Lab 10: Newton s Second Law in Rotation

Similar documents
Conservation of Angular Momentum

Lab #9: The Kinematics & Dynamics of. Circular Motion & Rotational Motion

Centripetal Force OBJECTIVE INTRODUCTION APPARATUS THEORY

Name. Date. Period. Engage Examine the pictures on the left. 1. What is going on in these pictures?

Section 26 The Laws of Rotational Motion

Physics 211: Newton s Second Law

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Physics 107 TUTORIAL ASSIGNMENT #8

Chapter 5 Force and Motion

Lab #4: Newton s Second Law

Chapter 5 Force and Motion

OSCILLATIONS AND GRAVITATION

Physics 4A Chapter 8: Dynamics II Motion in a Plane

To Feel a Force Chapter 7 Static equilibrium - torque and friction

Current Balance Warm Up

Physics 1114: Unit 5 Hand-out Homework (Answers)

Phys 201A. Homework 6 Solutions. F A and F r. B. According to Newton s second law, ( ) ( )2. j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j.

Magnetic Dipoles Challenge Problem Solutions

Chapter 13 Gravitation

DYNAMICS OF UNIFORM CIRCULAR MOTION

c) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?

AP Centripetal Acceleration Lab

Lecture 13. Rotational motion Moment of inertia

Dynamics of Rotational Motion

Chapter. s r. check whether your calculator is in all other parts of the body. When a rigid body rotates through a given angle, all

Magnetic Field. Conference 6. Physics 102 General Physics II

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018

SAMPLE QUIZ 3 - PHYSICS For a right triangle: sin θ = a c, cos θ = b c, tan θ = a b,

PHYS Summer Professor Caillault Homework Solutions. Chapter 5

Physics 201 Lecture 18

Circular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.

Electrostatics (Electric Charges and Field) #2 2010

AP Physics 1 - Circular Motion and Gravitation Practice Test (Multiple Choice Section) Answer Section

Rotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart

Multiple choice questions [100 points] As shown in the figure, a mass M is hanging by three massless strings from the ceiling of a room.

UNIT 13: ANGULAR MOMENTUM AND TORQUE AS VECTORS Approximate Classroom Time: Two 100 minute sessions

Quiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)

Between any two masses, there exists a mutual attractive force.

Chapter 7-8 Rotational Motion

Chapter 2: Introduction to Implicit Equations

MODULE 5 ADVANCED MECHANICS GRAVITATIONAL FIELD: MOTION OF PLANETS AND SATELLITES VISUAL PHYSICS ONLINE

Motion in One Dimension

Chapter 4. Newton s Laws of Motion

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

Algebra-based Physics II

Physics 111 Lecture 10. SJ 8th Ed.: Chap Torque, Energy, Rolling. Copyright R. Janow Spring basics, energy methods, 2nd law problems)

Experiment 09: Angular momentum

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

Physics 2001 Problem Set 5 Solutions

Circular motion. Objectives. Physics terms. Assessment. Equations 5/22/14. Describe the accelerated motion of objects moving in circles.

Flux. Area Vector. Flux of Electric Field. Gauss s Law

Rotational Motion: Statics and Dynamics

Physics 181. Assignment 4

Rotational Motion. Lecture 6. Chapter 4. Physics I. Course website:

Centripetal Force. Lecture 11. Chapter 8. Course website:

Motithang Higher Secondary School Thimphu Thromde Mid Term Examination 2016 Subject: Mathematics Full Marks: 100

PS113 Chapter 5 Dynamics of Uniform Circular Motion

B. Spherical Wave Propagation

Lecture 13 EXAM 2. Today s Topics: Rotational motion Moment of inertia. Tuesday March 8, :15 PM 9:45 PM

r cos, and y r sin with the origin of coordinate system located at

Uniform Circular Motion

History of Astronomy - Part II. Tycho Brahe - An Observer. Johannes Kepler - A Theorist

Physics 111 Lecture 5 Circular Motion

Nuclear and Particle Physics - Lecture 20 The shell model

06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )

Handout: IS/LM Model

Physics 161 Fall 2011 Extra Credit 2 Investigating Black Holes - Solutions The Following is Worth 50 Points!!!

Circular Orbits. and g =

b) (5) What average force magnitude was applied by the students working together?

