1-Jun-2015/3-Jun-2015: Simple harmonic motion

PURPOSE

The purpose of this experiment was to derive equations that modeled objects in simple harmonic motion. This allowed us to find the theoretical values for their periods, which we compared to the values we found experimentally.

PROCEDURES

Period of a Ring in Simple Harmonic Motion

Figure 1

Figure 2

For the first part of the experiment, we attempted to derive an equation for the period of a ring in simple harmonic motion. The first step we took in order to accomplish this task was set up the torque equation as shown in Figure 2 by using the diagram displayed in Figure 1. Using the moment of inertia of a ring rotating about its center of mass [I = 1/2 M(R² + r²), where R = 0.0696 m, r = 0.0579 m, and M does not matter because it ends up canceling out], we were able to find the moment of inertia of the ring when it was rotating about a pivot that was a distance Ravg = 0.06375 m (the average of the inner and outer radii of the ring) away from the ring's center of mass by using the parallel axis theorem (I = Icm + Md²). Next, we solved the equation for α and assumed that sinθ was approximately equal to θ for very small values of θ. We took these steps in order to put the equation in the general form of simple harmonic motion (α = -ω²θ). From the resulting equation, we were able to deduce that ω was equal to [(2gRavg)/(R² + r² + 2Ravg²)]^1/2 = 8.753 rad/s, which we plugged into T = 2π/ω to find the period. We found this value to be 0.7178 s,

Figure 3

Following the calculations, we set up the experiment as shown in Figure 3. We pulled the ring back at an angle and released it to let it oscillate. We captured the ring's period using a pre-programmed file in Logger Pro, which we found to be 0.72 s. This gave us a percent error of 0.306 percent, which is pretty much as accurate as an experiment can be. This was probably due to the fact that there wasn't much room for error since the only force affecting the ring's motion was gravity. The only factor that could have contributed to error in the results was the sway of the ring when we first released it from rest. However, this did not have a significant effect on the results because the ring stabilized itself after a few oscillations.

Period of a Triangle in Simple Harmonic Motion

Figure 4: Derivation of the moment of inertia's

We began the second part of the experiment by deriving the moment of inertia's of a triangle with the axis of rotation at its apex and a triangle with its axis of rotation at the center of its base. First, we found its moment of inertia with respect to its apex as shown in the top portion of Figure 4. We designated dM to be a thin bar with its center of mass a distance y away from the axis of rotation. Therefore, in order to find the moment of inertia of dM, we had to apply the parallel axis theorem (with d = y) on the moment of inertia equation of a thin bar rotating about its center of mass (I = 1/12 ML², where L = 2x in this case). We were able to find the moment of inertia of the entire triangle by integrating the resulting equation from 0 to H. We found this value to be I = 1/24 MB² + 1/2 MH². Then, we used the same dM to find the center of mass of the triangle. We did this by using the equation ycm = (∫dM*y)/(∫dM). We found the center of mass to be 2/3 H from the triangle's apex. From this value, we were able to apply the parallel axis theorem again (with d = 2/3 H) to find moment of inertia of the triangle with respect to its center of mass. We did this by turning the equation from I = Icm + md² into Icm = I - md². From this step, we were able to find the moment of inertia of a triangle rotating about its center of mass was I = 1/24 MB² + 1/18 MH². Then, using d = 1/3 H, we implemented the theorem a third time to find the moment of inertia of the triangle rotating about the center of its base. The resulting equation was I = 1/24 MB² + 1/6 MH².

Figure 5: Derivation of the period of the triangle rotating about its axis

Figure 6: Derivation of the period of the triangle rotating about the center of its base

After finding the moment of inertia of the triangle with respect to the two axes of rotation, we were able to derive equations for the period of the triangle in both orientations (Figures 5 and 6). Just as before, we first set up the torque equation and set everything equal to α. In addition, we assumed sinθ to be approximately equal to θ for very small values of θ. We did this to get the equation in the general form of systems in simple harmonic motion (α = -ω²θ). From the resulting equation, we concluded that ω was equal to the square root of (16gH)/(B² + 12H²) for the triangle rotating about its apex and the square root of (8gH)/(B² + 4H²) for the triangle rotating about the center of its base. Then, we found the period by dividing 2π by these values. The resulting equations were 2π[(B² + 12H²)/(16gH)]^1/2 and 2π[(B² + 4H²)/(8gH)]^1/2, respectively.

Figure 7: Theoretical values of the periods

After deriving the equation, we plugged in the appropriate values (B = 0.14 m, H = 0.147 m, and g = 9.81 m/s²) to find the theoretical periods. For the triangle rotating about its apex, we found the period to be 0.69081 s. On the other hand, for the triangle rotating about its base, we found the period to be 0.60238 s. The mathematical processes are displayed in Figure 7.

Figure 8: Rotating about its apex

Figure 9: Rotating about the center of its base

Following our calculations, we set up the triangle about its apex and released it from a slight angle to let it oscillate. Its period was measured using a pre-programmed set-up for a pendulum on Logger Pro. As it can be seen from Figure 8, the experimental value for the period was 0.695362 s, which was only 0.659 percent away from the expected value. Then, we set up the triangle about the center of its base and reran the experiment. The resulting data is shown in Figure 9. From Figure 9, it can be seen that the experimental period for the triangle rotating about its base was 0.607886 s. In this case, the percent error between the theoretical and experimental values was 0.551 percent. Once again, there was virtually no error in the results.

