Lab 18: Moment of Inertia of a Triangle

In Lab 16, we used a rotary device that pushes air through a pair of disks, minimizing frictional torque, and allowing the disks to spin for a long time, almost unimpeded. We were able to discover torque from the tension of the mass and angular acceleration, setting up the equation to solve for inertia, but we did not go through with it. In Lab 17, we used a disk that did not push air through it, so we had to solve for frictional torque. Here, we will use the air disk device again to simplify calculations, and use measured angular acceleration to actually solve for inertia of the disks, and any object attached to them. In this case, we use a uniform metal triangular plate rotated about its center of mass.

There are two ways to affix the triangle on the rotating mechanism. We will call the first the "vertical" orientation:

And this we call the "horizontal" orientation:

Our model presents this triangle very graciously.

Before we experiment, it is possible to solve for the inertia of the triangle analytically. We find it easiest to solve for the inertia from one edge, then work backwards with the parallel axis theorem to find the inertia about the center of mass:

Whew! So, apparently, the inertia only depends on the mass and one of its dimensions. So let's get out those calipers and start measuring.

Length	0.149 m
Width	0.098 m
Mass	0.455 m

So analytically, our moments of inertia are:

I_vertical = 0.455 kg * 0.098² m² / 18
I_vertical = 0.000243 kg m²
I_horizontal = 0.455 kg * 0.149² m² / 18
I_horizontal = 0.000561 kg m²

Now that we have in mind what to look for, let's get the experiment underway. As before, we will first find the inertia of the disks by measuring angular acceleration with a mass hanging off a torque pulley. We need to make a few measurements:

Hanging Mass	0.025 kg
Pulley Radius	0.025 m

Everything is hooked up to Logger Pro. We let it record the rotational motion and use the slope of the velocity charts to figure out acceleration:

Disks by itself, without triangle.

Triangle in vertical orientation.

Triangle in horizontal orientation.

And as before, we average the descent and ascent acceleration to cancel out any remaining effects of friction that the disk has, even with the air blowing through it.

α (rad/s2)
Disk by itself.
down	2.1040
up	2.2780
average	2.1910
Vertical orientation.
down	1.9430
up	2.1020
average	2.0225
Horizontal orientation.
down	1.7350
up	1.9380
average	1.8365

Finally, we have all the pieces to calculate the inertia experimentally:

We can simplify this by replacing all the variables that remain constant:

Now, we can just substitute the angular acceleration we found earlier to find the total moments for each system. The inertia of the triangle by itself are then derived by subtracting the inertia of the disks.

The results are surprisingly similar. It's always daunting to set out to mathematically prove something, since any error either calls into question the entire concept in one's knowledge, or sends him in furious exercise to try to match what he knows as theory with reality. In this case, it worked out okay...

	Results (kg m2)		Error
	Analytical	Experimental	Absolute (kg m2)	Relative (%)
Ivertical	0.000243	0.000233	0.000010	4.3
Ihorizontal	0.000561	0.000540	0.000021	3.9