MAY-13-2015 Utilizing the Parallel Axis Theorem

Utilizing the Parallel Axis Theorem

(Finding the moment of inertia of a uniform triangle about its center of mass)

Purpose:  To determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle.

Procedure:  We first determined the moment of inertia for the object about its center of mass theoretically utilizing the parallel axis theorem.  We first determined its moment of inertia about its edge, then used Ip = Icm + Md^2 to find the moment of inertia about its center of mass.

The following are the measurements of the uniform right triangular thin plate:

Height = h = 14.80 cm

Base = b = 9.85 cm

Mass = m = 456 g

Depicted below are two theoretical derivations for the moment of inertia; the first with h in the vertical direction, and the second with h in the horizontal direction (two perpendicular orientations).

(Derivation of Icm in orientation 1)

(Derivation of Icm in orientation 2; evaluation omitted)

The value of Icm for the second derivation when evaluated is:

Icm = 5.55 X 10^-4 kg * m^2

In order to determine our experimental values of Icm for each orientation, we used methods similar to those in our "Angular Acceleration" lab.  Depicted below is the experimental setup for orientation 1:

(Experimental Setup for Orientation 1)

The first objective is to determine the moment of inertia of the system with the top steel rotating disk WITHOUT the triangle attached to the system.  Afterwards, we then added the triangular thin plate and calculated the new system's moment of inertia; the difference between these values yields the moment of inertia of the uniform triangular thin plate about it's center of mass.

Without Triangular Plate:

I = (mgr)/alpha - mr^2

Torque Pulley:
   Radius = 0.0126 m

Hanging Mass:
       Mass = .025 kg

We know the value of alpha from the following LoggerPro data:

(LoggerPro Data; no triangular plate)

This gives us:

alpha = (1.171 + 1.273) / 2 = 1.222 rad/s^2

Thus:

I of system w/o triangle = [(.025 *9.81 * .0126) / 1.222] - (.025)(.0126)^2
              = 2.52 X 10^-3 kg * m^2

Now that we have the moment of inertia for the system WITHOUT the triangular plate, we can attach it, recalculate the system's moment of inertia, and subtract our first result from the new value in order to attain the moment of inertia of the triangular plate about its center of mass for both orientations.

Triangular Plate; Orientation 1:

alpha = (1.072 + 1.174) / 2 = 1.123 rad/s^2

I of system with triangle @ orientation 1 = 2.75 X 10^-3 kg * m^2

Difference = I of triangle @ orientation 1 = 2.28 X 10^-4 kg *m^2

That result quite close to our theoretical calculation from above!

Triangular Plate; Orientation 2:

alpha = 0.9965 rad/s^2

I of system + triangle @ orientation 2 = 3.10 X 10^-3 kg * m^2

Difference = I of triangle @ orientation 2 = 5.77 X 10^-4 kg * m^2

Once again we're close to our theoretical value.

These results are summarized below:

(Summary of Theoretical v. Experimental Findings)

Conclusion:  I'm satisfied with how the experimental values turned out when compared with the theoretically calculated values determined earlier in the lab.  Sources for the minor differences between these values include measurements taken via vernier calipers, as well as the assumption of zero friction within the torque pulley system.