CONCLUSION

In this experiment, we found the moment of inertia of objects rotating about different points to set up torque equations. Then, we set everything equal to α to get the equations into the general form of simple harmonic motion (α = -ω²θ). This allowed us to figure out what ω was, which in turn allowed us to find the period of the motion. We then ran the actual experiment and compared the theoretical values to the experimental ones.

The theoretical and experimental values were almost identical. As mentioned before, the reason was because the only force affecting the ring's motion was gravity. Since the force of gravity is constant as long as we're close to the Earth's surface, there was not much room for error. One thing that was concerning at first was the sway of the objects when we first released them and let them oscillate. This ended up not having much of an affect on the results because the objects stabilized after a few oscillations. Another factor that I thought would affect the results was the friction in the pivot. This most likely did not alter the results significantly because the pivot had a smooth outer surface.

20-May-2015: Conservation of energy/conservation of angular momentum

PURPOSE

Figure 1

In this experiment, we attempted to predict a how high above the ground a system would reach after a meter stick rotating about a pivot collides with a piece of clay at rest (Figure 1). In order to do so, we employed a combination of conservation of energy and angular momentum equations.

PROCEDURES

Figure 2

Figure 3

For the first part of our calculations, we found the moment of inertia of the meter stick, which had a mass of 0.08 kg, rotating about an axis that was 0.475 m away from its center of mass (Figure 2). We used the parallel axis theorem (I = Icm + md²) and found this value to be 0.02472 kg*m². Next, we applied the conservation of energy equation to the meter stick rotating from the horizontal position to the vertical position as shown in Figure 3. We set the reference point as the meter stick's center of mass in its final position. Therefore, the meter stick had an initial gravitational potential energy, which eventually all turned into rotational kinetic energy. We set the initial gravitational potential energy (mgh) equal to the final kinetic energy (½Iω²) and solved for the final angular velocity (ω). We found this value to be 5.492 rad/s.

Figure 4

Afterwards, we applied the conservation of angular momentum equation to the meter stick-clay system before and after the collision. For the initial angular momentum, we found the product of the moment of inertia of the meter stick by itself and the angular velocity found earlier with the energy equation (ωo = 5.492 rad/s). On the other hand, the final angular momentum was equal to the product of the moment of inertia of both the meter stick and the clay, and the angular velocity right after the collision (ωf), which was unknown. We found the moment of inertia of the piece of clay by treating it as a particle and applying the equation Ic = mcd², as shown on the bottom of Figure 4. Then, we solved for ωf, which was 2.746 rad/s.

Figure 5

Figure 6

Then, for the final step of our calculations, we set up the conservation of energy equation by using the diagram shown in Figure 5. First, we established the pivot as the reference point and up as the positive y-direction. Then, we used trigonometry to figure out that the vertical distance between the pivot and the final positions of the meter stick and piece of clay were 0.475cosθ and 0.975cosθ, respectively. This resulted in the equation shown on the very top of Figure 6. Plugging in I = Isys = 0.02472 + 0.02472 = 0.04944 kg*m², ω = 2.746 rad/s, m1 = 0.08 kg (mass of meter stick), m2 = 0.026 kg (mass of clay), and g = 9.81 m/s², we were able to eventually solve for cosθ. We found this value to be 0.7001. We took the inverse cosine of this number to find the angle that the system made with the vertical, which we found to be 45.57°. We realized later that finding the angle was unnecessary because the value we really needed was cosθ.

Figure 7

From this angle, we used basic trigonometric relationships to find the vertical distance between the final position of the piece of clay and the ground, as shown in Figure 7. One thing to note is that we treated the piece of clay as a particle as we did throughout the rest of the calculations. Therefore, the dimensions of the clay piece were not considered in the derivations. From Figure 7, it can be seen that the theoretical maximum height of the stick-clay system was 0.2924 m.

Figure 8

Finally, we ran the actual experiment and found out how high above the ground the system actually reached after the collision. We did this by utilizing video analysis in Logger Pro. As it can be seen from Figure 8, We found the height to be 0.2904 m. This gave us a percent difference of 0.6840 percent. This accurate result was due to the fact that gravity was the only significant external force acting on the system throughout the entire process.

CONCLUSION

In this experiment, we broke down a collision into three different stages: motion before the collision, the collision itself, and the motion after the collision. We utilized conservation laws during these stages, which ultimately allowed us to estimate how high above the ground a meter stick and a piece of clay would rotate after a collision occurred between the two objects.

As mentioned earlier, the percent error was very low in this experiment because the only external force that had a considerable impact on the results was gravity. However, another reason the theoretical and experimental values were so close to each other was that we did not factor in the clay piece's dimensions into our calculations. If we had tracked the distance between the end of the meter stick and the ground (which is what our theoretical value actually represents) instead of the center of the clay piece, it would have been 1 to 2 cm smaller. Therefore, the percent error would have been approximately 3 to 7 percent higher. However, this is to be expected because there were other assumptions we made in our calculations, First, we did not account for the friction between the pivot and the meter stick as it was rotating. In addition, there was probably some angular momentum lost during the collision since it was not a perfectly head-on collision. This explains why the experimental value was smaller than the theoretical one.