calculated for the cart to travel 1 meter was  8.57 seconds. The values are pretty close but there is some error that could have came from the method that we came up to calculate the time theoretically. For example, when we were marking the point of one revolution the disk would take sometime it was hard to see the mark we made on the disk so we had to guess a little when the mark was at the point of revolution. this could have altered the acceleration a little. Also, the time we calculated experimentally of 7.98 seconds I feel was a little off we should have ran more trials to get a value that was more accurate.

Lab 17 Finding moment of inertia of a uniform triangle

Purpose: The purpose of this lab is to determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle.

The apparatus will be set up pretty much the same as the previous lab 16 except that this time there will be a triangle connected to the top.

The approach we took to accomplish our goal was by first determining the experimental value for the inertia of the system without the triangle and then the moment of inertia of the system with the triangle and subtracting the moment of inertia value with the triangle to the inertia value calculated without the triangle to get the inertia value of the triangle itself. We used the same method we used in the previous lab of angular acceleration where we measured the angular acceleration up and down, and got the average of the two to plug into the equation to find the inertia of the system. The equation was derived in the previous lab and is the following:

Where m is the mass of the hanging mass, g is the acceleration due to gravity, r is the radius of the torque pulley, and alpha is the average of the angular accelerations up and down.

Next we derived a formula for the inertia of the triangle with the axis of rotation being at one end so it is easier to integrate. We did this so that we could compare our experimental values for the moment of inertia with the theoretical values for the two orientations. We used the parallel axis theorem to get the moment of inertia from the center of mass of the triangle which is what we wanted.

To derive an equation for the moment of inertia of the triangle with the axis of rotation on one side, we had to take a horizontal piece dm of the triangle and find the moment of inertia of that piece and integrate the moment of inertia of that piece from 0 to h. Since the piece of the triangle was a rod being rotated about one end we knew the equation for the moment of inertia of such a rod to be 1/3ML^2 where M becomes the mass of the rod piece dm and L becomes the length x of the rod. so then we had to find a function to replace dm and we did so by making the proportions of the mass of the piece over the mass of the whole thing equal to the area of the piece over the area of the whole triangle and solving for dm. But then we had to change the x variable into y because we were integrating with respect to dy. So we found an equation for the line of the hypotenuse of the triangle and got x = -b/hy+b and plugged that into our equation. We integrated the equation and got our moment of inertia of the triangle to be:

Where M is the mass of the triangle and b is the length of the base.

We could then use the parallel axis theorem to find the moment of inertia at the center of mass of the triangle:

The derivation to find the moment of inertia of the center is the following:

As can be seen the moment of inertia for the center of mass is 1/18 M b^2 where M is the mass of the triangle and b is the length of the base.

The measurements that we took for this experiment were the mass of the hanging mass, the diameter of the torque pulley used, the mass of the triangle, the length of the base of the triangle, the length of the height of the triangle, and the angular acceleration up and down of the system. The reason we needed the mass of the hanging mass, the diameter of the torque pulley used and the angular accelerations up and down, was because we need it to plug into the first equation to find the moment of inertia of the system experimentally. the reason we needed the mass of the triangle, the length of the base, and the length of the height of the triangle is so that we could find the moment of inertia of the triangle theoretically. The measurements that we took were as follows:

We did the first trial to find the angular velocity up and angular velocity down and collected enough data to let the hanging mass go up and down three times. Then we took the derivatives of each slope to get the angular accelerations:

For the second trial we did the same thing except we put the triangle on the apparatus with the longer side facing vertically( not the hypotenuse). The data we collected was the following:

For the third trial we changed the orientation of the triangle so that it was perpendicular to the position it was in the second trial. The data we collected for this orientation was:

 The average angular accelerations were calculated for the system by itself, and with the triangle in both positions. The following shows the calculations:

 

Using this data the moment of inertia for the system alone, and with the triangle in different positions were calculated experimentally.

 

 

And the experimental moment of inertia of the triangle at the different positions was calculated to be:

 

The theoretical moment of inertia of the triangle at the different positions was calculated to be:

Conclusion:
According to the data the experimental values of .000436 for the triangle in the vertical position, and .000326 for the triangle in the horizontal position are a little different than the theoretical values of .000245 for the triangle in vertical position and .000565 for the triangle in the horizontal position. It is safe to say there is a small error either with the experimental data or rounding error in the calculations. The experimental data shows that the moment of inertia of the triangle in the vertical position is greater than the moment of inertia of the triangle in the horizontal position. While the theoretical data says the opposite. I think the theoretical data is correct and there must have been some error for the experimental data because the triangle in the horizontal position has a larger moment of inertia it is harder to move because its mass is over a larger diameter. For example, when you spin on a chair and hold your arms out you would go slower than if you were to have your arms in. The triangle in the horizontal position has a longer base than in the vertical position so it is harder to move in the horizontal position so the moment of inertia is greater.

