Conditions for equilibrium (both translational and rotational): 0 and 0

Similar documents
15 N 5 N. Chapter 4 Forces and Newton s Laws of Motion. The net force on an object is the vector sum of all forces acting on that object.

Physics Sp Exam #4 Name:

Physics 20 Lesson 28 Simple Harmonic Motion Dynamics & Energy

Work and Energy Problems

Physics 6A. Practice Midterm #2 solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

PHYSICS 151 Notes for Online Lecture 2.3

Physics 6A. Practice Midterm #2 solutions

DYNAMICS OF ROTATIONAL MOTION

AP Physics Momentum AP Wrapup

Application of Newton s Laws. F fr

PHYSICS 211 MIDTERM II 12 May 2004

Second Law of Motion. Force mass. Increasing mass. (Neglect air resistance in this example)

Practice Problem Solutions. Identify the Goal The acceleration of the object Variables and Constants Known Implied Unknown m = 4.

Physics 6A. Practice Final (Fall 2009) solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 11 HW #6 Solutions

SPH4U/SPH3UW Unit 2.3 Applying Newton s Law of Motion Page 1 of 7. Notes

Physics 20 Lesson 16 Friction

Seat: PHYS 1500 (Fall 2006) Exam #2, V1. After : p y = m 1 v 1y + m 2 v 2y = 20 kg m/s + 2 kg v 2y. v 2x = 1 m/s v 2y = 9 m/s (V 1)

Example 1: Example 1: Example 2: a.) the elevator is at rest. Example 2: Example 2: c.) the elevator accelerates downward at 1.

3. In an interaction between two objects, each object exerts a force on the other. These forces are equal in magnitude and opposite in direction.

Name: Answer Key Date: Regents Physics. Energy

Newton s Laws & Inclined Planes

Chapter 4 FORCES AND NEWTON S LAWS OF MOTION PREVIEW QUICK REFERENCE. Important Terms

Chapter 8 Torque and Angular Momentum

5.5. Collisions in Two Dimensions: Glancing Collisions. Components of momentum. Mini Investigation

t α z t sin60 0, where you should be able to deduce that the angle between! r and! F 1

= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5.

PHY 211: General Physics I 1 CH 10 Worksheet: Rotation

s s 1 s = m s 2 = 0; Δt = 1.75s; a =? mi hr

Solution to Theoretical Question 1. A Swing with a Falling Weight. (A1) (b) Relative to O, Q moves on a circle of radius R with angular velocity θ, so

PHY 171 Practice Test 3 Solutions Fall 2013

Physics 20 Lesson 18 Pulleys and Systems

( kg) (410 m/s) 0 m/s J. W mv mv m v v. 4 mv

Physics 30 Lesson 3 Impulse and Change in Momentum

1.1 Speed and Velocity in One and Two Dimensions

SPH3UW/SPH4UI Unit 2.4 Friction Force Page 1 of 8. Notes. : The kind of friction that acts when a body slides over a surface. Static Friction Force, f

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003

Conservation of Energy

2. REASONING According to the impulse-momentum theorem, the rocket s final momentum mv f

A 30 o 30 o M. Homework #4. Ph 231 Introductory Physics, Sp-03 Page 1 of 4

XI PHYSICS M. AFFAN KHAN LECTURER PHYSICS, AKHSS, K.

( ) rad ( 2.0 s) = 168 rad

All Division 01 students, START HERE. All Division 02 students skip the first 10 questions, begin on # (D)

Prof. Dr. Ibraheem Nasser Examples_6 October 13, Review (Chapter 6)

Work and Energy Problems

Name Period. What force did your partner s exert on yours? Write your answer in the blank below:

Pearson Physics Level 20 Unit III Circular Motion, Work, and Energy: Unit III Review Solutions

Practice Midterm #1 Solutions. Physics 6A

Physics Sp Exam #3 Name:

15 Newton s Laws #2: Kinds of Forces, Creating Free Body Diagrams

Elastic Collisions Definition Examples Work and Energy Definition of work Examples. Physics 201: Lecture 10, Pg 1

Physics 207: Lecture 26. Announcements. Make-up labs are this week Final hwk assigned this week, final quiz next week.

