May 13th, 2015: Lab 17: Moment of Inertia (at center of mass) of a Uniform Triangle

Purpose:

Determine I-cm of a triangle by using Parallel Axis Theorem and compare this to the experimental value obtained from the measured angular acceleration.

Experiment:

- Use the same tools from angular acceleration lab. Ideally, use the same kit from last lab so the measurements of rotational disks, hanging mass, and torque pulleys stay true for this lab as well.
- Measure the mass and dimensions of the triangle.

- Make sure the platform is leveled. Wrap the string-hanging mass around large torque pulley till the mass is at maximum height. On top of the pulley, screw in the metal rod, as shown below.

- Set up Loggerpro and other needed equipment just like angular acceleration lab. Make sure the sensor setting is 200 counts/revolution.
- Turn on the compressed air, so the the two disks turn independently. Release the disks and let them rotate. Collect data on Logger pro. Let the hanging mass move up and down three times so there's 3 α-up and 3 α-down. We're going to take average of these to omit any friction in the system.
- Repeat the same steps to taking data for system with the triangle in two different orientations as shown below,

               

            

Data and Analysis:

Theoretical inertia of the triangle:

- Coming up with an expression for an inertia of a triangle spinning around its edge is easier because we can take "dm" of the triangle and claim that each of that strip has dI = (1/3)*dm* x^2. Therefore we get the inertia to be (1/6)MB^2. The Parallel Axis Theorem allows us to move axis of rotation to the triangle's center of mass. Using equation for center of mass, we get the distance from center of mass of x-axis to be 1/3 of the base.

Calculation for center of mass of a uniform triangle

- Since it is a uniform triangle, we can use area density to express "dm".

- We can repeat the same process for another orientation although we have to keep the variables unchanged for each dimension. Using the derived equations, we could plug in the mass, base and height that we measure to get "I-cm" for both cases.

Experimental moment of inertia for the triangle:
Data:

Triangle's mass= 0.454 +/- 0.001  kg

base= 0.0987 +/- 0.00001 m

height= 0.14756 +/- 0.00001 m

hanging mass = 0.0245 +/- 0.001 kg

r-large torque pulley= 0.025 +/- 0.001 m

- Since the descending and ascending angular acceleration are not the same, we have to take their average so that we don't have to deal with frictional torque in the system.
- The way we obtain α is the same; take linear fit of omega vs. t graph.
- Due to the existing inertia in the rod used to support the triangle, we have to calculate it separately then subtract from the inertia of triangle+rod. α is also partially determined by the inertia of the spinning disk, so we don't have to calculate that inertia separately. The only inertia we're adding to the system is the triangle.
- Shown below, in order, α of system with rod only, horizontal, and vertical orientation:

- Take the average of these 3 cases give us,

Ascending α has (-) sign so we have to take absolute value

- We then use this obtained value for the equation derived below. 

       

- Moment of inertia is inversely proportional to α.

Conclusion:

Our error and uncertainty is rather large. Mass measurement would be one of the biggest contribution here since our equipment's uncertainty is 1 gram. As far as collecting α is concerned, friction due to the rotating disks is our most reasonable explanation for error because a slight change in α, even to 2 decimal places, change our error calculation by significant amount. Another source might also be from the amount of compressed air that was in the system. Too little air, indirectly, can cause α to be smaller. Nonetheless, within our range of error and uncertainty, our experimental calculations are close to calculations using Parallel Axis Theorem.