Swinging+Ball-+Lekha,+Joey,+and+Chase

__Title of Lab__: Swinging Ball

__Researchers__: Lekha, Joey, and Chase

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

There are two significant forces acting on the ball: the tension of the string it is attached to, and the force due to gravity acting on the ball. In this case, tension is acting as a centripetal force, which also means that it always acts perpendicular to the ball's path of motion. Therefore, tension does not do any work on the ball, regardless of the ball's displacement.

Because gravity is the only significant conservative force making the ball accelerate along its path of motion, the law of conservation of mechanical energy should hold. Namely, the gravitational potential energy relative to the change in height at the point of release should be equal to the kinetic energy at the ball's lowest point: mgh=1/2*m*v 2 ...where h is the change in height from the ball's lowest point to the point of release, and v is the speed of the ball at its lowest point. Solving and simplifying for v 2 would yield: v 2 =2gh Because h is a variable (not a constant) and the research question asked for the speed of the ball to be related only to the angle, it would be beneficial if we could get h as a function of the angle, theta, and any other constants, if applicable.

Using the above image and basic trigonometry: cos(theta)=(L-h)/L ...where L is both the length of the string and the radius of the ball's circular path of motion. Solving for h would yield: h=L-L*cos(theta) Because L is constant, we can substitute L-L*cos(theta) back into our previous equation to get: v 2 =2g( L-L*cos(theta)) After simplifying by distributing 2g out, we finally arrive at the end equation: v 2 =-2gL*cos(theta))+2gL ...which relates the velocity of the ball at its lowest point to the angle the ball was released at, and a couple other constants.

__Hypothesis__: We predict that the square of the speed of the ball at its lowest point will be inversely proportional to the cosine of the angle the ball was released from by a factor of 2gL. Represented by the equation v 2 =-2gL*cos(theta))+2gL

__Materials__: Pendulum, ball attached to string, laser sensor, meterstick.

__ Constants __: length of string, diameter of the ball, gravity

__Variables__: angle of release, time, change in height, speed of the ball

__Procedure__: 1. Attach ball to pendulum to metal tower 2. Measure length of string hanging 4. Attach sensor to the point on the metal tower where the center of ball passes through. Measure the diameter of ball. 5. Let the ball swing from a certain angle 6. Divide diameter of ball by the time it takes to pass through to find speed 7. Repeat steps 5&6 for various angles.

__Data__: length of string: 0.788 m diameter of ball: 0.0312m


 * Angle (°) || time (sec) || speed (m/s) ||
 * 0 || ∞ || 0 ||
 * 10 || 0.0628 || 0.497 ||
 * 15 || 0.0392 || 0.796 ||
 * 20 || 0.0292 || 1.068 ||
 * 25 || 0.0255 || 1.224 ||
 * 30 || 0.0216 || 1.444 ||
 * 35 || 0.0189 || 1.651 ||
 * 40 || 0.0166 || 1.880 ||
 * 45 || 0.0143 || 2.182 ||
 * 50 || 0.0135 || 2.311 ||
 * 55 || 0.0124 || 2.516 ||
 * 60 || 0.0115 || 2.713 ||

__ Data Analysis __ :

The slope is about -14.6 m 2 /s 2. Because it is negative, that means as the cosine of the angle increases, speed 2 decreases. Because as the angle increases as the cosine of the angle decreases in the domain we are concerned with (0 o < theta < 90 o ), this means that as the angle we release the ball from increases, the ball has a faster speed at its lowest point, which makes sense: the higher we drop the ball, the faster it'll be at the end. The line of best fit crosses the x-axis at about (1, 0m 2 /s 2 ). The cosine of the angle is 1 when the angle is at 0 o. When the angle is at 0 o, the ball starts at the lowest point from rest. So, the ball is moving at a speed of 0 m/s when the cosine of the angle is 1. The y-intercept is at (0, 14.7m 2 /s 2 ).This is almost opposite of the slope, which makes sense, considering the graph crosses the x-axis at (1, 0m 2 /s 2 ). The slope would equal the average rate of change from the y-intercept to the x-intercept: -14.6m 2 /s 2 =(y-0m 2 /s 2 )/(0-1) 14.6m 2 /s 2 =y

According to our hypothesis, the theoretical slope would be equal to -2gL = -15.46m 2 /s 2. This leads to a percent error of about 6%. The cause of this is most likely due to us neglecting friction and air resistance as the ball traveled through its path of motion. This error could also be attributed to either incorrect or imprecise measurements of the diameter of the ball and the angle we dropped the ball from. One interesting thing to note is that it would be incorrect to use the slope and the y-intercept of the line of best fit for our data in calculating the percent error. As stated previously, the x-intercept should be at (1, 0m 2 /s 2 ), regardless of the slope of the line, since if you start the ball at rest from the ball's lowest point (theta = 0 o, cos(theta)=1), the speed would be 0 at the ball's lowest point (v=0 and v 2 =0), since it started from that point at rest. Therefore, because speed^2 is inversely proportional to the cosine of the angle (as stated previously), the slope and the y-intercept should always be opposite of each other, regardless of the accuracy of your data (assuming the line of best fit still crosses at (1, 0m 2 /s 2 )). If we were to use the slope and the y-intercept (and not our theoretical slope of -2gL), the percent error for our data would only be about 0.68%. To reiterate, this percent error is incorrect.

__Conclusion__: Our hypothesis was supported by the experiment, since there was a negative linear relationship between cos(theta) and speed 2. Our experimental slope was -14.6, so there was a percent error of 6% which is a fair representation of the hypothesis. The experiment could be improved using a better tool for measuring the angle from which the ball was dropped. We could increase confidence in our results by taking more trials and gathering more data.