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics

3.2 Centripetal Acceleration

The Millikan Experiment: Determining the Elementary Charge

Chapter 8. Accelerated Circular Motion

PHYSICS 151 Notes for Online Lecture #20

Phys 201A. Homework 5 Solutions

Uniform Circular Motion

Inverse Square Law and Polarization

Sections and Chapter 10

7.2. Coulomb s Law. The Electric Force

Gravitation. AP/Honors Physics 1 Mr. Velazquez

PY208 Matter & Interactions Final Exam S2005

1121 T Question 1

Physics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 6: motion in two and three dimensions III. Slide 6-1

Section 8.2 Polar Coordinates

MAGNETIC FIELD INTRODUCTION

Solutions to Problems : Chapter 19 Problems appeared on the end of chapter 19 of the Textbook

( ) ( ) Review of Force. Review of Force. r = =... Example 1. What is the dot product for F r. Solution: Example 2 ( )

Physics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism

AP * PHYSICS B. Circular Motion, Gravity, & Orbits. Teacher Packet

Circular Motion. x-y coordinate systems. Other coordinates... PHY circular-motion - J. Hedberg

LIST OF EQUIPMENT SET UP AND PROCEDURE 3B SCIENTIFIC PHYSICS EXPERIMENT

Describing Circular motion

Our Universe: GRAVITATION

PHYSICS 1210 Exam 2 University of Wyoming 14 March ( Day!) points

Uniform Circular Motion

PHYSICS 220. Lecture 08. Textbook Sections Lecture 8 Purdue University, Physics 220 1

ω = θ θ o = θ θ = s r v = rω

Physics 11 Chapter 4: Forces and Newton s Laws of Motion. Problem Solving

Transcription:

Lab 10: Newton s Second Law in Rotation We can descibe the motion of objects that otate (i.e. spin on an axis, like a popelle o a doo) using the same definitions, adapted fo otational motion, that we have used fo objects that move in a staight line. That is, fo linea motion we defined position, velocity, and acceleation; fo otational motion we define position, angula velocity, and angula acceleation. Fo objects in linea motion, Newton s Second Law tells us that foces acting on the object cause its acceleation; fo an object in otational motion, Newton s Second Law tells us that toque acting on the object causes its angula acceleation. n Lab 4 we demonstated Newton s Second Law fo an object in linea motion: we attached a sting to a cat on an aitack and a hanging mass and allowed the system to acceleate. n Lab 10, we will attach a sting to a otating appaatus and a hanging mass; as the hanging mass falls, the tension of the sting will ceate a toque on the otating appaatus that will cause it to otate. Newton s Second Law in otation has exactly the same fom as fo linea motion. That is: Linea: Foce = mass acceleation Rotation: Toque = moment of inetia angula acceleation n symbolic notation, the latte equation is τ = α whee τ and α ae the Geek lettes tau and alpha. We use to epesent the moment of inetia, which plays the same ole in otational motion that mass does fo linea motion. The toque applied to a otating object is defined as: τ = F sinφ whee F is the foce applied to the object, is the distance fom the point of otation to the point whee the foce is applied, and φ is the angle between and F. Note that if a foce is applied at the point (o axis) of otation, the toque will be zeo (because is zeo.) Think about opening a doo: if you push on the side

with the hinge, will be zeo, and the doo will not move. This is why we push on the side of the doo opposite the hinge: bigge esults in bigge toque, making it easie to open the doo. Fo Lab 10, we can conside thee foces acting on the otating appaatus: gavity, nomal foce and the tension of the sting. Gavity and nomal foce each act at the cente of the appaatus, so is zeo fo each of these two foces; they exet no toque (i.e. gavity and nomal do not cause the appaatus to otate.) The tension acts at the edge of the clea hoizontal pulley at a distance equal to the adius of the pulley. The tension pulls along the edge of the pulley (i.e. in a tangential diection), so the angle between and F is 90 degees. Fom these obsevations, we can claim that the toque acting on the appaatus is: τ = T whee is the adius of the clea hoizontal pulley and T is the tension of the sting. Just as the toque acting on the object depends on the foces and whee those foces ae applied, the moment of inetia of a otating object is a measue of the mass of the object and how it is distibuted. Mass that is fathe fom the cente of otation (i.e. the axis of otation) is moe difficult to otate, because it moves in a lage cicle, than mass that is close to the axis. Fo this eason, the moment of inetia of an object is defined by its mass and its linea dimensions. The geate the mass and the geate the size of the object, the geate its moment of inetia. Lab 10 has two pats: Pat 1: Calculate the expected value of the moment of inetia of the otating appaatus. Pat 2: Measue the moment of inetia of the otating appaatus by applying a toque to the system (by the tension in the sting) and measuing the coesponding angula acceleation. Afte measuing the data fo Pat 2, we will ceate a gaph and fom the gaph we will be able to detemine the moment of inetia of the appaatus (i.e. how fast it actually acceleated elative to how much toque was applied... Newton s Second Law!) To accomplish this, we have to do a bit of algeba: Rotating appaatus: T = α (this is a esult of the definitions pesented above.) Hanging mass: mg - T = ma The sting and hanging mass have the same linea acceleation, and the sting has the same linea acceleation as a point on the edge of the clea pulley. The clea pulley also has the same angula acceleation as the otating appaatus. We can algebaically connect the linea acceleation and angula acceleation of a point on the edge of the clea pulley: a = α whee is the adius of the clea pulley

