Lab report 2

docx

School

University of Mississippi *

*We aren’t endorsed by this school

Course

213

Subject

Physics

Date

Apr 3, 2024

Type

docx

Pages

Uploaded by cadencetiller

Tiller, Cadence Partner: Treadway, Luke Moment of Inertia TA: Pandey, Kumar, Section #3 Date Performed: April 17, 2023 Objective : The objective of this experiment is to determine the moment of inertia of the rotating system by We manipulated the experiment by adding mass to the hanger and on the system. We then had to accurately determine the new moment of inertia. Moment of inertia is defined as the rotational analog of mass. Newton’s second law states that the acceleration of a body is directly proportional to the vector sum of all forces applied to the body (F= ma ). Moment of inertia is derived from this equation τ = Iα. When an object experiences a constant angular acceleration, it must be under the influence of a constant torque. This is similar to how constant linear acceleration implies a constant force acting on an object. In rotational motion, the relationship between torque, moment of inertia, and angular acceleration is given by τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. This equation can be used to determine the moment of inertia of a rigid body by measuring the torque applied and the resulting angular acceleration. Then, by rearranging the equation τ = Iα, you can solve for the moment of inertia I. The Moment of Inertia equation is I = t/  . The independent variable in this experiment are the masses added to the hanger and the rotating system, and the dependent variable is the moment of inertia. Data and Analysis:

In this experiment, we began by measuring the diameter of the axle ( D axle ) with the venier caliper. We found the radius of the axle ( r axle )by dividing the diameter by two. We then used a meter stick to measure the distance from the bottom of the mass hanger to the floor which represent our y floor. We found the R masses by measuring the distance from the center of the disk to the outer set of tapped holes. We used a stopwatch to measure the time it took for the hanger to fall to the floor. We found the average time by adding all the trials up and dividing by five. In part 1a, we measured the time it took for the hanger (0.05 kg ) to fall to the ground. In part 1b, we added an additional 100 g (0.1 kg ) to the hanger and measured the time it took for the hanger to fall the ground. In part 2, we added three masses to the disk which are 1.35 kg each. In part 2a, we used the mass of the hanger (.05 kg ), and in part 2b we added the additional 100 g (0.1 kg ) and measured the time it took for the hanger to reach the ground. From this we were able to calculate the linear acceleration, angular acceleration, tension, and torque. D axle : 0.056 m r axle : 0.028 m Y floor : 0.848 m R masses : 0.172 m Raw Data: Trial t-Part 1 (a) (s) t-Part 1 (b) (s) t-Part 2 (a) (s) t-Part 2 (b) (s) 1 12.08 ( s ) 6.74 ( s ) 26.90 ( s ) 15.48 ( s ) 2 11.88 ( s ) 6.95 ( s ) 26.47 ( s ) 15.55 ( s ) 3 12.42 ( s ) 7.21 ( s ) 27.20 ( s ) 15.22 ( s ) 4 12.36 ( s ) 7.03 ( s ) 26.49 ( s ) 15.73 ( s ) 5 11.74 ( s ) 6.63 ( s ) 26.81 ( s ) 15.73 ( s )

Average 12.24 ( s ) 6.91 ( s ) 26.78 ( s ) 15.54 ( s ) Calculated Data: Part 1 (a) Part 1 (b) Part 2 (a) Part 2 (b) a 0.011 m / s 2 0.035 m / s 2 0.002 m / s 2 0.007 m / s 2  0.392 rad / s 2 1.29 rad/s 2 0.07 rad/s 2 0.25 rad/s 2 T 0.489 N 1.46 N 0.49 N 1.46 N t 0.013 N * m 0.04 N * m 0.013 N * m 0.04 N * m Calculations: a = 2  y /t 2  = a / r T = m (g – a ) t = rT Part 1 (a): a = 2(0.848 m ) / (12.24 s ) 2 = 0.011 m / s 2  = 0.011 m / s 2 / 0.028 m = 0.392 rad/s 2 T = 0.5 kg (9.8 m/s 2 – 0.011 m/s 2 ) = 0.489 N t = 0.028 m * 0.489 N = 0.013 N * m Part 1 (b): a = 2 (0.848 m ) / (6.91 s ) 2 = 0.035 m/s 2  = 0.035 m/s 2 / 0.028 m = 0.392 rad/s 2 T = 0.150 kg (9.8 m/s 2 – 0.035 m/s 2 ) = 1.46 N

