LEC17 - Lecture notes 17 PDF

Title LEC17 - Lecture notes 17
Course Modern Mechanics
Institution Purdue University
Pages 28
File Size 1.8 MB
File Type PDF
Total Downloads 53
Total Views 180

Summary

weekly lecture notes...


Description

PHYS 172: Modern Mechanics Spring 2021

Lecture 17 Tuesday, March 23rd

MID TERM EXAM 2 • SEE EMAIL/ANNOUCEMENT IN Brightspace. • Exam will be administered via Examity and LON CAPA – Be sure to complete Examity Process Practice (see email)

• WHEN: 24 Hr. Window to access 2 Hr. Exam – Starting 9:00AM ET, Mon. 3/29, Closes 8:59AM ET, Tue. 3/30. – You must access the exam in Examity 2 hours prior to closing

• The exam will consist of up to 30 MC questions • No books, equations sheets, notes, human assistance, internet searches, chat rooms, etc. • Accommodations as per DRC letters (if you have not provided one, please do so ASAP!)

How to Prepare • Practice Exam posted in Brightspace – Solutions will be posted 12 Noon Wed (March 24th)

• ‘What You Should be Able To Do’ – Review this sheet (formula sheet neither allowed nor provided on exam)

• Review : Problem Quiz 3 & 4, HW (11-18), Recitation (6-10), Lecture Quiz problems • Read relevant sections of text, analyze example problems

EXAM 2 • FORMAT SIMILAR TO EXAM 1 : Multiple Choice • COVERS: ALL MATERIAL SINCE EXAM 1. • PREPARATION: – RECITATION PROBLEMS – LECTURE : LECTURE Qs AND PROBLEMS – LAB, QUIZZES, HW, EXAMPLES IN TEXTBOOK, END CH PROBLEMS

• PRACTICE EXAM 2 POSTED: YOUR MAIN RESOURCE! – TAKE IT LIKE A REAL EXAM (TIMED, WITH CALCULATOR, NO FORMULA SHEET, NO HELP FROM OTHERS) – CHECK YOUR ANSWERS & SOLUTIONS (POSTED 12 NOON WED. 3/24 ) : FIGURE OUT WHY YOU GOT QUESTIONS WRONG – MAKE SURE YOU UNDERSTAND HOW TO SOLVE THE PROBLEMS & HOW THEY CAN BE TWEAKED 4

RECAP: ENERGY PRINCIPLE

Revised

+

5

RECAP: MULTIPARTICLE KINETIC ENERGY Three modes of movement • Translation of Center of Mass (CM) • Rotation about axis passing through CM • Vibration around CM Total Kinetic Energy KE due to motion relative to Center of Mass (CM) 6

TRANSLATIONAL KINETIC ENERGY SYSTEM CM

ORIGIN

7

ROTATIONAL K.E Rigid rotator with multiple particles: 









  







MOMENT OF INERTIA Depends on mass of object & how it is distributed





PARALLEL AXIS THEOREM • Previous examples: – Axis passing through CM • When Axis Does NOT Pass Through CM X

Moment of Inertia through Axis of Rotation Moment of Inertia of object about axis thru CM

Axis of Rotation

CM

Shortest distance of CM from Axis of Rotation 9

APPLYING ENERGY PRINCIPLE POINT PARTICLE

EXTENDED 







 















 





Rotational Motion

= 0 because Point Particle





 



 





Moment of Inertia

10

POINT PARTICLE vs. EXTENDED

• Assume – entire mass is concentrated at CM – All forces act at CM. • Find Net Force: • Find displacement of CM: • Calc. Work by Net Force  • Calc. change in system’s Translational KE: 



 



• On diagram : Each Force, & its point of application. thru which • Find displacement each force acts. • Find Work Done by Each Force:  • Find Total Work done by all F’s:  • Set Up:  







 



Problem: Sliding Box

a) What is the speed of the box at that later time? b) As you pull the string through the tight hole in the box you expect the string and box to get warmer because of friction interaction between them. Estimate the change of the box-string system’s thermal energy. (String mass is negligible)

12

PROBLEM : BOX & STRING A box (in outer space) and its contents have a total mass . A string passes through a hole in the box, and you pull on the string with a constant force .

(a) Initially the speed of the box was . When the box has moved a long distance , your hand has moved an additional distance (total distance ), because additional string of length came out of the box. What is the box’s speed, , now? 13

PROBLEM : BOX & STRING (Cont’d)

(b) If we looked inside the box, we would see that the string was wound around a hub that turns on an axle with negligible friction. Three identical balls, each of mass , are attached to the hub at a distance from the axle. If the initial angular speed relative to the axle was , what is the final angular speed in terms of given quantities? 14

PROBLEM : Falling Yo-Yo You are playing with a Yo-Yo, mass 0.1 kg attached to a massless string. You pull the string with a constant force of 0.2 N through a distance of 0.3 m. In this time the Yo-Yo falls a distance of 0.4 m. (a) What is the approximate change in translational KE of the Yo-Yo? (b) What is the approximate change in rotational KE of the Yo-Yo? 15

Rolling vs. Sliding You pull the axle of a pair rolling wheels, mass through distance with constant force . Next, you pull a block of ice of the same mass on a frictionless surface through the same distance, applying the same force. If both objects started from rest, how will their final translational speeds compare?

