Conservation of Linear and Angular Momentum Nov. -13- 2014

Conservation of Linear and Angular Momentum

Objective:

   Show that the linear and rotational momentum of the system is conserved during this rotational collision.

Set Up:

   There are two set ups for this lab:

1) With a small downward ramp and a steel ball on top of a table, we will use the measurements of its final position to calculate velocity.

2)From there we have a rotating apparatus with a catch for the ball so we may use the found velocity to find the resulting rotation velocity.

Calculations:

We used what we know about the ball's moment of inertia and measurements of the ramp to the table to calculate the ball's theoretical velocity down the ramp without slip. From there we used the measurements of the ball's final location and solved the actual velocity of the ball which when compared to theoretical is smaller suggesting some slip. With this information we moved on to our second set up, but before doing so we have to calculate the inertia of the rotating apparatus in order to find the final momentum of the inelastic collision. To do so we used a hanging mass as a known source of torque then found the average value of alpha to determine the inertia of the apparatus. From there we placed the ball in a similar height on the ramp to have a consistent initial velocity and had it collide to the apparatus. With our calculations we found our final omega to be 1.87 rads/ s with a theoretical value of 2.43 rad/s.

Figure 1. Calculation for both set ups in the lab.

Conclusion:
  It appears that since there is slip of the ball on the ramp greatly affects the value of the actual compared to the theoretical.

Ruler Collision Activity Nov.- 18- 2014

Ruler Collision Activity

Objective:

   Derive an expression for finding the final height for an end of a swinging meter stick after an inelastic collision.

Set up:

We used a meter stick with a hole drilled in at the .2 m mark that can freely swing. We then have a clay piece with tape placed where it may collide with the meter stick and will get stuck to it.

Data Collection:

As a change in usual procedure, we performed the lab prior to doing the theoretical calculations. As seen in (fig. 1), we found our maximum height to be .657 m. With that data we measured the height of our zero mark relative to the ground which came to be .5 m, so according to this information the maximum height from our zero in  height is .157 m.

Figure 1. Our video capture set up and one data point at the maximum height after the collision.

To theoretically calculate our maximum height we split this problem into three phases:

I) Energy from GPE to KErot.

       This step was to find our value for omega just before collision.

II) Rotational Collision Lbefore = Lafter 

      This step assumes that the rotational momentum is conserved to find the resulting omega for both the clay and bottom of the meter stick. Since there is a fixed pivot point we do not need to find the new COM and perform the parallel axis theorem.

III) Energy from KErot to GPE since the final rotational velocity is zero

      At this final phase we determine what the final height of the meter stick will be.

In the figure below (fig 2.) we see that our calculated results came to be .131 m which is lower than our actual height by 16.5%.

Figure 2. This is my calculation work of the three phases and overall answer with the actual on the bottom.

Conclusion:
  Our calculated value was significantly lower than the actual which means there must have been a mistake in scaling during the video capture thus causing the data to read higher numbers than it should.

Moment of Inertia of a Uniform Triangle about its Center of Mass Nov.- 13- 2014

Moment of Inertia of a Uniform Triangle about its Center of Mass

Objective:

Figure 1. The Pasco set up with a triangular mass attached
whose inertia we have to find in different orientations.

Use the apparatus to calculate the inertia of the triangle at different orientations using the average angular acceleration from a hanging mass rotating the apparatus.                             Set up:                                                                    Using the apparatus similar to the set up of the previous angular acceleration lab, the only changes is a triangle and its attachment to a Pasco rotational sensor (as seen in fig 1.).

Data Collection:

 Initially we had to derive the moment of inertia for a triangle about its base center with the help of the parallel axis theorem resulting to:

I= M*b
      18
From there our theoretical values for our .455 kg (.15 m X .098 m) right triangle in both orientations are:

about COM with base b=.15 m
I= .184 kg*m^2

about COM with base b= .098 m
I= 2.428 * 10^(-4) kg*m^2

As we see, the smaller the length of the base of the triangle, the smaller its moment of inertia which coincides with our conceptual understanding of moment of inertia.

As seen in the figure above, we also used what we know about our given set up to derive how we would calculate our inertia for the triangle. Unfortunately, due to time we were only able to test for one of the two orientations of the triangle using a .4 kg weight. Using the known weight we used the mass as a source of a known torque to find our total inertia then subtract the known inertia for the disk apparatus using the value of the difference as our found triangle value of orientation about COM with b= .15 m with an I=.1843 kg*m^2

Conclusion:
With our results only ,1% off, it may be safe to assume that our derived formula may produce values very close to our theoretical values. Only sources of uncertainty may lie in friction in either pulley and the fact we did not check consistency with a second trial.

Moment of Inertia Nov. 4- 2014

Moment of Inertia Lab

Objective:

Calculate the object's mass and predict the rotational acceleration if a cart was attached to it on a string going down a frictionless slope.

