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. Then compare the experimental and theoretical results for the moment of inertia of the triangle in each of the two orientations.
Procedure: The apparatus consisted of two rotating disks adjacent to a Pasco rotational sensor. We attached a torque pulley on top of the two disks with a string wrapped around it. The other end of the string goes over another pulley and hangs off the edge with a hanging mass. A triangular thin plate was attached on top of the torque pulley with a mounted holder. Turning on the compressed air allowed the disk(s) to rotate freely without friction, which caused the hanging mass to oscillate up and down.
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| Lab apparatus, disk/holder/triangle system with hanging mass |
To approach this experiment, we had use our understanding of the parallel axis theorem, which states that
I(parallel axis) = I(around cm) + M*d^2, where d is the displacement from the original cm and M is the mass.
Because the limits of integration are simpler if we calculate the moment of inertia around a vertical end of the triangle, you can calculate that moment of inertia and then get the moment of inertia around the center of mass with
I(around cm) = I(around one vertical end of the triangle) + M*d^2, where d is the displacement from the original cm and M is the mass.
We had to derive the moment of inertia of the triangle around its center of mass and use the parallel axis theorem to find the moment of inertia around its new axis. The derivation for this is below.
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| Derivation of moment of inertia of triangle around its new axis |
The new moment of inertia of the triangle came out to
I = 1/18*M*R^2. In order to solve for the moment of inertia of the right triangular thin plate, we had to first measure the the triangle's mass M, horizontal length R1, and vertical length R2. These came out to M=0.455 kg, R1=0.09853 m, R2=0.14950. R would depend on the orientation of the triangle during the experiment.
Next was the experimental portion of the lab. The moment of inertia of the entire disk/holder/triangle system can be solved for by knowing the torque exerted by the tension from the hanging mass and the angular acceleration of the system produced by the torque. Newton's 2nd law relates these values with
Torque = Tension*radius = I*alpha
So, by finding the total moment of inertia of the entire system and subtracting it by the moment of inertia of just the disk/holder, we can get the moment of inertia of just the triangle. This moment of inertia will be our experimental value of I. We first recorded the angular acceleration and the moment of inertia of just the disk/holder, then added the triangle to the system and measured the angular acceleration and moment of inertia for the entire system.
Note: Because there is some frictional torque in the system (the disk is not completely frictionless and the frictionless pulley is not massless), the angular acceleration of the system is not the same when it is descending as when it is ascending. For this reason we had to derive an equation which took into account the frictional torque. The derivation is below
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| derivation of experimental moment of inertia |
We began recording the data of the angular acceleration of just the disk/holder system. Angular acceleration was found by finding the slope of the angular velocity vs. time graph below.
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| graph of angular velocity vs time for disk/holder system |
We then took the average angular acceleration going up and going down, then found the average of the two. Then using our derived equation, and our measured mass and radiuses, we were able to calculate the experimental moment of inertia of the disk/holder system.
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| experimental moment of inertia of disk/holder system |
Next, we added the triangle to the system in the vertical orientation and began recording the angular acceleration of the disk/holder/triangle system. Once again, we found the angular acceleration using the slope of the angular velocity vs time graph.
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| Graph of angular velocity vs time for disk/holder/triangle system (vertical) |
Taking the average acceleration, we were able to calculate the moment of inertia for the disk/holder/triangle system (vertical).
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| experimental moment of inertia of triangle (vertical) and theoretical moment of inertia |
By subtracting our disk/holder/triangle system inertia with the disk/holder system inertia, we were able to get the experimental moment of inertia of just the triangle which came out to 2.09x10^-4 kgm^2. Then by using the theoretical equation we derived earlier of I = 1/18*M*R^2, we were able to find the theoretical moment of inertia which was 2.44x10^-4 kgm^2. Our percent error came out to 14.3%
Lastly, we turned the triangle so that it was horizontal and again measured the system's angular acceleration using the slope of the angular velocity vs time graph below.
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| graph of angular velocity vs time for disk/holder/triangle system (horizontal) |
Then, using that angular acceleration and our measured values of M and the disk's r, we were able to calculate the triangles experimental moment of inertia (horizontal).
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experimental moment of inertia of triangle (horizontal) and theoretical moment of inertia
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Comparing the experimental value for moment of inertia of the triangle (horizontal) I =5.32x10^-4 with the theoretical value I=5.65x10^-4 gave us a percent error of 5.84%.
Conclusion/Uncertainty: Through this equation, we were able to derive an equation for the triangle's moment of inertia on using the parallel axis theorem. Using our data from measuring the disk's angular acceleration and dimensions and properties, we were able to calculate the experimental value for the triangles moment of inertia. Then, by comparing our experimental results for moment of inertia and the theoretical value using our derived equation, we were able to see how accurate we were able to get using our understanding of newton's second law and the parallel axis theorem. Our results, even though the percent error was a little larger than we had hoped, was still close enough to verify our experiment.
A few errors occurred during our experiment beginning with the fact that our vertical triangle was not perfectly vertical. We did not catch that until after we recorded our data and may be why our percent error for the vertical was large compared to that of our horizontal results. Other things such as the friction from not having a perfectly clean disk and slight error in measurements of the triangle may have skewed our results a bit.
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