Lab 3 ( kinematic free fall) PDF

Preview

Summary

it is a mandatory lab for Physics 181 students ...

Description

Experiment 3 Kinematics free fall

My Lab Partner:

TA:

I. Purpose The purpose of this experiment is to study the motion of an object in free fall, and to deduce from the measurements, a value for the acceleration due to gravity, g, which we will compare to the accepted value of 980.35 cm/s2.

1

II.

Data

2

Graph

III. Calculations According to the general theory of motion, the vertical position, y, and velocity, v, of an object falling freely with constant acceleration of magnitude g, are given by: a= g …………………………………….. (1) v= Vo+¿ …….…………………… (2) 1 ∆ y=Vot + g t2 ……….……….. (3) 2

In using the above equations for this experiment, following is assumed: 1. The vertical position, y, the velocity, v, and the acceleration, g, are measured positively in the downward direction. At time t = o

3

y = yo = 0 cm v= Vo Divide both side by t and yo can be removed. So our new equation would be, y 1 =Vo + >¿ t 2

As we find our equation, from our graph, The accepted value for g = 980.35 cm/s^2 The slope of the graph S = 518.452 (cm/s^2) The gravity of our experiment, g = 2* 518.452 = 1036.904 cm/s^2 = 1040 cm/s^2

Then, the error is, Sg = 2* 24.52719 cm/s^2 = 49.05438 cm/s^2 Round up= 50 cm/s^2 To find the percent difference, we can use the following percent difference equation.

% =

%=

− gaccepted |gexperimental |∗100 gaccepted

− 980.35 | 1036.904 |∗100 980.35 = 5.8%

4

IV. Results The acceleration of the objects due to the gravity is measured 1040 cm/s^2 +/- 50 cm/s^2. This solution had a percent difference of 5.8% when compared to the accepted value of 980.35 cm/s^2.

V.

Error Analysis/Discussion We previously stated that our experiment yielded the following results. g𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒= 1040 cm/s^2 𝑒 g = 50 cm/s^2 g𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 980.35 cm/s^2. We can then compare the experimental and accepted values. |g𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒− g𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒| = |1040−980.35| = 59.65 when comparing this value to the error in theta, 60cm/s^2 > 50 cm/s^2 Thus, |g𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 − g𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒> |>

g

Systematic errors cannot be ignored because the accepted value doesn’t falls within our bounds. In addition. To get better results, the sample size could be increased by taking more measurements. Systematic error can be difficult to identify and correct. In our experimental procedure and setup, it doesn't matter how many times we repeat and average our measurements; the error remains unchanged. No statistical analysis of the data set will eliminate a systematic error, or even alert you to its presence. We can minimize Systematic error with careful

5

analysis and design of the test conditions and procedure; by comparing our results to other results obtained independently, using different equipment or techniques; or by trying out an experimental procedure on a known reference value, and adjusting the procedure until the desired result is obtained.

VI. Questions 1. Whenever you have correlated data, you should graph it. So in Excel, you should have graphed y/t vs. t. So based on this graphical evidence, is acceleration a constant? Answer yes or no, and explain how you determined this.

Ans: from the Excel graph, the acceleration is constant. Constant acceleration refers to motion where the speed increases by the same amount each second. The most notable and important example is free fall. When an object is thrown or dropped, it experiences a constant acceleration due to gravity, which has a constant value of approximately 10 meters per second squared. In the reading of the graph tells us after each 1/60 second the acceleration is almost same value. That’s why that graph has a liner line, except for first two dots are far away from the line because that where the object started to coming down to ground From these value the rate of change is the same every time when the object is acceleration due to gravity. Technically, free fall is the best example of constant speed (neglecting the air resistance) but in that case from the graph the acceleration is not constant.

2. Using your vo and g from your two-dimensional statistical linear analysis, compute the instantaneous velocity v (Eq. 2) of the falling object at t = 15.5/60 seconds.