Physics 140 D100 Midterm Exam 2 Solutions 2017 Nov 10

Rolling without slipping Angular Momentum Conservation of Angular Momentum. Physics 201: Lecture 19, Pg 1

ROTATIONAL MOTION FROM TRANSLATIONAL MOTION

Uniform Acceleration Problems Chapter 2: Linear Motion

What Are Newton's Laws of Motion?

Linear Momentum. calculate the momentum of an object solve problems involving the conservation of momentum. Labs, Activities & Demonstrations:

AP Physics Charge Wrap up

Physics 4A Solutions to Chapter 15 Homework

KEY. D. 1.3 kg m. Solution: Using conservation of energy on the swing, mg( h) = 1 2 mv2 v = 2mg( h)

1.3.3 Statistical (or precision) uncertainty Due to transient variations, spatial variations 100%

ME 375 EXAM #1 Tuesday February 21, 2006

Tarzan s Dilemma for Elliptic and Cycloidal Motion

PHYSICS - CLUTCH CH 05: FRICTION, INCLINES, SYSTEMS.

NEWTONS LAWS OF MOTION AND FRICTIONS STRAIGHT LINES

For a situation involving gravity near earth s surface, a = g = jg. Show. that for that case v 2 = v 0 2 g(y y 0 ).

Assessment Schedule 2017 Scholarship Physics (93103)

Definition of Work, The basics

Discover the answer to this question in this chapter.

ME 141. Engineering Mechanics

Motion Along a Line. Readings: Chapter 6 (2. nd edition)

The Lagrangian Method vs. other methods (COMPARATIVE EXAMPLE)

Simple Harmonic Motion

PHYSICS 2210 Fall Exam 4 Review 12/02/2015

Kinetics of Rigid (Planar) Bodies

Physics 101 Lecture 12 Equilibrium and Angular Momentum

Part A Here, the velocity is at an angle of 45 degrees to the x-axis toward the z-axis. The velocity is then given in component form as.

Page 1. t F t m v. N s kg s. J F t SPH4U. From Newton Two New Concepts Impulse & Momentum. Agenda

v 24 m a = 5.33 Δd = 100 m[e] m[e] m[e] Δd = 550 m[e] BLM 2-6: Chapter 2 Test/Assessment Δd = + 10 s [E] uuv a = (10 0) s uuv a = (20 0)s

Physics 2210 Fall smartphysics 20 Conservation of Angular Momentum 21 Simple Harmonic Motion 11/23/2015

Chemistry I Unit 3 Review Guide: Energy and Electrons

Page 1. Physics 131: Lecture 22. Today s Agenda. SHM and Circles. Position

3pt3pt 3pt3pt0pt 1.5pt3pt3pt Honors Physics Impulse-Momentum Theorem. Name: Answer Key Mr. Leonard

Momentum. Momentum. Impulse. Impulse Momentum Theorem. Deriving Impulse. v a t. Momentum and Impulse. Impulse. v t

Today s s topics are: Collisions and Momentum Conservation. Momentum Conservation

Practice Problems Solutions. 1. Frame the Problem - Sketch and label a diagram of the motion. Use the equation for acceleration.

Phys101 Lectures 13, 14 Momentum and Collisions

Sample Problems. Lecture Notes Related Rates page 1

Lesson 27 Conservation of Energy

Chapter 5, Conceptual Questions

Test 2 phy a) How is the velocity of a particle defined? b) What is an inertial reference frame? c) Describe friction.