A. B. C. D. Not enough Info. 16

Rolling vs. Sliding System 1:

System 2:

• 2 wheels + axle (total mass ) • Pulled by a thread looped around axle. • Rolls without Slipping.

• Ice cube (mass ) • Pulled by a thread. • Negligible Friction

Initially at Rest

Initially at Rest



Which moves faster after being pulled the same distance

?

Rolling vs. Sliding : Explanation #1 

Energy Principle to Point Particle System 1:

Friction. So:

  

Energy Principle to Point Particle System 2:

No Friction. So:

  

18

Rolling vs. Sliding : Explanation #2



Energy Principle to Extended (Real) System 1:

Δr for frictional force is 0 because the point of contact between wheel and ground is momentarily at rest where frictional force is acting.

Energy Principle to Extended (Real) System 2:

Only left with Ktran 19 in which we can apply the result from before.

ROLLING WITHOUT SLIPPING: Bicycle Wheel • Speed of CM (i.e. axle): • Speed of Rim Relative to Axle: • Speed of Rim at TOP: • Speed of Rim at BOTTOM:

ROLLING WITHOUT SLIPPING : Point of contact with ground 0 does not move 

20

ROLLING WITHOUT SLIPPING: Bicycle Wheel

• Total Kinetic Energy:

21

Two Wheels Wheel 1 of mass rolls down a slope. Wheel 2 of the same mass slides down the same slope with negligible friction. Which wheel will have the larger total kinetic energy when it reaches the bottom? Energy Principle to A. Wheel 1 wheel-Earth system: B. Wheel 2 C. Both the same (Both wheels)

1

2

22

Two Wheels

1 Wheel 1 of mass rolls down a slope. Wheel 2 of the same mass slides down the same slope with negligible friction. Which wheel will reach the bottom first? 2 • Same total energy, A. Wheel 1 but energy of rolling B. Wheel 2 wheel is divided into C. Both the same translational and rotational energy. • The sliding wheel’s larger translational kinetic energy means that it is moving faster during each step and will reach the bottom first.

Two Wheels : Detailed Analysis Wheel 1: Roll without slipping 1 1 1 1 2 2 2 2     Mv cm Mv Mgy Mv Mvrel ,i rel , i cm, i cm, f , f  Mgycm , f 2 2 2 2 1 2 1 2  Mgycm,i  Mvcm  Mvcm ,f , f  Mgycm, f 2 2 since vcm  vrel , the rolling constraint

 v2cm, f  g y cm

SMALLER

Wheel 2: Slide without friction, so no rotation

1 1 2 2 Mvcm Mgy Mvcm   ,i cm, i , f  Mgy cm, f 2 2 2  vcm , f  2 g y cm

LARGER 24

Wheel & Hoop A hoop and a disk with the same radius R and mass M are released at the same time to roll without slipping down the same inclined plane. Which one will reach the bottom first? • Same total energy, but it is divided into A. Hoop translational and rotational energy. • Object with greater moment of inertia will B. Disk have greater rotational energy, with less C. Both the same translational energy. • Hoop has greater rotational inertia (all mass concentrated at circumference)  smaller translational E  Slower

RECALL: PARALLEL AXIS THEOR • Previous examples: – Axis passing through CM • When Axis Does NOT Pass Through CM X

Moment of Inertia through Axis of Rotation Moment of Inertia of object about axis thru CM

Axis of Rotation

CM

Shortest distance of CM from Axis of Rotation 26

Problem: Rod Rotating on Axle A uniform rod of mass m rotates at constant angular speed ω on an axle through one end. What is the total kinetic energy of the rod? ω

For a rod: 

L



Center of mass

27

Be Sure To… • Complete Lecture Quiz 17 by 11:59 PM Tuesday, March 23rd. HW 17 due 11:59 PM Wednesday, March 24th. Lecture Quiz 18 by 11:59 PM Thursday, March 25th. HW 18 due 11:59 PM Friday, March 26th. REVIEW QUIZ 4 in 24-hour window 9:00AM, Friday, March 26th – 8:59 AM, Saturday, March 27th – Recitation 10 by 12:00 Noon Sat, March 27th. – Lab 09 + Lab 06-Part 4 by 11:59 PM Sat, March 27th. – – – – –

• Read relevant sections in text as per syllabus, before next lecture SEE YOU NEXT LECTURE! 28...


Similar Free PDFs