Set up:

First we are given a large metal disk on a central shaft(it has a known overall mass) and a caliper to measure its volume. Second, we have a track and a cart on a light string tied to the shaft of the apparatus.

The Approach:

  We approached this problem assuming that the disk has a uniform distribution of density, and that the shaft has negligible friction. From there find the volume of the apparatus by separating it into the sections (two small cylinders that stick out of a large disk) as seen in the figure below. Once we find their volumes, we find what ratio the large disk is compared to the small cylinders since density is evenly distributed by the same ratio.

M = V               M,V= values for large disk

m     v                m,v= values for small cylinders (each)

Since V= 13.18v, then compared to 2v, V is 6.59 times larger in comparison to the sum of the cylinders. Thus, the disk's mass is 6.59 times larger resulting to mass of 3.915 kg from the total 4.615 kg.

 Now that we have the mass values, we calculate each of their inertia and add them together for a system inertia of:

(Big) + 2(small)= (.02 kg*m^2)+ 2(3.99*10^-5 kg*m^2)= 2.008*10^-2 kg*m^2

Figures 1A and B. A picture of the rolling disk apparatus and Lab work for each part including a small diagram of the
second set up for this experiment.

 With the total inertia known, there was a need to derive the predicted angular acceleration  for a cart of a mass (m) moving down a frictionless track while attached to the shaft of the rolling disk. We start off as treating this problem as a dynamic problem or unequal torque is the tension of the string times the force on the cart parallel to the track which turns out to be the sine value of the force of gravity on the cart. From there we had the tangential acceleration become terms of angular acceleration and radius of the turn shaft. The result is the following:

α= mgsinθ*r       Where m is the mass of the cart, r is radius of the shaft, θ is the angle of the slope,
  (m(r^2)-Itot)         and Itot is the total inertia of the apparatus.

We set our surface at 25 degrees with a .4 kg cart attached to the shaft. Before beginning the trial we predicted with out derived formula that the α of the cart would be -1.253 rads/s^2 and our actual came to be -1.266 rads/s^2, only 1.04% off. 
Conclusion:
It seems our derived equation is accurate enough to predict the angular acceleration of our cart set up. Any apparent source of error would source from the friction of the disk, friction of the surface and friction of the axle. 

28- Oct. - 2014 Rotation Lab

Rotational Acceleration  Lab

Objective:

Figure 1. Lab apparatus for this set up.

  To see what factors affect angular acceleration α.

Set- Up:

 As seen in (fig 1.) we have a Pasco rotational sensor with a hanging mass, 2 steel disks and an aluminum one, two torque pulleys and a Lab Pro. We exchange different parts of this set up to change different variables to see what affects angular acceleration.

Data Collection:

Figure 2. The derivation of finding the inertia of the apparatus
as well as what affects angular acceleration. 

First we derived an equation for angular acceleration (α). We approached this problem as a dynamics problem with :

unequal net torque= (inertia of the system) *α

Since our torque will be known as well as the radius of the torque pulley (r), and inertia of our system the only unknown we would have is the average angular acceleration αavg.

We found this to be:

αavg = αup + αdown = MB*(g-asys.)r

                2                  Isys.

With this we find that if we get the sum of the absolute value of each (αup,  αdown) and average it, then we get the average angular acceleration. This is useful to compare resulting α's when we altar the main variables found in the other equation that can alter angular acceleration:

• Hanging Mass
• Radius of the torque pulley
• Inertia of the apparatus

With each of these in mind we performed 6 trials each changing one of these factors (seen in fig 2.). The first three trials were purely change in mass with a single steel disk as the system's inertia. The fourth trial altered only the radius of the torque pulley. The fifth altered both radius and system's inertia by using a lighter aluminum disk of same dimensions. The last altered the radius and doubled the inertia of a steel disk by having the hanging mass overcome 2 steel disks stacked and stuck together.

Figure 2. Our results for the 6 trials of our lab with recorded angular accelerations and αavg.

Conclusion:
From this we see that the apparent that the factors we found to affect angular acceleration all seem to have interesting relationships. When one looks at the results, doubling or even tripling the mass or radius has the same result as having half or a third of the inertia showing just like in our derived equation that they are inversely related. Any source of uncertainty comes down to the apparatuses air distribution and cleanliness of the disks as well as small imperfections of reading the rotations by the machine.

9- Oct.- 2014 Momentum/ Impulse Lab

Momentum Lab

Partner: Adam Moro

Objective: 

  Show that in each scenario (both elastic and inelastic collisions) that momentum is conserved .

Set Up:
There are three set ups:

Figure 1. A set up very similar the first set up (except the extra mass)

1) We use a track and a cart with a force sensor attached to the top of it, another cart that is stationary with a springy bit extended. A motion sensor is on the opposite end of the track reversing the settings in the motion sensor so it read motion toward it as positive. You must collide the movable cart with the stationary one and record your data.

