Forces of Rolling. 1) Ifobjectisrollingwith a com =0 (i.e.no netforces), then v com =ωr = constant (smooth roll)
|
|
- Britton Grant
- 5 years ago
- Views:
Transcription
1 Physics 2101 Section 3 March 12 rd : Ch. 10 Announcements: Mid-grades posted in PAW Quiz today I will be at the March APS meeting the week of th. Prof. Rich Kurtz will help me. Class Website:
2 Forces of Rolling 1) Ifobjectisrollingwith a =0 (i.e.no netforces), then v =ωr = constant (smooth roll) if constant speed, it has no tendency to slide at point of contact no frictional forces a = 0 α = 0 τ = 0 f = 0 a = αr I α = τ net τ net = R f sinθ 2) If object is rolling with a 0 (i.e. there are net forces) and no slipping occurs, then α 0 τ 0 static friction needed ddto supply torque! kinetic friction - during sliding (a αr) static ti friction - smooth rolling (a = αr)
3 Rolling down a ramp : Energy considerations An object at rest, with mass m and radius r, roles from top of incline plane to bottom. What is v at bottom and a throughout. ( KE ) final 0 KE rot +trans,com 1 mv 2 2 COM ΔE mech = 0 ΔKE tot = ΔU [ ] ( ) init = ( 0 ) final ( mgh ) init I COMω 2 = mgl sinθ Using v = ωr h L sinθ = h θ Using v 2 = v aL v COM = 2gL sinθ 1+ I mr 2 Using 1 D kinematics (const accel) a v2 gsinθ = = 2L 1+ I mr 2 Speed at Bottom Constant Acceleration
4 Rolling down a ramp: acceleration Acceleration downwards Friction provides torque Static friction points upwards Yo Yo Newton s 2 nd Law Linear version x ˆ : f s F g sinθ = ma y ˆ : N F g cosθ = 0 Tension provides torque Here θ = 90 Angular version z ˆ : I α = τ = Rf s a = R 2 ( mgsinθ ma ) I Al Axle: R R 0 a = αr α = a R = Rf s I gsinθ a = 1+ I mr 2 g a = 1+ I 2 mr 0
5 a Rolling Down a Ramp Consider a round uniform body of mass M and radius rolling down an inclined plane of angle θ. We will calculate the acceleration a of the center of mass along the x-axis using Newton's second law for the translational and rotational motion. R Newton's second law for motion along the x-axis: f Mgsin θ = Ma (eq. 1) Newton's second law for rotation about the center of mass: s τ = Rf s = I a a α = We substitute α in the second equation and get Rfs = I R R a fs = I (eq. 2). We substitute f from equation 2 into equation 1 2 s R a g sinθ I = a = Mgsinθ Ma 2 R I 1+ 2 MR α
6 a a = g sinθ I 1+ 2 MR Cylinder Hoop 2 MR 2 I1 = I2 = MR 2 g sin θ g sin θ a1 = a2 = 1 + I / MR 1 + I / MR g sinθ g sinθ a 1 = a = MR /2MR 1 + MR / MR gsinθ gsinθ a1 = a2 = 1+ 1/ gsinθ gsinθ a1 = = (0.67) gsin θ a2 = = (0.5) gsinθ 3 2