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

t = 0.028 m * 1.46 N Part 2 (a): a = 2(0.848 m ) / (26.78 s ) 2 = 0. 002 m/s 2  = 0.002 m/s 2 / 0.028 m = 0.07 rad/s 2 T = 0.05 kg (9.8 m/s 2 – 0.002 m/s 2 ) = 0.49 N t = 0.028 m * 0.49 N = 0.013 N*m Part 2 (b): a = 2 (0.848 m ) / (15.54 s ) 2 = 0.007 m/s 2  = 0.007 m/s 2 / 0.028 m = 0.25 rad/s 2 T = 0.150 kg (9.8 m/s 2 – 0.007 m/s 2 ) = 1.46 N t = 0.028 m * 1.46 N = 0.04 N * m We were given the Experimental I 0 from the slope of the Part 1 graph that compared angular acceleration vs. torque. From this we were able to calculate the Theoretical I new . The Experimental I new was found from the Part 2 graph comparing the angular acceleration vs. torque. Theoretical I new = I 0 + 3MR 2 Theoretical I new = 0.033 kg * m 2 + 3(1.35 kg )(0.172 m ) = 0.153 kg * m 2 Experimental I new = 0.158 kg * m 2 Percent Difference I new =  I new, exp. – I new,theo.  / ( I new, exp. + I new, theo. ) /2 * 100 Percent Difference I new =  0.158 kg * m 2 – 0.153 kg * m 2  / (0.158 kg * m 2 + 0.153 kg * m 2 ) /2 *100 = 3.22% Conclusion: The objective of this experiment was to determine the moment of inertia of the rotating system, then altering the system by adding additional masses to predict the new moment of

inertia. In this experiment, when we added additional mass, it had an effect on the angular acceleration. The angular acceleration increased with an increase in mass. Overall, this affected the moment of inertia of the system. Angular acceleration and moment of inertia are proportional (  = net t / I ). The greater the angular acceleration the smaller the moment of inertia, and the smaller the angular acceleration the larger the moment of inertia. Newton’s second law states that F=ma showing that mass and acceleration are proportional. The greater the mass the slower the acceleration, and the smaller the mass the faster the acceleration. This can be proven when observing the different acceleration of the trials. In this experiment, we were successfully able to find the moment of inertia and predict the new moment of inertia after altering the system. A few sources of failure for our experiment include using the stopwatch to accurately time the hanger hitting the ground. In this experiment, we failed to have the string wrapped around our axle in the right direction, which could have altered the time it took for our hanger to reach the ground. This overall could have affected our acceleration of the system. Some improvements for this lab could be to in the instructions state which way the string is supposed to be wrapped around the axle. Also, address how to properly time the hanger hitting the ground and stopping the stopwatch to reduce error in this experiment. Questions: 1. What are the units for Torque, Moment of Inertia, and Agular Acceleration? Show all work. - Torque- t = N*m , Newton’s * Meters - t= rfsin  , r=m, f=N - Moment of Inertia - I=kg * m 2 , Kilograms * meters squared

- I=mr 2 , m= kg , r 2 = m 2 - Angular acceleration -  = rad/s 2 , Radians/ seconds squared -  =  w/  t, w= rad/s, t=s 2. If the Torque applied to a rigid body is doubled, what happens to the moment of Inertia? - t=I  - Torque is proportional to inertia so if the torque is doubled the moment of inertia is doubled. 3. Why did you need to calculate acceleration to determine I 0 ? Could you have calculated the I 0 without running any trials? - 4. Were any torques ignored in this experiment? What are they? Do you believe they may have significantly altered your results?

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Related Documents

FMO1 Final Kress.docx

2DC Final Kress.docx

E1 Report.pdf

Lab 15 Asteroids and Kuiper Belt Objects question 2 answers.png

Final Review - Comprehensive.pdf

Midterm Review - Chapters 21-28.pdf

06-07_task.docx

Laboratory 3.docx

Lecture 16 Assignment.docx

Lecture 18 Assignment.docx

Phys 132 Lab 8.pdf

PHYS172_S24-Exam2-PRACTICE-SOLUTIONS-TWEAKS.pdf

Recommended textbooks for you

Glencoe Physics: Principles and Problems, Student...

Physics

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Glencoe/McGraw-Hill

Physics for Scientists and Engineers: Foundations...

Physics

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Cengage Learning

Physics for Scientists and Engineers with Modern ...

Physics

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Cengage Learning

College Physics

Physics

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:OpenStax College

University Physics Volume 1

Physics

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:OpenStax - Rice University

College Physics

Physics

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Cengage Learning

SEE MORE TEXTBOOKS

Recommended textbooks for you

Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning