13-May-2015: Moment of Inertia of a Uniform Triangle

The purpose of this lab was to find the moment of inertia of a triangle about its center of mass using physical methods.

Part 1: Deriving our Theoretical Value:
To begin the lab, we first wanted to find an expression for the moment of inertia of a uniform triangle by derivation. To do this, we took two approaches.

In the first approach, we considered the triangle to rotate about the edge of its base. We found an expression for this moment of inertia and then used the parallel axis theorem(I_cm=I_{around one edge} + M(d_{parallel axis displacement})²) to find its moment of inertia about its center of mass. (see Fig. 1).

Fig. 1
Although the formula uses "b" as its term for the side perpendicular to the axis of rotation, this is actually the "height" of the triangle.

In the second approach, we considered the triangle to rotate about the edge of its height. We used the same method as we did in the first approach. (see Fig. 2)

Fig. 2
Using the same method as before, we altered the axis at which the triangle rotated.

Once we had derived from both approaches, we found that our expression for the moment of inertia about the center of mass of the triangle was going to be I= (1/18)Mb^2, where "b" is the base perpendicular to the axis of rotation.

^{Part 2: Physically Finding the Moment of Inertia:Once we had our expression, it was now time to actually find the moment of inertia using the apparatus from our previous angular acceleration lab}. Using the formula I=(mgr/a)-mr^2. (a=angular acceleration), we could find the moment of inertia of the apparatus in three separate trials. We used a hanging mass of 25.0 g, and a torque pulley of 2.51 cm radius. The only information we needed was the mass that was being dropped and the radius of the torque pulley, given that LoggerPro would give us the angular acceleration of the system. 

The idea is that if we find the moment of inertia of the system by itself, and then the moment inertia of the system and the triangle (orientated either way), we could find the moment of inertia of the triangle by taking the difference of the two. 

We ran three trials: One without the triangle (moment of inertia of just the system), one with the triangle long-ways up (moment of inertia of the system and vertical triangle- Fig. 3), and one with the triangle long-ways down (moment of inertia of the system and horizontal triangle- Fig. 4). In each setup, we wrapped a string around the torque pulley and allowed the system to free fall. Using LoggerPro, we were able to find the angular acceleration of the system by taking the slope of the angular velocity vs time graph. However, since there is some friction in the system, we had to take the average value of angular acceleration (going up and down) as our usable value  (Fig. 5)

Fig. 3
We placed the triangle vertically up, giving us a specific value for the moment of inertia.

Fig. 4
The moment of inertia of the triangle would be the difference of its combined inertia with the system and the inertia of the system by itself.

Fig. 5-1
We needed to find the average angular acceleration in order to attempt to cancel out the torque due to friction.

Fig. 5-2

Once we had all of our angular accelerations, it was time to begin calculating the moments of inertia of each setup. (see Fig. 6) 

Fig. 6

We were left with a final moment of inertia of the center of mass of a triangle rotating about its vertical height being .00023902 kg*m^2. The moment of inertia of the center f mass of a triangle rotating about its horizontal base was .0005586 kg*m^2.

Once we had our theoretical values, it was time to find the percent error from the actual values. With our equations from before, we measured the height of the triangle to be 14.936 cm, its base to be 9.844 cm, and its mass to be 456 g. (see Fig. 7)

Fig. 7
The low % error showed that our expression for the moment of inertia was legitimate. 

We found that our percent error was less than 3% for both cases, which indicated that the experiment in all was a success and our expression for the moment of inertia of a uniform triangle was good in both cases.

Conclusion:
By deriving an expression for the moment of inertia of a uniform triangle around some axis and then using the parallel axis to find it about its center of mass, we were able to find a good prediction of a real physical attribute of this rotating triangle. Considering that the main source of uncertainty would be in the case of friction altering our angular acceleration, I believe that our % error being less than 3% is very reasonable.