7 Prob NON smooth rolling motion A bowler throws a bowling ball of radius R along a lane. The ball slides on the lane, with initial speed v,0 and initial angular speed ω 0 = 0. The coefficient of kinetic friction between the ball and the lane is μ k. The kinetic frictional force f k acting on the ball while producing a torque that causes an angular acceleration of the ball. When the speed v has decreased enough and the angular speed ω has increased enough, the ball stops sliding and then rolls smoothly. a) [After it stops sliding] What is the v in terms of ω? Smooth rolling means v =Rω b) During the sliding, what is the ball s linear acceleration? From 2 nd law: (linear) ˆ x : f k = ma ˆ y : N mg = 0 But f k = μ k N = μ k mg So a = f k m = μ k g c) During the sliding, what is the ball s angular acceleration? From 2 nd law: τ = Rf k ( ˆ z ) But f k = μ k N (angular) Iα( ˆ z ) = τ = Rf k ( ˆ z ) = μ k mg So Iα = Rf k = R( μ k mg) α = Rμ kmg I d) What is the speed of the ball when smooth rolling begins? When does v =Rω? v = v 0 + a t v = v 0 μ k gt From kinematics: ω = ω 0 + αt t = ω α = Iω Rμ k mg I v R v = v 0 μ k g Rμ k mg e) How long does the ball slide? v = t = v 0 v μ k g v 0 1+ ImR 2 ( )
8 Torque & Angular Momentum ω Conservation of: Linear momentum (closed, isolated system) KE (elastic collisions) Mechanical energy (only conservative forces) Total energy Angular momentum (closed, isolated system)
9 Angular Momentum (single particle) A Linear momentum when particle passes through point it has linear momentum: p = mv Angular momentum with respect to point O the angular momentum is defined as: = r p = m r v ( ) where r is the position vector of the particle with respect to O. with ponents in x y plane Units kg m 2 /s ponents in z direction only! ( to x y plane) just as in τ, has meaning only with respect to a specified point (axis)
10 Newton s 2 nd Law in Angular Form (single particle) Relationship between force and linear momentum Newton s 2 nd Law in linear form F net = F i = ma = d p dt Relationship between torque and angular momentum τ net = τ i = d dt The vector sum of all the torques acting on a particle is equal to the time rate change of the angular momentum of that particle. τ net = d dt has no meaning unless the net torque τ net, and the rotational momentum, are defined with respect to the same origin
11 Example #1 A cannon ball, with mass m, is shot directly upward with velocity v 0. What is the angular momentum and torque of the ball about a point a distance x 0 from the cannon? The only force on the in flight cannon ball is gravity: The vector position and velocity of the in flight ball is: x ˆ : x(t) = x 0 + v 0,x t = x 0 y ˆ : y(t) = y 0 + v 0,y t + 1 at 2 = v 2 0 t 1 gt 2 2 F g = mgˆ j r = ( x 0 )ˆ i + ( v 0 t 1 gt 2 2 )ˆ j v = ( v 0 gt)ˆ j The vector angular momentum about the man of the in flight ball is: i ˆ ˆ j k ˆ l = m r v = m x 0 v 0 t 1 gt = mx 0 ( v 0 gt)ˆ k 0 v 0 gt 0 The vector torque about the man of the in flight ball is: ˆ ˆ ˆ τ = r F i j k = x 0 v 0 t 1 gt 2 0 = x 2 0 mgk ˆ 0 mg 0 Check τ = d l dt = d [ mx 0 ( v 0 gt)ˆ k ] dt = mx 0 g k ˆk
12
13
14 Total angular momentum Newton s 2 nd Law in Angular Form System of particles L = l 1 + l 2 + l l n = Relationship between torque and angular momentum for system of particles n l i i=1 τ net = d L dt τ net represents the net external torque.
