5/13/15: Moment of Inertia of a Uniform Triangle Lab 17

Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. We will compare an experimental value of moment of inertia to a theoretical value of moment of inertia. 

Procedure / Set Up:

The set up consists of 2 stacked disks that are placed adjacent to a Pasco rotational sensor. On top of the disks is a torque pulley with a string wrapped around it as well as triangle mounted on a holder. The string is attached to a hanging mass which goes over another pulley at the edge of the apparatus. When the compressed air is turned on, the disks rotate independently and the hanging mass moves up and down.

Approach - Theory:

We want to calculate the moment of inertia about the center of mass of a thin triangular plate, but because the limits of integration are simpler if we calculate the moment of inertia around a vertical end of the triangle, we can calculate that moment of inertia and use the parallel axis theorem to then get the moment of inertia around the center of mass.

The parallel axis-theorem is give by:

Since we will be calculating the moment of inertia of the triangular plate in two different orientations, upright and sideways, the tension in the string will exert a torque on the pulley-disk combination and by measuring the angular acceleration of the system we can determine the moment of inertia of the system. We will accomplish this by measuring α of the (disk + holder), α of (disk + holder + triangle), determining I of (disk + holder), I of (disk + holder + triangle), and the difference between the two moments of inertia will give us the moment of inertia of the triangle. These will be our experimental values for I. In total, we will compare four different values.  

Before beginning the experiment, we derived and expression for (I around cm), which will be our theoretical value. By using the parallel axis theorem, I = 1/18 ML^2
This equation for I will be the same for both the upright and sideways orientations. 
where M = 0.455 g, L = 0.0983 m (upright), L = 0.1493 m (sideways)

Because of frictional torque in the system, the rotating disk isn't truly frictionless and there is mass is the frictionless pulley. We derived an expression for the the moment of inertia of the system following this approach back in the angular acceleration lab. The derivation is shown below. The boxed equation on the bottom left is the one to notice.

Now we can measure α of the (disk + holder), α of (disk + holder + triangle), and determine I of (disk + holder) and I of (disk + holder + triangle).

Upright Triangle:

How the setup should look at this point for α of the (disk + holder).

α of the (disk + holder) was determined by allowing the mass to rise up and down while Logger Pro generated the angular velocity graph.The slope of ω vs. t would give us the α. 

An average was taken of α up and α down in order to find α of the system. I of (disk + holder) was then calculated as shown below. I = 9.237 x 10^-4 kg*m^2

Using the same logic as above, I of (disk + holder + triangle) was calculated as shown below.

Graph of ω vs. t for (disk + holder + triangle) to find α

Using α to find I (disk + holder + triangle) calculation.

The difference between the two moments of inertia will give us the experimental moment of inertia of the upright triangle. I = 2.482 x 10^-4 kg*m^2. Using the theoretical derivation I = 1/18 ML^2, we concluded I = 2.44 x 10^-4 kg*m^2. Comparing these two values, we get a percent error of 1.721 %.

Sideways triangle: 

How to set up should look:

Once again α of the (disk + holder) was determined by allowing the mass to rise up and down while Logger Pro generated the angular velocity graph.The slope of ω vs. t would give us the α. I (disk + holder) is the same value as was in the upright triangle I = 9.237 x 10^-4 kg*m^2.

The same logic was applied as from the upright triangle. α of the system (disk + holder + triangle) was determined by taking an average of α up and α down from the slope of the ω vs. t graph.

The difference of I of (disk + holder) and I of (disk + holder + triangle) would give us the experimental value of I for the sideways triangle. Below are the calculations.

By comparing the experimental value I = 5.587 x 10^-4 kg*m^2 and the theoretical value I = 5.634 x 10^-4 kg*m^2, we se we have a percent error of  0.834 %.

Conclusion: 

Our goal was to determine the moment of inertia of a triangular plate in two different orientations. We were able to achieve this by using the parallel axis theorem. We derived an expression for the moment of inertia, measured α of the system, calculated I of the system, and lastly compared an experimental value to a theoretical value. Our percent error for both orientations was relatively low, so it is safe to say the experiment was a success. An error we encountered at first was that our percent error for the sideways triangle was relatively high, but it was easily fixed by remeasuring the length of one of the sides of our triangle.