Monday, November 28, 2016

Lab 17 Finding the moment of inertia of a uniform triangle

LAB 17
Finding the moment of inertia of a uniform triangle 

Name: Carlos Hernandez
Partners: Bmaya, Mohammed

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

Background Info and Approach: To be clear, our physics utensils that are needed to do the lab include, comprehension of torque and the parallel axis theorem. The Parallel axis theorem was used on a notebook problem to find the moment of inertia around a triangle from the center of mass. 

Procedure: What we would like to reach at the end of the lab is a comparison of the inertia of a triangle between our experimental data(through our apparatuses) and our formula for the triangle which we have previously found to be 1/18*M*B^2. The set up to retrieve our experimental data is almost exactly the same as Lab 16. A photo of it below. 
-The main and only difference of this set p is the ability to put the triangle as shown above, directly on the center of mass. 

We will collect data 3 times. The first without the triangle, the second as shown above with the longest side going up with its 90 degrees, and the third time with the longer side now parallel to the table. Below will be the data collected with the 3 experiments. 


 The same process was used to get the omega. Finding the slope of the graph. Both going up and down and averaging them





After having our omega values, the first step was to calculate the value for inertia for all 3 trials. Later on the Inertia of the system with the triangles will be subtracted from the system without a triangle, to find the inertia of just the triangle and that's what we will use to compare. Below is a photo of the calculation of the systems inertia for all 3 cases.

Now we are able to compare. For the experimental we will subtract the triangle system inertia by the solo system. And that shall give us the inertia of the triangle. That was done for both of the triangles positions. Then we compared it to the inertia by the formula 1/18MB^2 Derived by the parallel axis theorem. Below is the work and values for each triangle and both experimentally and through the Inertia formula. 
As one might see the values are actually very very close.
Triangle long side parallel to the ground : Exp-.00053 , IFormula-.00056
Triangle short side parallel to the ground : Exp- .00023, IFormula-.00024

Conclusion: Hurray! The Experiment is a success, the numbers are at around 5% error, I believe there are very few places to blame for error as there is very few places for humans to manipulate the actual system. A place for error might be that the system is completely friction-less or the other is the calculation of omega, we did average out that calculation. Nonetheless, Our comparisons were great and our understanding of inertia is ever more clear. Cheers. 

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