15 Example #2 (Problem 11 27) #1 Two particles are moving as shown. a) What is their total moment of inertia about point O? b) What is their total angular momentum about point O? c) What is their net torque about point O? #2 a) The total moment of inertia about point O is found from: I tot = I i = m i r i = (6.5 kg)(1.5 m) + (3.1 kg)(2.8 m) = 38.9 kg m b) The total angular momentum about point 0 is: r L = = ri r tot l i m i ( r i v i ) = m 1 ( r 1 v 1 sinθ 1 )( ẑ) z) + m 2 ( r 2 v 2 sinθ 2 )(+ẑ) )(+z) L tot = [ (6.5 kg)(1.5 m)(2.2 m /s) + (3.1 kg)(2.8 m)(3.6 m /s)]ˆ z = 9.8 kg m 2 /s z ˆ c) The net torque about point 0 is: dl τ tot net = = 0 closed, isolated system tot dt
16 Angular Momentum of a Rigid Body Rotating about a Fixed axis dl I α d ω τ net = Iα dt = = I dt L = Iω Example #3 (Problem 11 39) Three particles are fastened to each other with massless string and rotate around point O. What is the total angular momentum about point O? I tot = md ( ) 2 + m( 2d) 2 + m( 3d) 2 = 14mdd 2 L 2 tot = Iω = 14mωd ẑz [kg m 2 /s]
17
I 2 comω 2 + Rolling translational+rotational. a com. L sinθ = h. 1 tot. smooth rolling a com =αr & v com =ωr
Rolling translational+rotational smooth rolling a com =αr & v com =ωr Equations of motion from: - Force/torque -> a and α - Energy -> v and ω 1 I 2 comω 2 + 1 Mv 2 = KE 2 com tot a com KE tot = KE trans
More informationKinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)
Kinematics (special case) a = constant 1D motion 2D projectile Uniform circular Dynamics gravity, tension, elastic, normal, friction Motion with a = constant Newton s Laws F = m a F 12 = F 21 Time & Position
More informationRotational Dynamics continued
Chapter 9 Rotational Dynamics continued 9.4 Newton s Second Law for Rotational Motion About a Fixed Axis ROTATIONAL ANALOG OF NEWTON S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS I = ( mr 2
More informationPhysics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1
Physics 201 p. 1/1 Physics 201 Professor P. Q. Hung 311B, Physics Building Physics 201 p. 2/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia All elements inside the rigid
More information= 2 5 MR2. I sphere = MR 2. I hoop = 1 2 MR2. I disk
A sphere (green), a disk (blue), and a hoop (red0, each with mass M and radius R, all start from rest at the top of an inclined plane and roll to the bottom. Which object reaches the bottom first? (Use
More informationIf rigid body = few particles I = m i. If rigid body = too-many-to-count particles I = I COM. KE rot. = 1 2 Iω 2
2 If rigid body = few particles I = m i r i If rigid body = too-many-to-count particles Sum Integral Parallel Axis Theorem I = I COM + Mh 2 Energy of rota,onal mo,on KE rot = 1 2 Iω 2 [ KE trans = 1 2
More informationChapters 10 & 11: Rotational Dynamics Thursday March 8 th
Chapters 10 & 11: Rotational Dynamics Thursday March 8 th Review of rotational kinematics equations Review and more on rotational inertia Rolling motion as rotation and translation Rotational kinetic energy
More informationω = ω 0 θ = θ + ω 0 t αt ( ) Rota%onal Kinema%cs: ( ONLY IF α = constant) v = ω r ω ω r s = θ r v = d θ dt r = ω r + a r = a a tot + a t = a r
θ (t) ( θ 1 ) Δ θ = θ 2 s = θ r ω (t) = d θ (t) dt v = d θ dt r = ω r v = ω r α (t) = d ω (t) dt = d 2 θ (t) dt 2 a tot 2 = a r 2 + a t 2 = ω 2 r 2 + αr 2 a tot = a t + a r = a r ω ω r a t = α r ( ) Rota%onal
More informationChapter 8- Rotational Kinematics Angular Variables Kinematic Equations
Chapter 8- Rotational Kinematics Angular Variables Kinematic Equations Chapter 9- Rotational Dynamics Torque Center of Gravity Newton s 2 nd Law- Angular Rotational Work & Energy Angular Momentum Angular
More informationHandout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum
Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion Torque and angular momentum In Figure, in order to turn a rod about a fixed hinge at one end, a force F is applied at a
More informationTranslational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work
Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational
More informationRolling, Torque, Angular Momentum
Chapter 11 Rolling, Torque, Angular Momentum Copyright 11.2 Rolling as Translational and Rotation Combined Motion of Translation : i.e.motion along a straight line Motion of Rotation : rotation about a
More informationCircular Motion, Pt 2: Angular Dynamics. Mr. Velazquez AP/Honors Physics
Circular Motion, Pt 2: Angular Dynamics Mr. Velazquez AP/Honors Physics Formulas: Angular Kinematics (θ must be in radians): s = rθ Arc Length 360 = 2π rads = 1 rev ω = θ t = v t r Angular Velocity α av
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Monday, 14 December 2015, 6 PM to 9 PM, Field House Gym
FALL TERM EXAM, PHYS 111, INTRODUCTORY PHYSICS I Monday, 14 December 015, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing. There
More informationLectures. Today: Rolling and Angular Momentum in ch 12. Complete angular momentum (chapter 12) and begin equilibrium (chapter 13)
Lectures Today: Rolling and Angular Momentum in ch 1 Homework 6 due Next time: Complete angular momentum (chapter 1) and begin equilibrium (chapter 13) By Monday, will post at website Sample midterm II
More informationReview questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right.
