Ball+on+Incline-+Swathi+Ganesh+&+Anand+Rajagopal

= Title of Lab: Ball on Incline =

Researchers:
Anand Rajagopal and Swathi Ganesh

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

Research:
Assuming a frictionless surface, mgsin θ represents the force of the rolling ball (inertia)(as represented by the diagram below) and is equal to ma (because the sum of the forces equals mass multiplied by acceleration). By diving mgsin θ=ma by m on both sides, we get the equation gsin θ=a. You can substitute gsin θ for a in the kinematics equation Δx=vit+(1/2)at^2, resulting in the equation Δx=vit+(1/2) gsin θt^2. Since the initial velocity is 0 cm/s, the equation can be simplified to Δx=(1/2) gsin θt^2. After solving for t^2, the equation t^2= 2Δx/(gsinθ) can be derived.



**Hypothesis:** The time it takes for a ball to roll down an incline is directly proportional to the distance traveled (in the form of a square root relationship).

Materials:
tennis ball, meter stick, wooden plank, stopwatch, protractor with string with weight attached

Procedure:
(1) Make sure all your instruments are "calibrated" (the protractor set starting at 90 ⁰ and the stopwatch set at 0:00). (2) Set your plank at an angle of inclination of 20⁰ with one end touching the floor and the other end leaned on another surface. (3) Place the meter stick on the incline to measure the distance traveled by the ball. (4) Hold the ball on the plank and let it go so that it rolls for a certain distance. Make sure that you start the stopwatch as soon as you release the ball and stop it as soon as the ball crosses the given distance. (Tip: To reduce percentage error, have the same person who rolls the ball down the ramp, be the timer too). (5) Remember to reset your stopwatch and repeat Step 4 for a variety of distances.

**Data:**

**Data Analysis:** __// Time vs. Distance //__

__ //Time^2 vs. Distance// __

The time squared vs displacement graph has a slope of 0.01 s^2/cm, and using the equation g= 2/(slope*sin θ), we find the experimental g is equal to 5.85 m/s^2 as opposed to the accepted value of 9.81 m/s^2. This gave us a percentage error of 40.4%. This error could be due to the lack of precision of the stopwatch used, as well as the great amount of scope for human error in stopping the stopwatch at exactly the point the ball reached the given distance. Another possible contribution to our large amount of error could have been from our negligence to perform multiple trials over a greater number of distances. The significance of the y-intercept, in this case, 0 seconds, is that this is the only sure point on the graph in that it does not have an error because an object will not take any time to travel a distance of 0 cm. This can be proven with by plugging in 0 for t in the equation Δx=vit+(1/2)at^2 (in which Δx equals 0 as a result). We can use this point as a starting point for our line of best fit, which will help us to derive a more accurate best-fit line, and therefore, slope.

**Conclusion:** The hypothesis was not supported by the evidence. Although there was a positive linear relationship between time squared and distance, our experimental g, 5.85 m/s^2, did not match the theoretical g, 9.81 m/s^2. In order to improve our experiment, we would've used more precise instruments for measuring while also carrying out multiple trials to help increase the chance of obtaining accurate and precise results.