# Free Acceleration Due To Gravity Report Example

Type of paper: Report

Topic: Ball, Acceleration, Gravity, Doubt, Uncertainty, Value, Time, Length

Pages: 6

Words: 1650

Published: 2021/01/11

## REPORT

Reg.No.

## Introduction

The experimental objective for Lab 6: Acceleration Due to Gravity, was to find a precise value for the acceleration due to gravity using the method of free fall and the swinging pendulum method. For part A of the lab, a steel ball was dropped from an electromagnet landing on a reaction pad. While falling, a photogate was placed at certain heights recording the time over a distance in seconds and meters. The photogate measured the time from when the ball passed the photogate to when the ball landed on the reaction pad. For any free falling object, mass is negligible, thus the formula below is used to calculate an accurate value for the acceleration due to gravity.

g=2L-yt2

Here, g, is the acceleration due to gravity which is equivalent to the total length of the ball to the ground minus the distance from the photogate in meters, vertically, over the course of time in seconds. Taking the partial derivative of the equation above for each variable enables the calculation of the total uncertainty in acceleration due to gravity.

As for part B of the experiment, a pendulum was swung with a steel ball hung from a string at three different distances. A photogate was placed so that the ball passed right though it as the ball swung measuring the period of the pendulum system. Again, the acceleration due to gravity can be calculated as long as known values of length and period exist measured in meters and seconds. Thus, the formula:

g=4π2LT2

The acceleration due to gravity can be found for Part B different to that in Part A. Here, L stands for the length of the pendulum string, ranging from short to medium to long in length of meters. And T is the period measured precisely by the photogate.

## Initial Data & Preliminary Analysis

Initial Data:

In Meters.

In Meters.

Calculated Data:

PART A

PART B

Analysis

PART A

The calculations below are required in order to find the acceleration due to gravity while an object is in free fall. From initial data, excel helps to give an average value for the time the steel ball passes to the photogate to when the ball lands on the reaction pad. Similarly, excel helps to find the standard deviation and the fluctuation found by dividing the standard deviation by the square root of 10, since there were 10 trials. The total uncertainty in time can now be found using the fluctuation and the uncertainty in measurement in time.

σ=stdev1:1010

## Thus,

∆t=∆tpad-tgate=σ2+∆t2=0.0002981s2+0.001s2=0.001043s

## This process was used for all values of y in the experiment for Part A and for Part B.

G, or the acceleration due to gravity, was found while an object was dropped in free fall. The equation below is one way to find g in this circumstance. As well as finding the value for acceleration due to gravity, the uncertainty must also be found by deriving the equation below for each variable.

g=2L-yt2=21.027m-0.261m0.232s2=9.3837 m/s2

∆gt=∂g∂t=4L-yt32∆t=41.027m-0.261m(0.232s)320.001s=0.08090 m/s2

∆gL=2L-yL*t2∆L=21.027m-0.261m1.027m*0.232s20.0001m=0.01843 m/s2

∆gy=2L-yy*t2∆y=21.027m-0.261m0.261m*0.232s20.0001m=0.03655 m/s2

∆g=∆gt2+∆gL2+∆gy2=0.08090ms22+0.01843ms22+0.03655ms22=0.09067ms2

## In Meters/Second^2.

These calculations are done for all values of y in part A. Since there are five values of y, calculations must be done for all distances. When the photogate is moved down the slightly, the value for y becomes 0.320 meters.