Review questions Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. 30 kg 70 kg v (a) Is this collision elastic? (b) Find the
More informationz F 3 = = = m 1 F 1 m 2 F 2 m 3 - Linear Momentum dp dt F net = d P net = d p 1 dt d p n dt - Conservation of Linear Momentum Δ P = 0
F 1 m 2 F 2 x m 1 O z F 3 m 3 y Ma com = F net F F F net, x net, y net, z = = = Ma Ma Ma com, x com, y com, z p = mv - Linear Momentum F net = dp dt F net = d P dt = d p 1 dt +...+ d p n dt Δ P = 0 - Conservation
More informationChapter 10: Dynamics of Rotational Motion
Chapter 10: Dynamics of Rotational Motion What causes an angular acceleration? The effectiveness of a force at causing a rotation is called torque. QuickCheck 12.5 The four forces shown have the same strength.
More informationWork and kinetic Energy
Work and kinetic Energy Problem 66. M=4.5kg r = 0.05m I = 0.003kgm 2 Q: What is the velocity of mass m after it dropped a distance h? (No friction) h m=0.6kg mg Work and kinetic Energy Problem 66. M=4.5kg
More informationPhysics 207: Lecture 24. Announcements. No labs next week, May 2 5 Exam 3 review session: Wed, May 4 from 8:00 9:30 pm; here.
Physics 07: Lecture 4 Announcements No labs next week, May 5 Exam 3 review session: Wed, May 4 from 8:00 9:30 pm; here Today s Agenda ecap: otational dynamics and torque Work and energy with example Many
More informationTorque. Introduction. Torque. PHY torque - J. Hedberg
Torque PHY 207 - torque - J. Hedberg - 2017 1. Introduction 2. Torque 1. Lever arm changes 3. Net Torques 4. Moment of Rotational Inertia 1. Moment of Inertia for Arbitrary Shapes 2. Parallel Axis Theorem
More information31 ROTATIONAL KINEMATICS
31 ROTATIONAL KINEMATICS 1. Compare and contrast circular motion and rotation? Address the following Which involves an object and which involves a system? Does an object/system in circular motion have
More informationPhysics 4A Solutions to Chapter 10 Homework
Physics 4A Solutions to Chapter 0 Homework Chapter 0 Questions: 4, 6, 8 Exercises & Problems 6, 3, 6, 4, 45, 5, 5, 7, 8 Answers to Questions: Q 0-4 (a) positive (b) zero (c) negative (d) negative Q 0-6
More informationDo not fill out the information below until instructed to do so! Name: Signature: Student ID: Section Number:
Do not fill out the information below until instructed to do so! Name: Signature: Student ID: E-mail: Section Number: Formulae are provided on the last page. You may NOT use any other formula sheet. You
More informationQ1. For a completely inelastic two-body collision the kinetic energy of the objects after the collision is the same as:
Coordinator: Dr.. Naqvi Monday, January 05, 015 Page: 1 Q1. For a completely inelastic two-body collision the kinetic energy of the objects after the collision is the same as: ) (1/) MV, where M is the
More informationRotational Motion. Rotational Motion. Rotational Motion
I. Rotational Kinematics II. Rotational Dynamics (Netwton s Law for Rotation) III. Angular Momentum Conservation 1. Remember how Newton s Laws for translational motion were studied: 1. Kinematics (x =
More informationLecture 13 REVIEW. Physics 106 Spring What should we know? What should we know? Newton s Laws
Lecture 13 REVIEW Physics 106 Spring 2006 http://web.njit.edu/~sirenko/ What should we know? Vectors addition, subtraction, scalar and vector multiplication Trigonometric functions sinθ, cos θ, tan θ,
More informationRotational Dynamics. Slide 2 / 34. Slide 1 / 34. Slide 4 / 34. Slide 3 / 34. Slide 6 / 34. Slide 5 / 34. Moment of Inertia. Parallel Axis Theorem
Slide 1 / 34 Rotational ynamics l Slide 2 / 34 Moment of Inertia To determine the moment of inertia we divide the object into tiny masses of m i a distance r i from the center. is the sum of all the tiny
