May 22, 2017 Lab 17: Moment of Inertia of a Uniform Triangle About Its Center of Mass

Lab 17: Moment of Inertia of a Uniform Triangle About Its Center of Mass

By: Chris Ceron, John Choi, Amy Chung

May 22, 2017

Goal of Lab:
Determine the moment of inertia of a right triangular thin plate around its center of mass with two different orientation, one with the long side facing down and the other with the short side facing down (Fig. 1 and 2 show the set up and triangle).

The triangle we used.

Theory/Introduction:
In order to find the moment of inertia of the thing right triangle plate, we first find the inertia about the edge and use the parallel axis theorem to derive an equation for the inertia at the center of mass. The shift in distance is found using the x center of mass position:



the relation between x and y can be found using the linear line, y = -(H/B)x + H



the x center of mass equation is derived by:



and the inertia about the edge can be found using:



so using the parallel axis theorem we find that the inertia about the center of mass is:



In our experiment, there is a weight of the pin that holds the triangle, hence it has a inertia value. Since we want to isolate the triangle, we first find the moment of inertia of the system and subtract if by the moment of inertia of the pin, leaving us with the moment of inertia of the triangle.

Since the setup isn't like physics problems where pulleys are frictionless and massless, we need to take into account the frictional torque within the system. The derivation is done exactly the same as from previous lab on angular acceleration. The final derivation for Inertia came out to be:



Once we find the experimental and theoretical data, we can check the % error and discuss reasons for discrepancy in data.

Apparatus and Procedure:

The apparatus with just the pin.

The setup looks identical to lab 16 for angular momentum but the disk now has a pin that holds a thin triangular plate. We connect the system to LoggerPro and find the angular acceleration going up and going down by finding the slope of the angular velocity reading on the graph.
We first find the inertia value of just the pin since the mass that it has also contributes to total inertia. Then we find the angular acceleration up and down values on one side of the triangle, we repeat the process with the triangle on its other side and for each of the inertia values we calculate, we subtract the inertia of just the pin.

Data:
The triangle we used had height 149.2 mm and base 98.2 mm. The radius of the pulley was 49.8 mm.

The mass of the hanging mass was 0.02457 kg.

The mass of the triangle was 0.462 kg.

	|α down| (rad/s^2)	|α up| (rad/s^2)
Pin	6.536	5.967
Triangle Base	5.258	4.674
Triangle Height	4.223	3.784

Alpha down for pin.

Alpha up for the pin.

Alpha down for the triangle with base down.

Alpha up for the triangle with base down.

Alpha down for the triangle with height down.

Alpha up for the triangle with height down.

Calculated Data:

Calculation of Inertia:

	Pin	Triangle Base	Triangle Height
Inertia (kg*m^2)	0.00104	0.00131	0.00163

Triangle base = 0.00027

Triangle height = 0.00059

The theoretical inertia for base is found by:

and for the height side down:

Conclusion:

	Experimental	Theoretical	% Error
Triangle Base	0.00027	0.000248	8.87%
Triangle Height	0.00059	0.000571	3.3%

Looking at the % error we found, the results were accurate. Sources of error could be from small uncertainty values in our measurement of mass, base and height. Our graphs did not have good correlation values for angular acceleration. There is also some small friction present in the pulley that the string goes over that we ignored.