Swinging+Ball-+Anand+Rajagopal,+Sam+Shaunak,+Aaron+Floyd

Title of Lab: Swinging Ball

Researchers: Anand Rajagopal, Aaron Floyd, Sam Shaunak

Research Question: How does the initial angle with the vertical relate to the speed of a swinging ball at its lowest point?

Research: The conservation of mechanical energy equation is applied to this question, because all the components involved in the equation are present. Ei=Ef which is also written: mgh(f) + .5 mv(f) ^2= mgh(i) + .5mv(i)^2 Because h(f)= 0 and v(i)= 0, the equations simplifies to .5mv(f)^2= mgh(i) The mass divides out so the equation once again simplifies, resulting in the equation .5v(f)^2= gh(i) This turns into v(f)^2= 2gh(i)

To find h(i), we use the formula from the triangle created by the swinging of the ball. L= Lcos(theta) + h(i) --> h(i)=L- Lcos(theta)

Applying this to the simplified equation solved for v^2, the result is v(f)^2= 2g [L- Lcos(theta)] ---> v(f)^2= 2gL - 2gLcos(theta)

Hypothesis:

The relationship between the angle with the vertical and the velocity is related using the equation v(f)^2= 2gL- 2gLcos(theta).

Procedure:
 * 1) Gather materials: ruler (m), stand, photo gates, ball attached to string.
 * 2) Set up photo gates at either side of the stand so that the ball will travel through them without touching either one.
 * 3) Measure the distance between the two photogates.
 * 4) Measure the angle from which the ball will drop.
 * 5) Proceed to drop the ball at specified angle.
 * 6) Using the photo gates, record the time the ball takes to get from one photogate to the next.
 * 7) Record the speed using the formula s=d/t.
 * 8) Repeat steps 4-7 changing the initial angle by increments of 10 degrees each time.

Data: Length of String: 0.9 m (seconds) || Trial 2 Time (seconds) || Trial Average Time (seconds) || Distance (meters) || Speed (m/s) (Average Time Used ||
 * Angle (Degrees) || Trial 1 Time
 * 10 || 0.0942 || 0.0938 || 0.0940 || 0.05 || 1.00 ||
 * 20 || 0.0496 || 0.0477 || 0.0487 || 0.05 || 1.03 ||
 * 30 || 0.0338 || 0.0315 || 0.0327 || 0.05 || 1.53 ||
 * 40 || 0.0251 || 0.0219 || 0.0235 || 0.05 || 2.13 ||
 * 50 || 0.0201 || 0.0192 || 0.0197 || 0.05 || 2.54 ||
 * 60 || 0.0165 || 0.0166 || 0.0166 || 0.05 || 3.01 ||

Data Analysis:

Running a linear regression shows the slope of the (cos ** θ vs v^2) linear graph is equal to -18 **

Conclusion: An accurate method of seeing if the equation v(f)^2= 2gL- 2gLcos(theta) correctly describes the relationship between velocity and theta is to test if the slope of our data for (cos ** θ vs v^2) is equal to the slope of the equation which is -2gL. Since our slope of the data was -18 and the theoretical (-2gL=2x10x0.9) was -18, our percent error for the lab was 0%. Thus ** v(f)= square root (2gL- 2gLcos(theta)).