Ball+on+Incline+-+Charlie+Morris+&+Graydon+Yoder


 * Title of Lab**: Ball Rolling Down Incline Lab

**Researchers**: Charlie Morris, Graydon Yoder

**Research Question**: How does the time it takes for a tennis ball to roll down an incline relate to the distance it travels?

**Research**: Using the kinetics equation Δx = v i t + ½ a t 2 given that vi equals 0 (because we released it from not moving), we find that Δx = ½ at 2. Solving for t, the equation becomes t 2 = 2Δx / a.

We then use Newton's Second Law to find acceleration. Because the ball is rolling, force due to friction equals zero. Therefore, mg sinθ - 0 = ma. Cancelling the mass on both sides leaves a = (9.81 m/s^2)(sinθ).

Plugging this in for acceleration in the kinetics equation we obtained gives us t 2 = 2∆x ÷ (9.81 m/s)(sin θ).

**Hypothesis**: The time it takes the ball to travel is directly related to the distance traveled, such that t 2 = 2∆x ÷ (9.81 m/s)(sin θ).

**Materials**: wooden board (used for incline), tennis ball, protractor with string and weight attached (angle measuring tool), meter stick, iPhone as stopwatch

1. Set the board to an inclination of 18 ° to the horizontal. 2. Set the meter stick at the top of the inclination, having the 0 cm end of the meter stick at the starting point. 3. Select a distance to test, and position the ball with its center at the highest end of the incline. 4. Release the ball. Time how long it takes the ball to reach the indicated distance.The same person both drops the ball and operates the timer in order to have the most accurate times. 5. Record data. 6. Repeat steps 1 through 5 at various distances along the incline.
 * Procedure** :

Constants: Angle of inclination, force due to gravity Variables: Time, displacement

**Data**:
 * **Distance (cm)** || **Distance (m**) || **Time (s**) ||
 * 10 || 0.10 || 0.2 ||
 * 20 || 0.20 || 0.3 ||
 * 30 || 0.30 || 0.4 ||
 * 40 || 0.40 || 0.5 ||
 * 50 || 0.50 || 0.6 ||
 * 60 || 0.60 || 0.7 ||
 * 70 || 0.70 || 0.7 ||
 * 80 || 0.80 || 0.8 ||
 * 90 || 0.90 || 0.9 ||
 * 100 || 1.00 || 0.9 ||
 * 110 || 1.10 || 0.9 ||
 * 120 || 1.20 || 1 ||

**Data Analysis**:

Time and displacement graph with no manipulation. This graph resembles a square root curve because at longer distances, acceleration has more time to increase velocity.

The time squared vs displacement graph has a slope of 0.86 s^2/m, and using the equation g= 2/(slope* sin θ), we find the experimental g is equal to 7.53 m/s^2. According to the line of best fit, the y-intercept is -0.0563 s^2. **Conclusion**: We found the experimental value of g to be equal to 7.53 m/s^2. Compared to the accepted value of 9.81 m/s^2, our experiment resulted in a 23.2% error. The y-intercept theoretically should be (0,0) because it takes 0 seconds squared to travel 0 centimeters, but our experimental y-intercept was -0.0563 s^2, representing error in our data. Our experiment had much experimental error due to the short time intervals and low-accuracy stopwatch. This could be improved by using a more accurate stopwatch, or using time gates. With such a large percent error, our hypothesis was not supported by our data.