Lab 16 angular acceleration

Purpose: The purpose of this lab for part 1 is to see what factors affect angular acceleration of system. For part 2 we are going to use the known torque and measured angular acceleration to determine the value for the moment of inertia.

The purpose of the apparatus used is to measure the angular velocity so that we then can get the angular acceleration.

Procedure:
First thing we did in this experiment was measure the diameters and the masses of the disks and pulleys and the mass of the hanging mass. Then we set up the apparatus on logger pro and changed the equation settings to 200 counts per rotation. We attached the air hose to the apparatus so that the top disk could move but the bottom disk wouldn't. We ran the trials and collected data on logger pro for the angular velocity versus time for each trial. We took the derivatives of the angular velocity graphs (for each part whether it was ascending or descending) to get the angular acceleration up or down. Then we took the average of the angular acceleration up and down to get the average acceleration overall for the trial. For the second part of the experiment we used the data from the first part of the experiment to determined the experimental values for the moments of inertia for the different combination of disks used in each trial.

The diameters and masses of the various disks and pulleys were as follows:

For the first trial we used the hanging mass, the small torque pulley and the top disk was the steel one. The data we got on logger pro is as follows:

The measured value that we got was in rotations per second squared so we multiplied it by 2pi to convert it to radians per second squared. We did this for the first three trials then for trials 4,5 and 6, we added an equation into logger pro to convert the angular acceleration to rad per sec squared for us.

For the second trial we doubled the hanging mass using the same small torque pulley and same top steel disk and collected the following data:

I did not take a pic of the third trial, but it is safe to assume it looks similar to the trials above. For the third trial we tripled the hanging mass and used the small torque pulley and the steel disk on top. The acceleration down for the third trial is 1.33 rad/sec^2, and the acceleration up is 1.43 rad/sec^2.

For the fourth trial we went back to the original hanging mass by itself and changed the torque pulley to the larger pulley and kept the top disk the steel disk. The data we collected was as follows:

For the fifth trial we kept the hanging mass the same the torque pulley was the large pulley, but the top disk was the aluminum disk. The data we collected for this trial is as follows:

 For the sixth trial we kept the hanging mass the same as the previous trial and kept the large torque pulley, but we changed the apparatus to allow the two disks to move and we used both steel disks on top and bottom. The data collected for this trial was as follows:

We copied the accelerations up and down from the computer and got the average for each trial and put everything on the following chart in our lab manual.

For the second part of the experiment we derived an equation in class to get the moment of inertia if the disk that is spinning using newton's second law(derivation is in the lab manual) and we got the equation:

Where m is the mass of the hanging mass, r is the radius of the torque pulley, g is acceleration due to gravity, and alpha is the average of the angular acceleration up and angular acceleration down.

Using the equation above we calculated the moment of inertia for the disk or disks combination for each trial.

Conclusion:
According to the data in the first part of the lab we can see that changing the hanging mass by making it bigger and keeping torque pulley and the top disk the same makes the average angular acceleration increase from .454 rad/sec^2 in trial 1 to .926 rad/sec^2 in trial 2 and 1.38 rad/sec^2 in trial 3. Changing the radius of the torque pulley from the small one in trial 1 to the larger one in trial 4 makes the acceleration increase. In trial 1 the average acceleration was .454 rad/sec^2 and in trial 4 the average acceleration was .8719 rad/sec^2. This makes sense because a larger radius equals a larger torque force. The effect of changing the rotating mass in trial 4  just letting the top steel disk spin the average angular acceleration is .8719 rad/sec^2, while in trial 5 making the top disk the aluminum disk and only allowing that disk to spins gives an angular acceleration of 2.4885 rad/sec^2. In trial 6 allowing both steel disk on top and steel disk on the bottom to move gives an angular acceleration of 1.033 rad/sec^2. It looks like making the top disk a lighter mass makes the average acceleration higher, but when it comes to trials 4 and 6 the two disks spinning together makes the acceleration value greater than just the top steel disk moving alone. For the second part of the experiment we calculated the moment of inertia for the different trials. The first three trials the moment of inertia was pretty much the same for all three trials being .0135,.0132, and .0133. Which makes sense because the top steel disk spinning was the same for all three trials. For the fourth trial the moment of inertia was a little larger than the first three trials being .0150 which should be so because we change the torque pulley from the smaller to the larger adding a little more mass to the rotating system. For the fifth trial the moment of inertia has the smallest value being .00522 which makes sense because mass of the spinning disk was the smallest. And for trial 6 the moment of inertia was .0127 which is the smallest value out of all the trials which makes sense because the mass was the greatest with both steel disks spinning. some sources of uncertainty that could have effected our calculations were the fact that we could have measured the wrong diameter on the large vernier caliper because sometimes it was hard to read the correct line that aligned the best. There also could have been sources of uncertainty from the machine itself because it is a very old machine and worn down from multiple use over the years.