More informationAngular Momentum. Physics 1425 Lecture 21. Michael Fowler, UVa
Angular Momentum Physics 1425 Lecture 21 Michael Fowler, UVa A New Look for τ = Iα We ve seen how τ = Iα works for a body rotating about a fixed axis. τ = Iα is not true in general if the axis of rotation
More informationω = 0 a = 0 = α P = constant L = constant dt = 0 = d Equilibrium when: τ i = 0 τ net τ i Static Equilibrium when: F z = 0 F net = F i = ma = d P
Equilibrium when: F net = F i τ net = τ i a = 0 = α dp = 0 = d L = ma = d P = 0 = I α = d L = 0 P = constant L = constant F x = 0 τ i = 0 F y = 0 F z = 0 Static Equilibrium when: P = 0 L = 0 v com = 0
More informationExam 3 Practice Solutions
Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at
More informationChapter 8. Rotational Motion
Chapter 8 Rotational Motion Rotational Work and Energy W = Fs = s = rθ Frθ Consider the work done in rotating a wheel with a tangential force, F, by an angle θ. τ = Fr W =τθ Rotational Work and Energy
More information16. Rotational Dynamics
6. Rotational Dynamics A Overview In this unit we will address examples that combine both translational and rotational motion. We will find that we will need both Newton s second law and the rotational
More information. d. v A v B. e. none of these.
General Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibrium Oct. 28, 2009 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show the formulas you use, the essential
More informationAssignment 9. to roll without slipping, how large must F be? Ans: F = R d mgsinθ.
Assignment 9 1. A heavy cylindrical container is being rolled up an incline as shown, by applying a force parallel to the incline. The static friction coefficient is µ s. The cylinder has radius R, mass
More informationPhysics 201, Lecture 18
q q Physics 01, Lecture 18 Rotational Dynamics Torque Exercises and Applications Rolling Motion Today s Topics Review Angular Velocity And Angular Acceleration q Angular Velocity (ω) describes how fast
More informationAngular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion
Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for
More informationNotes on Torque. We ve seen that if we define torque as rfsinθ, and the N 2. i i
Notes on Torque We ve seen that if we define torque as rfsinθ, and the moment of inertia as N, we end up with an equation mr i= 1 that looks just like Newton s Second Law There is a crucial difference,
More informationMoment of Inertia & Newton s Laws for Translation & Rotation
Moment of Inertia & Newton s Laws for Translation & Rotation In this training set, you will apply Newton s 2 nd Law for rotational motion: Στ = Σr i F i = Iα I is the moment of inertia of an object: I
More informationSolution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved:
8) roller coaster starts with a speed of 8.0 m/s at a point 45 m above the bottom of a dip (see figure). Neglecting friction, what will be the speed of the roller coaster at the top of the next slope,
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.6 The Action of Forces and Torques on Rigid Objects Chapter 8 developed the concepts of angular motion. θ : angles and radian measure for angular variables ω :
More informationPleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 2
Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 1 3 problems from exam 2 6 problems 13.1 14.6 (including 14.5) 8 problems 1.1---9.6 Go through the
More informationPHY131H1S - Class 20. Pre-class reading quiz on Chapter 12
PHY131H1S - Class 20 Today: Gravitational Torque Rotational Kinetic Energy Rolling without Slipping Equilibrium with Rotation Rotation Vectors Angular Momentum Pre-class reading quiz on Chapter 12 1 Last
More informationPhysics 131: Lecture 21. Today s Agenda