5.4 Conservation of Momentum in Two Dimensions

Chapter 8 continued. Rotational Dynamics

Momentum. Momentum and Energy. Momentum and Impulse. Momentum. Impulse. Impulse Increasing Momentum

Physics 218 Exam 3 Fall 2010, Sections

Physics 100A, Homework 12-Chapter 11 (part 2)

26 Impulse and Momentum

For more Study Material and Latest Questions related to IIT-JEE visit

Physics 20 Lesson 17 Elevators and Inclines

Transcription:

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Lat tie we began dicuing rotational dynaic. We howed that the rotational inertia depend on the hape o the object and the location o the rotational ai. We alo learned that the torque change the tate o rotation. Rotational Equilibriu We reeber that or an object to reain at ret, the net orce acting on it ut be equal to zero. (Newton irt law.) However, that condition i not uicient or rotational equilibriu. What happen to the object to the right? The orce have the ae agnitude. Condition or equilibriu (both tranlational and rotational): 0 and 0 The obedient pool. and ake the pool roll to the let, 4 to the right, and 3 ake it lide. Proble-Solving Step in Equilibriu Proble (page 74). Identiy an object or yte in equilibriu. Draw a diagra howing all the orce acting on that object, each drawn at it point o application. Ue the center o gravity (CM) a the point o application o any gravitational orce.. To apply the orce condition, chooe a convenient coordinate yte and reolve each orce into it - and y-coponent. 3. To apply the torque condition, chooe a convenient rotation ai generally one that pae through the point o application o an unknown orce. Then ind the torque due to each orce. Ue whichever ethod i eaier: either the lever ar tie the agnitude o the orce or the ditance tie the perpendicular coponent o the orce. Deterine the direction o each torque; then either et the u o all torque (with their algebraic ign) Leon, page

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- equal to zero or et the agnitude o the CW torque equal to the agnitude o the CCW torque. 4. Not all proble require all three equilibriu equation (two orce coponent equation and one torque equation). Soetie it i eaier to ue ore than one torque equation, with a dierent ai. Beore diving in and writing down all the equation, think about which approach i the eaiet and ot direct. There are any good eaple worked out or you in the tet. See page 8-89. Eaple: What i the allet angle a ladder can ake o that it doe not lide? W g N Solution: We will ue the condition or rotational equilibriu 0 We can chooe any ai about which to take torque. The ai I chooe i where the ladder touche the loor. The lever ar or the noral orce and the rictional orce will be zero and their torque will alo be zero. Recall that the torque i r I the ladder ha length L, the lever ar or the weight i the hort horizontal line below the loor in the diagra. The lever ar i the perpendicular ditance ro the line o the orce to the point o rotation. Here it i L r co The lever ar or the orce o the wall puhing againt the ladder i r Lin Leon, page

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Uing the condition or rotational equilibriu The condition or tranlational equilibriu are 0 g 0 L W Lin g co 0 0 and 0 y Reerring to the BD, the -coponent give W W 0 0 Again looking at the BD, the y-coponent N g 0 N g Oh, no! our equation: Ue the torque relation y 0 L W Lin g co 0 W W N g N L Lin g co 0 L W Lin g co g W in co Leon, page 3

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- tan g W Subtitute the two orce equation ro Newton econd law Since thi i a tatic riction orce, tan g N W N tan N N tan tan tan N I the coeicient o tatic riction i 0.4, the allet angle i 5 o. Equilibriu in the Huan Body orce act on the tructure in the body. Eaple 8.0: The deltoid ucle eert on the hueru a hown. The orce doe two thing. The vertical coponent upport the weight o the ar and the horizontal coponent tabilize the joint by pulling the hueru in againt the houlder. There are three orce acting on the ar: it weight (g), the orce due to the deltoid ucle () and the orce o the houlder joint () contraining the otion o the ar. Leon, page 4

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Since the ar i in equilibriu, we ue the equilibriu condition. To ue the torque equation we ue a convenient rotation ai. We chooe the houlder joint a the rotation ai a that will eliinate ro conideration. (Why?) r g g in5r r g 0 g 0 g 0 r 0 g rg (30 N)(0.75) r in5 (0. )in5 66 N To upport the 30 N ar a 70 N orce i required. Highly ineicient!! The Iron Cro. Here i an intereting video: http://www.youtube.co/watch?v=sdljygi_4 Becaue o the yetry, hal o the gynat weight i upported by each ring. Conider the BD above. 0 0 0 w Wrw r Leon, page 5