collision lab with ball and pendulum

Purpose: the purpose of this laboratory is to calculate the velocity of the ball using conservation of momentum and energy. And show that the values calculated are within reasonable error by calculating the propagated uncertainty.

The purpose of this apparatus is to measure the angle that the block rises after the ball is shot into it so that we can calculate the initial velocity of the ball .

Procedure:
In this experiment we used the apparatus to launch a ball into block and saw the angle in degrees that the block rose. Measuring the length of the string that the block is hanging from, and measuring the mass of the ball and the mass of the block, we will use conservation of momentum and conservation of energy to find the initial velocity. We will see if our value calculated for the initial velocity is allowed within the uncertainty of our instruments.

The first thing that we had to do was find a relationship to calculate the velocity initial. To do this we see that momentum is conserved before the collision and an instant after the collision before the block rises. So we use the equation for conservation of momentum.

The problem is that we are going to have two variables V initial and V final. Because there is kinetic energy in the beginning of the collision and potential energy at the end and energy is conserved we could use conservation of energy equation.

The V initial in the conservation of energy equation is equal to the V final in the conservation of momentum equation so we solve for V initial in the conservation of energy equation and we get the expression:

Where h is:

where L is the measured length of the string holding the block, and theta is the angle the block elevates. so the final expression we get is.

We measured everything we needed to measure and got:

m1 (mass of the ball) : .00763 kg
m2 (mass of the block) : .0809 kg
L (length of the string) : .2 m
theta (angle of elevation) : 16.5 degrees

plugging the measurements into the equation and solving for the initial velocity we get:

The initial velocity that we get is 4.66 m/s +/- the uncertainty.

Next, we have to find the propagated uncertainty to see if our value is within the allowed error. To do this we take partial derivatives for each variable measured and multiply them with their respective uncertainty in measurement and add them all up. The propagated uncertainty was calculated for this lab:

The data suggests that the initial velocity of the ball before the collision was 4.66 m/s with a uncertainty ranging from +/- 1.204 from the instruments we used to measure.

Lab 15 Collisions in two dimensions

Purpose: the purpose of this laboratory is to look at a two dimensional collision and determine if momentum and energy are conserved.

The apparatus we used consisted of a board and a camera is clamped to it and recording the collision from above.

Procedure:

For this lab we set up the camera to take a video on the computer and then we recorded two collisions . The first collision that we did was between the marble and the steel ball. We recorded the collision with the camera and used the video analysis on logger pro we marked the points of the trajectories of each ball throughout the video. We then graphed the data on a position vs time graph on logger pro. Using the position and time graph we were able to get the velocities before and after the collision for the marble and the steel ball. We repeated the procedure for the second collision which involved two steel balls of equal mass. What we are trying to show is that the momentum is the same before and after the collision in the x direction and in the y direction. Also, that the energy is the same before and after the collision. To do this we will use the following equations:

Where M1 is the mass of the steel ball and Vi is the initial velocity if the steal ball and Vf1 is the velocity final of the steel ball. M2 is the mass of the Marble in the first collision and the mass of the steel ball in the second collision and Vf2 is the final velocity of the marble in the first collision, and the steel ball in the second collision.

But for the conservation of momentum equation we will see if momentum is conserved in the x direction and the y direction separately because that is what logger pro will give for velocity. In the conservation of energy equation we are going to find the magnitudes of the velocities in the x and y directions and plug the magnitude in for the velocity initial and final.

