The apparatus will be set up pretty much the same as the previous lab 16 except that this time there will be a triangle connected to the top.
The approach we took to accomplish our goal was by first determining the experimental value for the inertia of the system without the triangle and then the moment of inertia of the system with the triangle and subtracting the moment of inertia value with the triangle to the inertia value calculated without the triangle to get the inertia value of the triangle itself. We used the same method we used in the previous lab of angular acceleration where we measured the angular acceleration up and down, and got the average of the two to plug into the equation to find the inertia of the system. The equation was derived in the previous lab and is the following:
Where m is the mass of the hanging mass, g is the acceleration due to gravity, r is the radius of the torque pulley, and alpha is the average of the angular accelerations up and down.
Next we derived a formula for the inertia of the triangle with the axis of rotation being at one end so it is easier to integrate. We did this so that we could compare our experimental values for the moment of inertia with the theoretical values for the two orientations. We used the parallel axis theorem to get the moment of inertia from the center of mass of the triangle which is what we wanted.
To derive an equation for the moment of inertia of the triangle with the axis of rotation on one side, we had to take a horizontal piece dm of the triangle and find the moment of inertia of that piece and integrate the moment of inertia of that piece from 0 to h. Since the piece of the triangle was a rod being rotated about one end we knew the equation for the moment of inertia of such a rod to be 1/3ML^2 where M becomes the mass of the rod piece dm and L becomes the length x of the rod. so then we had to find a function to replace dm and we did so by making the proportions of the mass of the piece over the mass of the whole thing equal to the area of the piece over the area of the whole triangle and solving for dm. But then we had to change the x variable into y because we were integrating with respect to dy. So we found an equation for the line of the hypotenuse of the triangle and got x = -b/hy+b and plugged that into our equation. We integrated the equation and got our moment of inertia of the triangle to be:
Where M is the mass of the triangle and b is the length of the base.
We could then use the parallel axis theorem to find the moment of inertia at the center of mass of the triangle:
The derivation to find the moment of inertia of the center is the following:
As can be seen the moment of inertia for the center of mass is 1/18 M b^2 where M is the mass of the triangle and b is the length of the base.
The measurements that we took for this experiment were the mass of the hanging mass, the diameter of the torque pulley used, the mass of the triangle, the length of the base of the triangle, the length of the height of the triangle, and the angular acceleration up and down of the system. The reason we needed the mass of the hanging mass, the diameter of the torque pulley used and the angular accelerations up and down, was because we need it to plug into the first equation to find the moment of inertia of the system experimentally. the reason we needed the mass of the triangle, the length of the base, and the length of the height of the triangle is so that we could find the moment of inertia of the triangle theoretically. The measurements that we took were as follows:
We did the first trial to find the angular velocity up and angular velocity down and collected enough data to let the hanging mass go up and down three times. Then we took the derivatives of each slope to get the angular accelerations:
For the second trial we did the same thing except we put the triangle on the apparatus with the longer side facing vertically( not the hypotenuse). The data we collected was the following:
For the third trial we changed the orientation of the triangle so that it was perpendicular to the position it was in the second trial. The data we collected for this orientation was:
The average angular accelerations were calculated for the system by itself, and with the triangle in both positions. The following shows the calculations:
Using this data the moment of inertia for the system alone, and with the triangle in different positions were calculated experimentally.
And the experimental moment of inertia of the triangle at the different positions was calculated to be:
The theoretical moment of inertia of the triangle at the different positions was calculated to be:
According to the data the experimental values of .000436 for the triangle in the vertical position, and .000326 for the triangle in the horizontal position are a little different than the theoretical values of .000245 for the triangle in vertical position and .000565 for the triangle in the horizontal position. It is safe to say there is a small error either with the experimental data or rounding error in the calculations. The experimental data shows that the moment of inertia of the triangle in the vertical position is greater than the moment of inertia of the triangle in the horizontal position. While the theoretical data says the opposite. I think the theoretical data is correct and there must have been some error for the experimental data because the triangle in the horizontal position has a larger moment of inertia it is harder to move because its mass is over a larger diameter. For example, when you spin on a chair and hold your arms out you would go slower than if you were to have your arms in. The triangle in the horizontal position has a longer base than in the vertical position so it is harder to move in the horizontal position so the moment of inertia is greater.
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