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Wr r w W (0.60 ) (0.045)in 45 9.4W The orce eerted by the latiiu dori and the pectorali ajor on one ide o the gynat body i ore than nine tie hi weight! The tructure o the huan body ake large ucular orce neceary. Are there any advantage to the tructure? Due to the all lever ar, the ucle orce are uch larger than they would otherwie be, but the huan body ha traded thi or a wide range o oveent o the bone. The bicep and tricep ucle can ove the lower ar through alot 80º while they change their length by only a ew centieter. (p. 9) A video o equilibriu and how eaily it can be dirupted: http://www.youtube.co/watch?v=k6rxaei57c Rotational or o Newton econd law Very iilar to the other econd law. I Motion o Rolling Object A rolling object ha rotational kinetic energy and tranlational kinetic energy. K K tran K rot v CM I CM Why doe the object roll (and not lide)? rictional orce eert a torque on the object. Leon, page 6

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Eaple 8.3 The acceleration o a rolling ball. The rotational or o Newton econd law i The torque on the ball i due to riction So I r I r I I r We can ue Newton econd law to ind the linear acceleration o the ball. A we uually do, take the +-ai along the incline. gin a a Ue the epreion or the rictional orce to ind, gin a I gin a r But the acceleration o the ball i related to it angular acceleration, a = r. Leon, page 7

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- I gin a r Ia gin a r Ia gin a r gin a I r or a unior, olid phere, I = (/5)MR and or a thin ring, I = MR. Which ha the larger acceleration? Conider the eect o the rotational inertia on the acceleration. Eaple: A olid phere roll down a hill that ha a height h. What i it peed at the botto? Solution: Ue conervation o energy. Since the ball roll without lipping, the rictional orce doen t do any work. The diplaceent i zero in the deinition W = r co. U gy K U K 0 0 v I The tranlational peed o the ball i related to it rotational peed, v = r. gy gh v v v ( I gh I r I I( v / r) r ) v or the olid phere, I = (/5)MR gh I r gh ( 5) gh (7 5) 0 v 7 gh Thi i le than the anwer we ound when we ignored rolling, v = gh. Angular oentu We introduced the idea o linear oentu in chapter 7. We had dp dt Leon, page 8

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- A iilar epreion eit or rotational otion dl dt The net eternal torque acting on a yte i equal to the rate o change o the angular oentu o the yte. The angular oentu L I i the tendency o a rotating object to continue rotating with the ae angular peed about the ae ai. Angular oentu i eaured in kg /. I the net torque i zero, we have the conervation o angular oentu L 0 L i L I the rotational inertia o the yte change, it angular peed will change to copenate. Angular oentu i a vector. (So i the torque!) The direction i given by the righthand rule. Our eaon are a conequence o the conervation o angular oentu. Leon, page 9

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- We have copleted our tudy o rotational otion. Try to ee how it i analogou to (the ore ailiar) linear otion. Here i a uary: Decription Linear Rotational poition diplaceent Rate o change o poition v Rate o change o velocity a Average rate o change o poition v, av av t t Intantaneou rate o change o poition v li li t 0 t t 0 t Average rate o change o velocity v a, av av t t Intantaneou rate o change o velocity v a li li t 0 t t 0 t v v a t t Equation o unior acceleration v t i ( v i Leon, page 0 a ( t) v ) t v v a i Inertia I r i i i t i ( t) ( ) t Inluence that caue acceleration r r Newton econd law Work Kinetic energy y a a y i I W r co W v I i

Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Moentu p v L Iω dp dl Newton econd law dt dt Condition or conervation o oentu 0 0 Conervation o oentu p i p Li L Leon, page