Ans: Equation 2, V= Vo + gt V= 149.457 *(15/60) + 1036.904(15/60) = 296.59025 cm/s

Given, Vo = 149.457 t = 15/60 g = 1036.904 V= ?

3. Compute the average velocity in the interval ta = 15/60 to tb = 16/60 seconds using values of y from your data for these times. The average velocity is given by: changein t changein y = (t -t ) ( y - y ) v = b a avg 16/ 60 15/ 60

6

Ans:

¿

Vavg =

76.5 −69.7 ∨¿ 0.27−0.25

given,

yb = 76.5 cm

= 340 cm/s

Ya = 69.5 cm 4. If the acceleration is constant, the two values computed in the last two questions should be the same. Taking into account errors in measurements, are they basically the same? Ans: the acceleration is constant, taking into account error measurements there is slightly different between two values. From the 2nd equation we got approximately 300 cm/s^2 and from 3rd question we got 340 cm/s^2. If we look at our measurement value and graph, the first two measurements are far away from our liner graph and these two value makes our Sslope, slope, and intercepts less accurate. Besides this two points, if we calculate these two value will be same. And also, point (15/60) and (16/60) are vertically far way from each other what gives us a greater velocity. We can say, the acceleration is constant, even though the value are not exactly same due to the error measurement. t s

y cm 8.6 12.5

y/t cm/s^2 0.05 172 178.571 0.07 4

16.3

0.08

20.5

0.1

24.8

0.12

29.5

0.13

34.4

0.15

39.6

0.17

45.1

0.18

50.9

0.2

56.6

0.22

63.1

0.23

69.7 76.5

0.25 0.27

203.75 205 206.666 7 226.923 1 229.333 3 232.941 2 250.555 6 254.5 257.272 7 274.347 8 278.8 283.333

N 18 slope (units) 506.372 8 intercep t (units) 152.226 6 R2 0.98251 6 Sy (units) 6.18299 7 Sslope (units) 16.8875

7

83.6

0.28

91

0.3

98.7

0.32

106.5

0.33

3 298.571 4 303.333 3 308.437 5 322.727 3

1 Sintercept (units) 3.54972 6

5. The square of the linear correlation coefficient, 𝑒 2 , is an important guide when judging relationships between two quantities. If you preformed your linear 2D Stats on the (𝑒, 𝑒) data, not the linearized (𝑒, 𝑒 𝑒 ) data, explain how 𝑒 2 might help you spot such a mistake. Ans: In general, the square of the linear correlation coefficient, R^2 is an important guide when judging relationships between two quantities. R-squared is a statistical measure of how close the data are to the fitted regression line. It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. The definition of R-squared is fairly straight-forward; it is the percentage of the response variable variation that is explained by a linear model. R-squared is always between 0 and 1. 0 indicates that the model explains none of the variability of the response data around its mean. 1 indicates that the model explains all the variability of the response data around its mean. In general, the higher the R-squared, the better the model fits your data. However, In our measurement, for measurement error we got better R^2 value for (y,t) date than our (y/t, t). If there is no measurement error we would get a better R^2 value for (y/t, t) than (y,t) what would us that in our measurement system something is wrong because the value will be far way from 1 for ( y,t) date. This way we can spot our mistake if we take the wrong variable in our date.

y cm

t s 2.6 5.6 8.6 12.5 16.3 20.5 24.8 29.5 34.4 39.6

N 20 0.02 slope 0.03 (units) 0.05 0.00299 0.07 intercept 0.08 (units) 0.1 0.035068 0.12 R2 0.13 0.983209 0.15 Sy 0.17 (units)

8

45.1 50.9 56.6 63.1 69.7 76.5 83.6 91 98.7 106.5

0.18 0.013105 0.2 Sslope 0.22 (units) 0.23 9.21E-05 0.25 Sintercept 0.27 (units) 0.28 0.005212 0.3 0.32 0.33

6. Preform a hand drawn graph of the y/t vs. t data following the course’s criteria.

9

10

11...