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. |
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.


No comments:
Post a Comment