g=2L-yt2=21.027m-0.320m0.207s2=9.3565 m/s2

∆gt=∂g∂t=4L-yt32∆t=41.027m-0.320m(0.207s)320.001s=0.09040 m/s2

∆gL=2L-yL*t2∆L=21.027m-0.320m1.027m*0.207s20.0001m=0. 002062m/s2

∆gy=2L-yy*t2∆y=21.027m-0.320m0.320m*0.207s20.0001m=0.003694 m/s2

∆g=∆gt2+∆gL2+∆gy2=0.09040ms22+0. 002062ms22+0.003694ms22=0.09050ms2

Here, y=0.408m

g=2L-yt2=21.027m-0.408m0.171s2=9.6001 m/s2

∆gt=∂g∂t=4L-yt32∆t=41.027m-0.408m(0.171s)320.001s=0.1123 m/s2

∆gL=2L-yL*t2∆L=21.027m-0.408m1.027m*0.171s20.0001m=0.002529 m/s2

∆gy=2L-yy*t2∆y=21.027m-0.408m0.408m*0.171s20.0001m=0.006931 m/s2

∆g=∆gt2+∆gL2+∆gy2=0.1123 ms22+0.002529ms22+0.006931ms22=0.1125ms2

And, y=0.520m

g=2L-yt2=21.027m-0.520m0.136s2=9.2387 m/s2

∆gt=∂g∂t=4L-yt32∆t=41.027m-0.520m(0.136s)320.001s=0.1359 m/s2

∆gL=2L-yL*t2∆L=21.027m-0.520m1.027m*0.136s20.0001m=0.003119 m/s2

∆gy=2L-yy*t2∆y=21.027m-0.520m0.520m*0.136s20.0001m=0.004383 m/s2

∆g=∆gt2+∆gL2+∆gy2=0.1359ms22+0.003119ms22+0.004383 ms22=0.1360ms2

Finally, Height is equal to 0.712m in trial five.

g=2L-yt2=21.027m-0.712m0.078s2=9.4567 m/s2

∆gt=∂g∂t=4L-yt32∆t=41.027m-0.712m(0.078s)320.001s=0.2425 m/s2

∆gL=2L-yL*t2∆L=21.027m-0.712m1.027m*0.078s20.0001m=0.005502 m/s2

∆gy=2L-yy*t2∆y=21.027m-0.712m0.712m*0.078s20.0001m=0.005502 m/s2

∆g=∆gt2+∆gL2+∆gy2=0.2425ms22+0.005502ms22+0.005502ms22=0.2426ms2

## PART B

As for part A, the average time is found by averaging the ten periods that the pendulum make while passing the photogate. Excel is then used to also find the standard deviation, the fluctuation, and finally the total uncertainty in time. This was done on excel for three trials while in each trial the length is shorter significantly.

The acceleration due to gravity can now be calculated. With only the knowledge of the period, the acceleration due to gravity, or g, can be rearranged and manipulated, algebraically, in formal (6-5):

T=2πLg

## Where,

g=4π2LT2

The formula above is used to calculate the acceleration due to gravity and can be derived to also calculate the uncertainty in measurement. While deriving this formula, with respect to each variable, the other variables become constants. Thus,

∆gT=8π2LT3∆T

∆gL=4π2T2∆L

## Therefore, the total uncertainty can be calculated below.

∆g=∆gT2+∆gL2

When the length of the string of the pendulum is shortest, it is at 0.314 meters long.

g=4π2LT2=4π20.315m1.1416s2=9.5421 m/s2

∆gT=8π2LT3∆T=8π20.315m1.1416s30.001s=0.01672 m/s2

∆gL=4π2T2∆L=4π21.1416s20.0001m=0.003030 m/s2

∆g=∆gT2+∆gL2=0.01672ms22+0.003030ms22=0.01700ms2

As for the medium length of the string, 0.571m, the calculations are done below for the uncertainty in the acceleration due to gravity.

