PURPOSE:
We are trying to find a relationship between mass and period for an inertial pendulum. Furthermore, we want to use this information to create an equation that will accurately predict the mass of any object we place on top of this pendulum.
INTRODUCTION:
Our task is to accurately measure the mass of a given object. The only caveat is that we are not allowed to use a spring scale, balance, or any other measuring device that utilizes the Earth's gravitational pull in order to determine mass, Therefore, we must make use of indirect methods to acquire this measurement. More specifically, we must make use of an inertial balance to measure inertial mass by comparing an object's resistances to changes in motion. Firstly, we will use the inertial pendulum to directly measure the periods of a set of pre-determined masses. Secondly, we will relate these quantities using a power-type equation. Thirdly, we will utilize Logger Pro to help us determine any unknown variables in our equation. Lastly, we will use this mathematical model to calculate the mass of an unknown object.
APPARATUS:
1) We used a C-clamp to secure the inertial pendulum (tray) to our table. Then, we put a thin piece of tape on the end of the tray.
2) We set up a photogate so that the thin piece of tape would pass completely through the beam of the photogate whenever the tray oscillates. The photogate is what allows us to measure the period of the oscillating pendulum.
3) We set up LabPro on our computer so we could record the period in real-time.
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| the fully assembly apparatus |
PROCEDURE:
1) We recorded the period with no mass on the tray.
2) We recorded the period with 100 g on the tray. We went all the way to 800 g, in 100 g increments. Here is a picture of one of our measurements:
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| Graph of period measurements with respect to time |
3) After each measurement, we recorded our data in a table:
4) Next, we related mass and period using the power-type equation:
T = A(m+Mtray)^n (We used m+Mtray because the mass of the tray is always present.)
This gives us three unknowns we need to solve for: A, Mtray, and n.
5) We took the natural log of both sides of the aforementioned equation to obtain:
ln T = n ln(m +Mtray) + ln A
Since it looks very similar to the linear form y = mx +b, we can plot it in LabPro as such.
6) We made a plot of ln T vs. ln(m +Mtray). This gave us a line of slope n and a y-intercept equal to log A (assuming our value of Mtray is correct).
Unfortunately, our group's initial value for Mtray was not so exact. In order to determine the correct Mtray, we needed to do a linear fit of our data and try different values for Mtray until we achieved a perfectly straight line plot. To get a straight line, we needed to make sure that the correlation coefficient was as close to 1.000 as possible. The following two graphs display linear fits of the aforementioned equation whose values of Mtray are consistent with a correlation value ~1.000.
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| Graph of ln T vs. lm (m +Mtray) -- Lower Bound |
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| Graph of ln T vs. lm (m +Mtray) -- Upper Bound |
Based on these graphs, our group was able to discern that there was actually a range of values for which the correlation coefficient would be very close to 1. This would seem to suggest that there are a number of plausible values for Mtray--when in reality, the mass of the tray is constant. This indicates that our mathematical model has some inherent uncertainty that could affect the accuracy of our calculations when we attempt to determine the masses of our unknown objects.
EXTENSION:
Using the values of A and n that correspond to our upper and lower bounds of Mtray, we finished constructing our mathematical model and used the resulting equation:
e^[(ln T - ln A)/n] - Mtray(1,2)
to determine the mass of two objects: a smartphone and a stapler. Below is a table of the information we collected from the graphs. Mtray1 represents the lower bound while Mtray2 represents the upper bound. The measured mass is the mass of the objects recorded from a scale. We used these values as a point of reference for the accuracy of our calculated mass.
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| ln A and n values collected from linear fit graphs (top). Period (T) and mass (g) collected for each object (bottom). |
CONCLUSIONS:
Overall, it appears that our calculations for the mass of the two objects were not as accurate as we were hoping for. For example. the lower bound values yielded a calculated mass for our first object (smartphone) that was about 6 grams higher than the gravitational mass we measured on the scale. In other words, the actual mass of the phone would not even be within the range of values we calculated. Moreover, a similar trend emerged when we calculated a range of values for the mass of the second object--a stapler. In fact, according to our calculations, the mass of the stapler ranged anywhere from 374.0 g to 375.8 g. However, when we went to determine the mass on the scale, it read 369.0 g.This uncertainty could be the result of several factors. For instance, we could have recorded a more accurate measurement for the period of the object instead of leaving it at one significant figure. Moreover, this error could have also resulted from the inconsistent placement of objects on the tray. Not every object is built the same way, thus, the weight of said object may not always sit evenly distributed on the tray. For instance, a majority of the stapler hung off the sides of the inertial pendulum because it was far too large to fit inside the tray itself. This could lead to inaccurate measurements for the period of the object, and--by extension--the mass of the object as well. Furthermore, as was previously stated, we utilized a range of values for Mtray, which had some inherent uncertainty in our calculations for the masses of the objects we chose,
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