Physics 131: Lecture 21 Today s Agenda Rotational dynamics Torque = I Angular Momentum Physics 201: Lecture 10, Pg 1 Newton s second law in rotation land Sum of the torques will equal the moment of inertia
More informationPhysics 131: Lecture 21. Today s Agenda
Physics 131: Lecture 1 Today s Agenda Rotational dynamics Torque = I Angular Momentum Physics 01: Lecture 10, Pg 1 Newton s second law in rotation land Sum of the torques will equal the moment of inertia
More informationRolling, Torque, and Angular Momentum
AP Physics C Rolling, Torque, and Angular Momentum Introduction: Rolling: In the last unit we studied the rotation of a rigid body about a fixed axis. We will now extend our study to include cases where
More informationRotation review packet. Name:
Rotation review packet. Name:. A pulley of mass m 1 =M and radius R is mounted on frictionless bearings about a fixed axis through O. A block of equal mass m =M, suspended by a cord wrapped around the
More informationPhysics 121, March 25, Rotational Motion and Angular Momentum. Department of Physics and Astronomy, University of Rochester
Physics 121, March 25, 2008. Rotational Motion and Angular Momentum. Physics 121. March 25, 2008. Course Information Topics to be discussed today: Review of Rotational Motion Rolling Motion Angular Momentum
More informationLecture 6 Physics 106 Spring 2006
Lecture 6 Physics 106 Spring 2006 Angular Momentum Rolling Angular Momentum: Definition: Angular Momentum for rotation System of particles: Torque: l = r m v sinφ l = I ω [kg m 2 /s] http://web.njit.edu/~sirenko/
More informationPHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011
PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this
More informationChapter 8. Rotational Equilibrium and Rotational Dynamics
Chapter 8 Rotational Equilibrium and Rotational Dynamics 1 Force vs. Torque Forces cause accelerations Torques cause angular accelerations Force and torque are related 2 Torque The door is free to rotate
More informationLecture 7 Chapter 10,11
Lecture 7 Chapter 10,11 Rotation, Inertia, Rolling, Torque, and Angular momentum Demo Demos Summary of Concepts to Cover from chapter 10 Rotation Rotating cylinder with string wrapped around it: example
More informationRotational motion problems
Rotational motion problems. (Massive pulley) Masses m and m 2 are connected by a string that runs over a pulley of radius R and moment of inertia I. Find the acceleration of the two masses, as well as
More informationChap10. Rotation of a Rigid Object about a Fixed Axis
Chap10. Rotation of a Rigid Object about a Fixed Axis Level : AP Physics Teacher : Kim 10.1 Angular Displacement, Velocity, and Acceleration - A rigid object rotating about a fixed axis through O perpendicular
More informationAngular Displacement. θ i. 1rev = 360 = 2π rads. = "angular displacement" Δθ = θ f. π = circumference. diameter
Rotational Motion Angular Displacement π = circumference diameter π = circumference 2 radius circumference = 2πr Arc length s = rθ, (where θ in radians) θ 1rev = 360 = 2π rads Δθ = θ f θ i = "angular displacement"
More informationPhysics 7A Lecture 2 Fall 2014 Midterm 2 Solutions. November 9, 2014
Physics 7A Lecture 2 Fall 204 Midterm 2 Solutions November 9, 204 Lecture 2 Midterm 2 Problem Solution Our general strategy is to use energy conservation, keeping in mind that the force of friction will
More informationChap. 10: Rotational Motion
Chap. 10: Rotational Motion I. Rotational Kinematics II. Rotational Dynamics - Newton s Law for Rotation III. Angular Momentum Conservation (Chap. 10) 1 Newton s Laws for Rotation n e t I 3 rd part [N
More informationConservation of Angular Momentum
Physics 101 Section 3 March 3 rd : Ch. 10 Announcements: Monday s Review Posted (in Plummer s section (4) Today start Ch. 10. Next Quiz will be next week Test# (Ch. 7-9) will be at 6 PM, March 3, Lockett-6
