Lab 17 Moment of Inertia of Uniform Triangle about Center of Mass

Partners:
Date of Lab: 2 November 2016 

Lab 16 Angular Acceleration

Mission Statement:To determine the moment of inertia of a uniform right triangular thin plate around its center of mass for two perpendicular orientations of the triangle. Then to compare the experimental and theoretical results for the moment of inertia of the triangle for each of the two orientations. 

Theory and Experimental Procedure: 
 The apparatus was composed of two spinning disks next to a Pasco rotational sensor. We connected a torque pulley to the top of the two disks with a string strung around it. The opposite end of the string slides over another pulley and dangles off the edge of the table with a hanging mass. The aforementioned triangle was then mounted to the top of the torque pulley by slotting into a protruding holder rod. Activating the compressed air permits the disks to rotate without friction, in turn prompting the hanging mass to oscillate up and down.

In essence, this experiment tested our understanding of the parallel axis theorem, which states the following:

I(parallel axis) = I(axis through center of mass) + M*d^2, where d is the displacement from the original cm and M is the mass of the object in question.

Since the limits of integration are more straightforward if we first find the moment of inertia about a vertical edge of the triangle, we can calculate said moment of inertia and then obtain the moment of inertia around the center of mass by rearranging the relationship above into the relationship below.

I(around cm) = I(around one vertical end of the triangle) + M*d^2

With this in mind, we needed to derive the moment of inertia of the triangle around its center of mass and then utilize the parallel axis theorem to identify the moment of inertia around its new axis. My  derivation for this is shown below.

 The desired moment of inertia of the triangle resulted in being I = 1/18*M*R^2. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. These came out to be 0.455 kg, 0.098 m, and 0.14950 m, respectively. The base and height length would depend on the  orientation of the triangle.

Now commenced the experimental portion of the lab. The moment of inertia of the whole system was solved for by knowing the torque exerted by the tension from the hanging mass together with the angular acceleration of the system which was produced by said torque. The connection between these qualities is shown below. 

Torque = Tension*radius = I*alpha  

Hence, by obtaining the total moment of inertia of the system and then proceeding to subtract from it  the moment of inertia of the disk/holder subsystem, we can yield the moment of inertia of just the triangle. This moment of inertia was henceforth our experimental value of I.

However, since there was some unknown amount of  frictional torque in the system, namely due to the the disk not being completely frictionless or the presumed-to-be-frictionless pulley nevertheless not being massless, the angular acceleration of the system during the descent of the hanging mass would not be equal to the alpha of ascension. Hence we were obliged to derive an equation which identified this mysterious frictional torque. The derivation is shown below

 Armed with this information, we proceeded to document the data of the angular acceleration for only the disk/holder system. Much like in prior labs, the angular acceleration was equivalent to the the slope of the angular velocity vs. time graph, shown below.

 We first finding the average upward angular acceleration, then the corresponding downward counterpart, and then the average of the two. Putting our derived equation into practice, making sure to use the proper mass and base/height values, I was thus able to calculate the experimental moment of inertia of the disk/holder system, shown below.

 Next, I positioned the triangle while oriented vertically and commenced the experiment anew, with the respective omega v time graph shown below.

Getting the average acceleration, I was able to calculate the moment of inertia for the vertical system, shown below.

Following the professor's suggestion, by subtracting my disk/holder/triangle system inertia from the disk/holder system inertia, I was able to obtain the experimental moment of inertia of only the triangle which resulted in being 2.09x10^-4 kg*m^2. After using the theoretical equation we derived earlier, we were able to say that the theoretical moment of inertia should be 2.44x10^-4 kg*m^2. Our percent error came out to 11.3%

Finally, I flipped the triangle so that the longer side was horizontal and rinsed and repeated.

Using the above average angular acceleration, I was able to calculate the triangle's experimental moment of inertia when horizontal, shown below.

Contrasting the experimental value for moment of inertia of the triangle (horizontal) I =5.32x10^-4 with the theoretical value I=5.65x10^-4 yielded a percent error of 6.84%.

Conclusion: Equipped with the Parallel Axis Theorem, I was able to derive an equation for the triangle's moment of inertia around choice axes of rotation. Using my hand-measured (and thus susceptible to human error) data of the disks' and triangle's dimensions, I was able to calculate the experimental value for the triangle's moment of inertia. Then, by juxtaposing my experimental results for moment of inertia with the theoretical value obtained through my derived equation, I was able to determine how precise we were able to get using our understanding of newton's second law.. My results, even though the percent error was larger than desired, were still close enough to verify our experiment.

Errors occurred during our experiment beginning with the fact that our vertical triangle was not Truly vertical.  Other things such as the friction from not having a perfectly clean disk, or non-uniform disk, and slight error in measurements of the triangle may have skewed our results a bit.