The first collision we did was with the marble and steel ball and the trajectory was the following:

The position vs. time graph of this collision looks like:

Where X and Y are the x and y positions for the steel ball that hits the marble ball and X2, Y2 are the x and y position for the marble after the collision. To get the velocity initial we took the derivative of the X and Y part of the graph before the collision. And to get the velocity final we took the derivative of the X, Y, X2, Y2 parts of the graph after the collision.

The velocities that we got are:

X V initial : -.005486 m/s
Y V initial : .3151 m/s
X V final : .02057 m/s
Y V final : .2043 m/s
X2 V final : -.08640 m/s
Y2 V final : .2783 m/s

The mass each ball:

mass of marble M2 : 19.8 g
mass of steel ball M1 : 66.6 g

Plugging in the information above into the equations  of conservation of momentum and energy and calculating the initial and final of each :

For the second collision with the steel ball the trajectory of this collision looks like:

The position vs. time graph for this collision looks like:

 

Velocities:

X V initial : .02400 m/s
Y V initial : .5796 m/s
X V final : .2190 m/s
Y V final : .3305 m/s
X2 V final : -.1914 m/s
Y2 V final : .2070 m/s

Mass of steel balls:

M1 : 66.6 g
M2 : 66.6 g

Plugging the data into the equations we get:

The data and calculations tell us the momentum and energy are conserved because the initial and final momentum values are very close to each other. The energy is also conserved for the same reason. Although, in the second collision the values are a little bit off but this had to be because of some error. The fact that the surface is not completely frictionless could be a source of error and also the tools that we used to video could be causing a little bit or error. But we learned that momentum and energy are conserved in an inelastic collision.

Lab 14 Impulse-Momentum activity

Purpose: The purpose of this laboratory is to test the idea that impulse is equal to the change in momentum of an object in a collision.

The apparatus used in this experiment was used to measure the velocity and the force of the cart over a certain period of time.

The impulse momentum theorem states that the amount of momentum change for the moving cart is equal to the amount of the net impulse acting on the cart. This is shown by the relationship:

To test this idea. We are going to measure the are under the curve of the Force vs. time graph for the collision. Also, we are going to calculate the change in momentum of the cart by knowing the mass of the cart and its velocity before and after the collision. And the two values should be equal to show that this theorem to be true.

The impulse momentum theorem can also be written as:

Where J represents the impulse, m represents the mass of the cart, Vf represents the final velocity, and Vi represents the initial velocity.
The equation to find impulse using Force is :

Where the integral of the force over the time interval that the force is acting on the cart will give the impulse J.

In experiment 1 we saw if the impulse momentum theorem was true for a nearly elastic collision.
Where the incoming cart collides with the still cart (that is clamped and has its spring extended), and bounces back after the collision.

The graphs for this collision are:

Using the velocity vs time graph we chose a point on the graph right before the collision to get the initial velocity and a point right after the collision to get the final velocity. On the force and time graph we integrated the graph when the collision is happening to get the impulse (J).

Velocity initial : .183 m/s
Velocity final : -.160 m/s
mass of the cart : .715 kg

plugging into the right equations:

For the second experiment we added 500g to the cart and repeated the experiment.

The graphs that we got are as follows:

The information that we got from these graphs:

Velocity initial : .334 m/s
Velocity final : -.319 m/s

Measured mass of the cart: 1.215 kg

Plugging into the proper equations like above:

Taking the integral we get the impulse to be -.827 and the change in momentum is -.793.

For the third experiment we examined the impulse momentum in an inelastic collision. where we attach a nail to the moving cart, and making the cart with the nail collide into a wooden block with clay attached to it. and seeing if the impulse momentum thermo works for this kind of collision. We left the same mass on the cart so that the mass of the cart was the same as in experiment 2.

We did the experiment and the graphs that we got for the experiment were:

The information that we collected from the graph:

Initial velocity: .301 m/s
Final velocity: 0 m/s

measured mass of the cart: 1.233kg

Plugging the information in the proper equations gives us the following:

The impulse given from taking the integral of the collision is -.408. The change in momentum calculated is -.371.

The data tells us that the impulse momentum theorem is true. The impulse in experiment 1 was -.2268 while the change in momentum was -.2542. In experiment 2 the impulse was -.827 while the change in momentum was -.793. In experiment 3 the impulse was -.408 and the change in momentum was -.371. The values are really close but not exact due to the fact that there are friction forces and air resistance that could have caused error in the data. But the information supports the idea that impulse is equal to the change in momentum of a collision.