2)Repeat the first experiment, but add a 500 g weight to the movable cart.

Figure 2. Our second set up has an extra half a kg
to increase the amount of impulse in the system

3) Replace the stationary cart with a platform with molding clay on it. Attach a nail to the end of the force sensor and collide.

Data Collection:
  With each experiment with its own objectives each had a very similar approach. For each we had to find the impulse or change in momentum (p) which means there are things we have to consider:

• Since friction in this case is negligible and the surface is level, the amount of net force just before collision in each case is 0N
• The point when the cart is experiencing maximum linear force is during impact at instantaneous rest
• There is no net force after impact acting on the cart
• As we will see later, since m*vf - m*vo=  F*t where t is the how long a collision took place, it varies in different scenarios

As seen below in (fig 4) we use data such as initial and

final velocity from the collision, collision time interval

Figure 3. Our cart now has a nail at the front meant to stick
into a wad of modeling clay for an inelastic collision.

and maximum force experienced during the collision. We then compared the theoretical calculations to the actual results with calculated error for each in percent. Where we got the numbers we used for these calculations will be on the next few figures.

Figure 4. Lab Calculations for each experiment with theoretical answers compared to
actual using known mass and given velocities from our results.

7- Oct.- 2014 Magnetic Conservation of Energy lab

Magnetic Conservation of Energy Lab

By: Jordan Fuentes

Partners: Adam Moro, Jonathan Cole, Jorge Gonzales, and Andrew Tek

Objective:

  Show that energy is conserved in this set up at differing angles.

Set up:

  We use a frictionless cart with a strong magnet on one end approaches a fixed magnet of the same polarity. The track it is on would need to be leveled first then tilted at small increasing angles and a motion detector with its direction reversed so towards it is positive. 

Figure 1. The overall set up of the lab.

Data Collection:
Using the set up as described above we tried to find if the system's energy is conserved. To do so, we had to identify what sources of energy are going into the system:

• Kinetic Energy
• Gravity Potential
• Magnetic Potential

The first half of the Lab is to derive a formula for magnetic potential. For every form of potential energy there tends to be a value of displacement interacting with a force which we found to be the sine value of gravity. To find the approximate behavior of our magnetic potential we used the set up to see changes in angle of the track result to what sort of change in the distance between the cart and the magnet as the cart reaches an instantaneous stop. We then recorded each result and graphed it (as seen in fig 3.) as well as used a power fit of Ax^B due to the assumption that the graph had that sort of relationship. Our values of A and B show to be (.0002133± .0002803) and (-1.903 ± .3145) respectively. We then assume that the initial position of the cart is far enough to not be affected by the magnet and that change in gravitational potential energy is negligible. We then integrate our found power fit from infinity to the value of our separation distance (or when the value of MPE is at its maximum). We chose to integrate because of the non- constant force from the magnet. However, as shown in (fig 2.), we see that after integrating we have a Magnetic potential formula of:

A*(position of the cart - (initial position = 0))^(B+1)
                                    [B+1]

Figure 3. Our force vs. distance the carts stops (dr) with our curve fit where trial 6 (6th point) follows the power rule best and is used to test our derived formula for MPE.

Conservation Test:

Figure 4A. This is a graph of a trial at a random small angle
and each values of energy is graphed, as kinetic energy
approaches 0, magnetic potential reaches its max value.

  Our next step was to test our derived formula by using the same set up, but place at an unknown angle and graphing the energy results to see if it is conserved. After our trial, we set calculated columns for kinetic energy and slightly altered our derived equation by having its value be positive at all times to check for consistency in total energy and also put into account the separation from our motion sensor to the end of the magnet (.405m apart).

(.0002133)*(position of the cart - .405)^(-.903)
                   absolute value [-.903]

Figure 4B. Numerical evidence of what was described in
figure 3A.

 Magnetic Potential peaked at the exact point where kinetic was found to be zero at the 1.25s mark which is a sign that the system is conservative. However, the total energy in the system seemed to increase then decrease during repulsion of the cart thus suggesting a non-conservative system.

Conclusion/ source of Error:
Seeing as though the potential for gravity was ignored, it may have greatly affected the results to show that this system is in fact conserved. Behavior- wise, the graphs of kinetic and magnetic potential energy did show to be consistant to expected. Uneveness of air flow in the track and a possibly unbalanced cart may have cause friction in some parts more than most.

14- Oct.- 2014 2D Collision Lab

2-D Collision Lab

Objective: Show that momentum is conserved for a 2-D collision.

Set Up:

Figure 1. A screenshot of motion tracking
 one of our trials.