We can eliminate a fom these thee expessions: α T = mg T = mα And then eliminate T: mg α = mα We will measue seveal values of m and the coesponding value of α fo each. We would like to be able to gaph these two vaiables as a staight line, but the equation above is fa too complicated to even hope fo a staight line with α and m. Fotunately a little bit of algeba can help. Fist, we ll eaange the equation to solve fo α: mg = mα + α α = mg m + / This will not wok fo a staight line gaph, as m appeas in the numeato and denominato on the ight side. So we use a cleve algebaic tick: invet both sides. m + / O: α = α = + mg g mg α = m g + g n this final expession, we have α -1 on the left and m -1 on the ight. f we gaph α -1 vs m -1, we expect to get a staight line with a slope of / g. Pat 1 We will neglect the contibution of the clea plastic pulley to the moment of inetia of the system. The mass of the pulley is elatively small and it is elatively close to the axis of otation, so the effect of the pulley on the inetia of the otating system should be negligible. The moment of inetia of the otating system should include only the long, thin metal od and the two bass weights (at eithe end of the od.)

To find the expected value of the moment of inetia: Caefully measue M od and L, the mass and length of the od. Measue M b, the mass of both bass weights togethe. Measue d, the distance fom the cente of the od to the cente of one of the bass weights. Calculate the moment of inetia fo the od and bass weights sepaately, and then add the esults to get the total moment of inetia of the appaatus: od = 1 12 M od L 2 bass = M d b 2 Show these calculations in you Excel sheet. Expess the esult of each calculation to thee significant figues. You total moment of inetia should be aound 50,000 g-cm 2. Pat 2 Using the calipe, measue the diamete of the clea pulley that will hold the black sting. Measue the diamete with and without the sting and aveage these two measuements. Recod the adius of the pulley by taking half of the aveage diamete. Set up the otating appaatus so that the sting between the appaatus and the black pulley is hoizontal and passes ove the black pulley staight. You will have to set up the pulley at an angle to the appaatus; see the example. Place 30 gams on the mass hange. The hange itself is 5 gams, so the total hanging mass is 35 gams. Afte connecting the Rotay Motion Senso to the black box and the black box to the compute, open Data Studio. Select the Rotay Motion Senso fom the input menu. Choose only angula velocity fom the Measuements menu and change the units to ad/sec. Fom the display menu choose Gaph.

Make sue the sting is wapped aound the coect clea plastic pulley. Release the appaatus, allowing it to otate. Click START on Data Studio; the softwae will display points on a gaph of angula velocity vs time. Click STOP befoe the sting is completely unwound; stop the otating appaatus. Use the FT menu at the top of the Data Studio gaph to display the stats fo the line. The slope of the line on the Data Studio gaph is the angula acceleation; ecod it in you data table. Ceate a data table of fou columns: m (in gams) α (in ad/s 2 ) m -1 (in g -1 ) α -1 (in s 2 /ad) Repeat the above pocedue fo m = 35, 45, 55, 65, 75, 85, 95. Ceate a gaph of α -1 vs m -1 and find the equation of the best fit line. Using the slope of the best fit line, calculate the measued value of the moment of inetia. Fom the equation deived above, the slope should be equal to / g. So: meas = g( slope) =... Note that all of ou measuements ae in gams and centimetes, so you should use 980 cm/s 2 fo g. Compae you measued value to the expected value fom Pat 1 (i.e. calculate the pecent diffeence.) f you measued caefully, you should find a diffeence of less than 1%. We will look fo evidence of andom and systematic eo in the class esults (i.e. does eveyone have a small positive diffeence, o ae thee some positive and some negative diffeences.)