11/14/16 - Moment of Inertia of a uniform triangle (Lab 17)

GROUP MEMBERS: Christian R. and Haokun Z.

PURPOSE: 
To find the moment of inertia of a uniform triangle about its center of mass, for two perpendicular orientations of the triangle.

THEORY: 
The parallel axis theorem states that Iparallel axis = Iaround cm + Md^2.  Since the limits of integration are much simpler if we calculate the moment of inertia around the vertical edge of the triangle, we can calculate the moment of inertia in this case, and then solve for Icm from Iaround cm = Iparallel axis - Md^2.
APPARATUS:
Note: This is the same setup that we used for the Rotational Acceleration Lab.
The apparatus shown below is attached to a Lab/Logger Pro setup connected to a computer.

Triangle - Long side up

Triangle - Long side down

EXPERIMENTAL PROCEDURE:
For each orientation of the triangle, we used the Pasco rotational sensor to measure the angular acceleration of the system (just like in the Rotational Acceleration Lab). In a similar fashion, we used this information to calculate the moment of inertia for each case: with triangle - long side up, with triangle - long side down, and without triangle.

3 cases - Moment of Inertia of the System

DATA/GRAPHS:
mass of triangle = 456.9 +/- 0.1 g
mass hanging = 24.6 +/- 0.1 g

long side of triangle = 149.5 +/- 0.1 mm
short side of triangle = 98.5 +/- 0.1 mm

diameter of pulley = 49.7 +/- 0.1 mm

Trial 1 - Angular Acceleration Graph

Angular Acceleration Data

ANALYSIS:
After measuring the angular acceleration of the system for three different cases, we were able to calculate the moment of inertia of the system with the triangle oriented in two different ways and without the triangle. The equation we used for the experimental moment of inertia was derived in part 2 of the Rotational Acceleration Lab. Moreover, the reason we measured the moment of inertia of the system without the triangle was because the peg that held the triangle in place also has some rotational inertia. Therefore, the difference between the I of the system with the triangle minus the I of the system without the triangle (peg only) will yield the experimental I of the triangle. Furthermore, in order to find an expression for the theoretical I of the triangle, we decided to use the parallel axis theorem, which yielded Itriangle = (1/18)MB^2. The beauty of this equation is its simplicity because if we want to find the moment of inertia of the triangle with a 90 degree rotation, then we merely change the value for our base.

Triangle - cm and moment of inertia around vertical edge

Triangle - Moment of inertia around cm

Sample Calculation - Rotational Inertia of Triangle

Results - Rotational Inertia of Triangle

CONCLUSION:
Overall, our lab was a success. We were able to mathematically express the moment of inertia of a uniform triangular plate using calculus and experimental means as well. For the triangle - long side down, we found that our experimental and theoretical values  were 5% different from one another. For the triangle - long side up, we discovered that there was a 14% difference between our theoretical and experimental values. This experimental error was most likely due to frictional torque in our rotating system. This frictional torque indicates that our angular acceleration will be different when the hanging mass is ascending and descending. Therefore, the average acceleration we used in our calculation is merely an approximation and adds some inherent uncertainty to our rotational inertia calculation. Another possible source of error was slight misreadings in the measurements of critical components, such as the mass of the triangle, the radius of the pulley, the length and width of the triangle.