With a camera, smooth surface, and a camera set up above looking down (as seen in fig 1)we take 3 round balls 2 of which are made of steel and are similar masses while the third mass is made of aluminum with a smaller amount of mass than the steel balls but relatively the same size. We do two trials, one with two steel and another trial with one steel and the other aluminum.

Data Collection:
As we finished video recording the collisions, we tracked their path in each frame as it was graphed (as seen in Fig 2A. and 3A.) From there we created linear fits for each mass before and after collision to get an average velocity in both ends of the trial. We then made calculated columns for momentum of the system in the x and y axis at any given time and then another calculated column for kinetic energy (seen  in fig 2B, and 3B).:

1) We found that the momentum in the x and y directions:

before collision: (mass 1)*(V1 in x or y)+ (mass 2)*(V2)=
after collision:     (Vf in x or y)*(sum of both masses)

2) We also found momentum of the system to calculate average overall velocity for each mass for any given time during and after collision:

p^2 = px^2 + py^2
m*v^2 = m*vx^2 + m*vy^2

3) Then used the found velocity to find total Kinetic Energy at any given time:

KE= .5*mass 1*(v^2) + .5*mass 2*(V^2) = 5*mass 1*(vx^2 + vy^2) + .5*mass 2*(Vx^2 + Vy^2)

Our first trial with the two steel balls show that resulting velocities after collision are close to parallel since they are the same mass (seen in Fig 2A.). However, possibly due to spin on the ball, it does not seem to show much consistency or conservation in the Kinetic energy column. Nevertheless, it does show consistency in in each momentum column (as seen in Fig 2B.). In trial two since the aluminum ball was initially in motion in this collision, it makes little change to the velocity of the steel ball it collides with and does not seem to have a great change in velocity (seen in 3A.). Numerically, it seems to show much more conservation in kinetic energy compared to trial one while still maintaining consistency in momentum of the system (seen in Fig 3B.).

Figure 2A. Trial one positions before and after collision with linear fit graphs to
find each of the average velocities for calculating momentum in both x and y directions.

Figure 2A. Our numerical results for our first trial for momentum in both the x and y directions as well as our calculated kinetic energy at any given time.

Figure 3A. Trial two positions before and after collision with linear fit graphs to 
find each of the average velocities for calculating momentum in both x and y directions.

Figure 2B. Our numerical results for our second trial for momentum in both the x and y directions as well as our calculated kinetic energy at any given time

Conclusion:

This lab for the most part shows that in both systems there is a conservation in momentum, but conservation in kinetic energy is only found in trial two. Initially it was suspected it was due to misreadings of masses, but we remeasured our masses and found them to be the same once again. We believe that any source of uncertainty would be from spin of the ball, imperfections of the ball, imperfections of the surface and approximated motion tracking credited to human error.

2- Oct.- 2014 Total Energy Lab

Total Energy Lab

Objective:

  Show that energy in a system is always conserved when one adds all forms of energy during each point in time and the sum stays consistent.

Set- up:

  In this set up we used a mass of 600g attached to an end of a hanging spring (as seen in fig. 1a- 1b). Underneath the hanging spring we put a motion detector with its settings set to zero when the spring with the mass is at a relaxed position and its readings reversed so toward the sensor is positive.

Data Collection:

Figure 1a. Our mass on the hanging spring

We set the weight at a height where the spring's length was the same as it was at a relaxed state without a weight attached to it. From there we recorded the data for position, velocity and acceleration for a trial of  seconds. From there we used that information to numerically calculate different components of what we thought was what made up the system's energy. As a clsass we observed that there were 5 forms of energy in the system that is not negligible (as seen in fig 2.):

Kinetic Energy of the Mass (m): .5*m*v(mass)^2

Gravity Potential of the Mass (m): g*m*(position)
where mass (m)=  .6 Kg

Elastic Potential of Spring: .5*k*(position- relaxed)^2

we calculated spring constant k to be :
F=k*(stretch)  where stretch= x = 25 cm

Figure 1b. Our motion sensor with appropriate
settings placed directly under the hanging mass
set up. 

Ms*g= k*x where Ms = .1 Kg
k= m*g = 23.52 spring constant
        x
Kinetic Energy of Spring (Ms)= (Ms)*V^2
                                                           6
Gravity Potential of Spring (Ms) = Ms*g*position*(.5)

To prove that energy of this system is conserved, we add up all the forms of energy and we then see if the sum is consistent. Numerically it is apparently close and evidence of this can also be seen in fig 3. where each form of energy is graphed and the orange graph on the top is the sum.

Conclusion:
Energy is in fact conserved in this system. My level of uncertainty is around 3% seeing as not the entire spring was uniform, but behaved consistent enough during the trial to complete the objective.

Figure 2. Each of our calculated values for different forms
of energy found in the set up with an approximated sum
of all the energy.

Figure 3. Graphed values our our calculated energies with the
sum of the energy values for each point in time graphed at the
top in orange with consistency.