Formal Lab 4 Final PDF

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Summary

This was lab report 4(One of 2 formal reports)...

Description

Physics 1501Q Section 001

Formal Lab 1 Abstract: In this lab we looked at a spinning apparatus, which was composed of a hanging mass and a rubber stopper, to understand and test the centripetal force associated with that object. To attest this, we attached masses to one end of a string, and a rubber stopper to the other ends, and then swung the rubber stopper in such a way that it sung in a constant acceleration. We found that though our data was quite close to the theoretical values but it still was not within the upper and lower bounds. Intro: It is commonly known that a body moving at a constant velocity will continue in the same straight line at a constant speed, unless it is acted upon by a outside force. This is known as Newton’s First Law. If a body in motion is acted upon by a force that is non-parallel to it velocity, the object would move along a curved path. It is assumed in that friction is reliable. It is also assumed that gravity is constant. The spring connected to the hanging mass is also massless. For our lab, we first calculated the theoretical time that it would take using equation 1. We then calculated the uncertainty for the time in the positive and negative direction using equations 2 and 3 respectively. We then preformed the experiment connecting two masses to a string and spinning one of them. You can see the free body diagram of both of the two masses by looking at figure 1. F=

mrs ×V 2 R

Where:

Distance=2 πr

Time panel = T 2 πr T M H =Hanging Mass

V = velocity = M rs=Mass Rubber Stoper FT =cosθ FT =M H × g m 2 πr 2 F= rs × T R 2 2 πr F=m rs × ×R T 2 πr 2 R=Lcosθ ∴ mrs × × Lcosθ T 2 2 πr FT × cosθ =m rs × × Lcosθ T 2 πr 2 ∴ F T =m rs × ×L T

( ) ( )

( ) ( ) ( )

( )

2 πr 2 ×L T M rs (2 π )2 L 2 ∴T = × MH g M H =M rs ×

Let M =

M rs M (2 π )2 L ∴T 2= g MH

=

4 π 2 × LM … … … … … … … 1 g

To calculate the uncertainty… 2

4π ×( L+ SL)( M + SM ) …………………2 g ¿ 4 π2 −¿ 2 ×(L−SL)( M −SM )… … … … … … … 3 T = g ¿

+¿

2

T =

Figure 1: FBD of the Experiment

Procedure:     

 

First weigh and record all of the masses of the rubber stoppers. Attach one rubber stopper to the top of the apparatus. Add a couple weights on the end of the apparatus; this is the hanging mass. Swing the top of the apparatus in such a way that the rubber stopper moves with a constant acceleration. Have one of the group members start the timer and record the time that it takes for 10 revolutions. o Have another group member be ready to place their hand quickly on the top of the apparatus after the 10th revolution. Then using a ruler, find and record the length of the string from the apparatus to the metal hook. Do this for each of the stopper masses. o You might find it helpful in order to put a piece of tape or mark the string in order to make sure that the stopper is moving with a constant acceleration.

Data

T^2 versus Mass 0.8

T^2(-) Linear (T^2(-)) T^2 + Linear (T^2 +)

0.7 0.6

T^2

0.5 0.4 0.3 0.2 0.1 0 0.3

0.35

0.4

0.45

Mass

0.5

0.55

0.6

Data for the Experiment:

M 0.348333333 0.4025 0.434166667 0.568666667 0.4765

T^2 (-) 0.148293519 0.132556494 0.357463422 0.655482256 0.434420222

T^2 (+) 0.151697087 0.136333826 0.362397567 0.662695567 0.440004216

Theoretical T^2 Experimental T^2 0.1499925 0.169744 0.13444192 0.178084 0.359927 0.451584 0.659084335 0.760384 0.437208384 0.512656

Analysis:

In our experiment, we compared the experimental time value that we got, to the approximate time that we had calculated. We were also told to calculate the uncertainty of the time, an example of which is shown below. We found that for the for the 20.9g rubber stopper, the theoretical time calculated was 0.1499925 seconds and the actual experiment time was 0.169744 seconds. The range for the time was 0.148293519 seconds  0. 151697087 seconds, making the experimental value time fall outside the range. We found that for the for the 32.2g rubber stopper, the theoretical time calculated was 0.13444192 seconds and the actual experiment time was 0.178084 seconds. The range for the time was 0. 132556494 seconds  0. 136333826 seconds, making the experimental value time fall outside the range. We found that for the for the 52.1g rubber stopper, the theoretical time calculated was 0.359927seconds and the actual experiment time was 0.451584 seconds. The range for the time was 0.357463422seconds  0.362397567 seconds, making the experimental value time fall outside the range. We found that for the for the 85.3g rubber stopper, the theoretical time calculated was 0.659084335 seconds and the actual experiment time was 0.760384 seconds. The range for the time was 0.655482256 seconds  0.662695567 seconds, making the experimental value time fall outside the range. We found that for the for the 95.3g rubber stopper, the theoretical time calculated was 0.437208384 seconds and the actual experiment time was 0.512656 seconds. The range for the time was 0.434420222 seconds  0.440004216 seconds, making the experimental value time fall outside the range. Sadly every single one of our experimental values was outside of the acceptable range. There are many reasons for our error. The most obvious one could be that the rubber stopper was not actually moving with a constant velocity. Another area where our error could have occurred would be when we were stopping the velocity after the 10th rotation in order to measure the length of the string. We might have stopped it incorrectly, causing the string to shorten or lengthen, thereby messing with the data. One of the last sources of error would be because of whoever was working with the stopwatch. If their reaction time was not great, that could also mess up the results. Let M =

2 M rs M (2 π ) L ∴T 2= MH g

=

4 π2 × LM … … … … … … … 1 g

To calculate the uncertainty… 4 π2 ×(L+ SL)(M +SM ) …………………2 g ¿ 2 4π −¿ 2 ×(L−SL)(M −SM ) … … … … … … … 3 T = g ¿

+¿

T 2=

δL =0.001 m δ Mrs δ MH δM = M + MH MH

(

)

∴ For the First Mass as an example… +¿

T 2=

4 π 2 ×(0.107 + 0.001)(0.34833 + 0.000696667) 9.81 ¿

= 0.151697087

2

−¿

2

T =

4π ×(0.107 −0.001 )( 0.34833 −0.000696667) = 0.148293519 9.81 ¿

General: 4.2) It is the force that is being acted upon the body is what keeps changing with time, causing the acceleration of circular motion. 4.4) The hanging mass must have a greater mass then the rubber stopper because it is the hanging mass that makes sure that the centripetal force doesn’t allow the rubber stopper to pull the string all the way out. 4.5) The marked segment on the sting moves faster as you spin the rubber stopper faster. This is because the rubber stoppers centripetal force is getting stronger, causing the hanging mass to rise up higher. If the rubber stopper is spun slower, then the centripetal force is less, allowing the hanging mass to pull more of the string down. You can see this using equation 4.6 because as the force increases, so does the radius. Conclusion: Throughout the course of the experiment we found that experimental values were well +¿ 2 T and −¿T 2 value range. Looking at the graph we can clearly see this, as all outside the ¿ ¿ of the experimental points fell outside the trend lines. One way the experiment could have been

improved would be if there was some sort of locking mechanism on the top of the apparatus that held the string in place after the revolutions so that we could accurately measure the string. Another thing that could have been changed would be maybe finding a way to make sure that the rubber stopper was actually moving at a constant velocity with circular motion....