g=4π2LT2=4π20.571m1.5193s2=9.7658 m/s2

∆gT=8π2LT3∆T=8π20.571m1.5193s30.001s=0.01286 m/s2

∆gL=4π2T2∆L=4π21.5193s20.0001m=0.001710 m/s2

∆g=∆gT2+∆gL2=0.01286 ms22+0.001710ms22=0.0130ms2

## Finally, the longest length of the string was 0.943m and thus,

g=4π2LT2=4π20.943m1.9568s2=9.7225 m/s2

∆gT=8π2LT3∆T=8π20.943m1.9568s30.001s=0.009937 m/s2

∆gL=4π2T2∆L=4π21.9568s20.0001m=0.001031 m/s2

∆g=∆gT2+∆gL2=0.009937ms22+0.001031 ms22=0.009426ms2

## Final Result

PART A

## Acceleration due to gravity is calculated for trial 1 as a possible range of values:

g=9.38 ±0.08ms2;∆gg=0.0086 (0.86%)

## As for trial 2:

g=9.35±0.09ms2;∆gg=0.01 (1%)

## Trial 3:

g=9.65±0.11ms2;∆gg=0.01 (1%)

## Finally, trial 4 and trial 5:

g=9.23±0.14ms2;∆gg=0.02 (2%)

g=9.46±0.24ms2;∆gg=0.03 (3%)

PART B

The range of possible values in part B can be calculated from the three trials taken at different lengths of string.

Trial 1, L=0.315m:g=9.54±0.02ms2;∆gg=0.002 (0.2%)

Trial 2, L=0.571m:

g=9.77±0.01ms2;∆gg=0.001 (0.1%)

And Trial 3, L=0.943mg=9.72±0.01ms2;∆gg=0.001 (0.1%)

## Conclusion

PART A

Displayed above, represents the range of possible values for the acceleration due to gravity. Since each value of g for all five trials has an uncertainty of more or less, 0.01m; all values of g agree with each other and are within each values possible range. The results for the lab are fairly accurate in that the uncertainty for each trial is from 1% to 2% error. The actual value for the acceleration due to gravity is 9.81m/s^2 and from the graph is it shown that the slope represents the average g for the lab, 9.71m/s^2; approximately 0.1m/s^2 of error. It was common that for most of the lab the uncertainty came to be about 1% but for the values that came up to be 2% represents where the most error took place in the lab.

In part A for trials 4&5 one can see the uncertainty is 2%. Thus, error has taken place in this part of the lab which is most significant to the final results. While one can see in trial 5 that g=10.13m/s^2, this value is the farthest from the actual value of g. Therefore, calculated, the experimental uncertainty is also the highest at 0.23m/s^2. The reason for most systematic error is part A is the error that takes place within the photogate and the steel ball as it is falling. When the ball passes the photogate it records the time at that moment. But, when we magnify the view of this moment, one can view that the front of the steel ball his the photogate infrared beam first. Thus, if the ball is dropped differently each time, the steel ball passes the photogate possibly one the side of the ball and therefore recording the time when the side of the ball passes the photogate rather than the front. This process is inconsistent and gives different values for times which will, in effect, skew the range of possible values for the result.

PART B

The three values of acceleration due to gravity that was found in the experiment are very close but not experimentally in agreement. The range of possible values do not contain the first trial in which the value for g was 9.58m/s^2 for the length of 0.314m. This throws off the date significantly giving a slope to the graph of 9.55m/s^2. It is possible that the steel ball just skimmed the infrared beam of the photogate which will obviously cause error in the experiment. As for the two other trials, the numbers are consistent for the most part as the values for g can be considered precise. Furthermore, most of the numbers were consistent and very close to the actual value of g. The experiment was accurate and precise for the most part. Possible errors that cannot be accounted for are the friction that exists while any object is falling or swinging freely. Though the error seems very small, it adds up.

While the length of the string that swung the ball was the smallest, at 0.314m long, the highest uncertainty is seen here. The uncertainty of the tension of the string is the biggest factor here because the when the string is the shortest it is also swinging the quickest and produces the most tension force on the ball causing systematic error. Finally, just from the data collected and from the graphs made, the best way to calculate the value for the acceleration due to gravity comes from the process in part A. Five trials were done and the average comes out to be the closest to the actual value of g.

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