December- 4 2014 Semicircle Physical Pendulum

Semicircle Physical Pendulum 

Objective:
  Find the period of a half circle mass acting as a physical pendulum at two different orientations.

Set up:
  As seen in the photo to the right, we (each group) had to cut semicircles our of a light material and use calipers to measure its dimensions and weight its mass with scales. We then attached light rings to the edge of orientations and record its motion at a small angle.

Data Collection:
  Firstly we had to find the moment of inertia of the center of mass of the object, but in order to do that we had to find the moment of inertia about the flat side of the semicircle and the location of the Center of Mass to apply the parallel axis theorem:
Iflat= ICOM + m(d^2)
ICOM= Iflat  - m(d^2)

As seen below, we found the center of mass first which is at the middle of the x axis parallel to the flat side and (4R/3pi) to the y direction toward the curved side. Then using what we found our moment of inertia to be about the flat side and subtracting according to parallel axis theorem we found the moment of inertia about its center of mass. We used this to then find our moment of inertia about the middle of the curved edge of the semicircle. We found that about the flat side has less inertia than id the semicircle were rotating about the curved side.

From there we set up our scenario as a dynamics problem of unequal non- constant torque to meet the form of simple harmonic motion assuming a small theta or angle. We found that our theoretical calculation for period if it were turning on its flat side is less than 1% off as is expected for using a taylor series on the sine value to assume it to be just theta. However in the actual for the curved side rotating, the period is exactly the same as if it were to be about its flat side which the theoretical calculations did not suggest however we were only 1.9% off.

Conclusion:
It seems friction or damping of the system may have altered the results. Another possibility is that we may have had a too large of a theta during our test for the curved side of the circle trial otherwise the sensor may not have read such a high average for its period.

Mass Spring Oscillations Lab Nov.- 20- 2014

Mass Spring Oscillations Lab

Objective: Find the relationship between period and spring constant as well as with the amount of mass that is oscillating.

Set up: There were a total of four groups each with a different spring, the appropriate mass so that way each group would have the same amount of oscillating mass determined by the equation (mosc= mhang+ mspring/3). We used a motion sensor and lab Pro to graph our results and place an appropriate fit.

Data Collection: The first step for our group to do is to find the spring mass of our given spring. To do so we weighed the spring for its mass, We then hung the spring to measure its length at a relaxed state, then once more with a .1 kg mass hanging at the end without oscillation:

Fnet= 0N= mg - kxstretch               xstretch= x- xo = .067 m

                                                   m= 26 g

mg= k(x -xo) => k= mg              k= 14.64 N/m

                              (x- xo)

We then went on to derive our equation for omega and, eventually, for our period using Newton's second Law as seen in (fig. 1).

Figure 1. All my derivations and notes from the set up as well as comparisons to our actual
unfortunately there is a mistake in the percentage comparison on theoretical period to actual.

Figure 2. Results of our 4 trials to find the average period for the set up.

We now had our theoretical equation for period of our set up and we tested with 4 trials for consistency. We counted the periods with in a 5 second trial and found that our theoretical calculation was almost 100% accurate to our actual. When we graphed our results compared to others and graphed them with an appropriate curve fit we find something interesting. First, how we analytically found the curve fit to be a power fit was this observation:

Our period came to a form of:

T= 2pi*(m/k)^.5

It seems to follow the power fit seeing as there is a constant followed by a value with a variable (either k or m) multiplied to a certain power. Units -wise is simple, since the constant is unitless or in radians so all that remains is to show that what's in the parenthesis needs to be s^2:

T= C*[(kg )]^.5 => [(kg*m) ]^.5        =>  [1 * (s^2)]^.5 = s

          [N/m ]            [kg*m/(s^2)]             [1       1   ]

Seeing this sort of relationship with mass and spring constant we can predict as k=> infinity with a constant mosc T would approach a period of zero at an curve similar to y= (1/x)^.5 seeing as k being in the denominator. However, in that respect, if mosc => infinity with a constant k then T will also approach infinity at an inverse rate compared to the predicted behavior of T vs. k, y= (x)^.5 , seeing as m being in the numerator. According to the graphs below with appropriate curve fit it seems as though our predictions were correct.

Figure 2A. Our graph of T vs. oscillating mass with appropriate curve fit.

Figure 2B. Our T vs. k graph with appropriate powerfit.

Lastly, there was the regard to see how much a 5% spring constant error would affect a calculated period in which for our set up we found the error to be around only 3% difference compared to both actual and correctly calculated spring constant as seen in (fig 1.).

Conclusion:
We found the our source of any uncertainty would lie in the calculations of other groups as well as any inconsistencies in the spring seeing as it swung side to side shortly after the 5 seconds of our trials.

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.