More informationPhysics 5A Final Review Solutions
Physics A Final Review Solutions Eric Reichwein Department of Physics University of California, Santa Cruz November 6, 0. A stone is dropped into the water from a tower 44.m above the ground. Another stone
More informationChapter 8. Rotational Equilibrium and Rotational Dynamics
Chapter 8 Rotational Equilibrium and Rotational Dynamics Wrench Demo Torque Torque, τ, is the tendency of a force to rotate an object about some axis τ = Fd F is the force d is the lever arm (or moment
More informationMomentum. The way to catch a knuckleball is to wait until it stops rolling and then pick it up. -Bob Uecker
Chapter 11 -, Chapter 11 -, Angular The way to catch a knuckleball is to wait until it stops rolling and then pick it up. -Bob Uecker David J. Starling Penn State Hazleton PHYS 211 Chapter 11 -, motion
More informationPhys101 Third Major-161 Zero Version Coordinator: Dr. Ayman S. El-Said Monday, December 19, 2016 Page: 1
Coordinator: Dr. Ayman S. El-Said Monday, December 19, 2016 Page: 1 Q1. A water molecule (H 2O) consists of an oxygen (O) atom of mass 16m and two hydrogen (H) atoms, each of mass m, bound to it (see Figure
More informationChapter 10. Rotation
Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised
More informationPhysics 101 Lecture 12 Equilibrium and Angular Momentum
Physics 101 Lecture 1 Equilibrium and Angular Momentum Ali ÖVGÜN EMU Physics Department www.aovgun.com Static Equilibrium q Equilibrium and static equilibrium q Static equilibrium conditions n Net external
More informationFinal Exam Spring 2014 May 05, 2014
95.141 Final Exam Spring 2014 May 05, 2014 Section number Section instructor Last/First name Last 3 Digits of Student ID Number: Answer all questions, beginning each new question in the space provided.
More informationWrite your name legibly on the top right hand corner of this paper
NAME Phys 631 Summer 2007 Quiz 2 Tuesday July 24, 2007 Instructor R. A. Lindgren 9:00 am 12:00 am Write your name legibly on the top right hand corner of this paper No Books or Notes allowed Calculator
More informationKing Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 4 Section 10 A 1.50-kg block slides down a frictionless 30.0 incline, starting from rest.
More informationPhysics 141. Lecture 18. Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 18, Page 1
Physics 141. Lecture 18. Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 18, Page 1 Physics 141. Lecture 18. Course Information. Topics to be discussed today: A
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.4 Rotational Work and Energy Work to accelerate a mass rotating it by angle φ F W = F(cosθ)x x = rφ = Frφ Fr = τ (torque) = τφ r φ s F to x θ = 0 DEFINITION OF
More informationLesson 8. Luis Anchordoqui. Physics 168. Thursday, October 11, 18
Lesson 8 Physics 168 1 Rolling 2 Intuitive Question Why is it that when a body is rolling on a plane without slipping the point of contact with the plane does not move? A simple answer to this question
More informationUNIVERSITY OF TORONTO Faculty of Arts and Science
UNIVERSITY OF TORONTO Faculty of Arts and Science DECEMBER 2013 EXAMINATIONS PHY 151H1F Duration - 3 hours Attempt all questions. Each question is worth 10 points. Points for each part-question are shown
More informationPHYSICS 220. Lecture 15. Textbook Sections Lecture 15 Purdue University, Physics 220 1
PHYSICS 220 Lecture 15 Angular Momentum Textbook Sections 9.3 9.6 Lecture 15 Purdue University, Physics 220 1 Last Lecture Overview Torque = Force that causes rotation τ = F r sin θ Work done by torque
More informationRotation and Translation Challenge Problems Problem 1:
Rotation and Translation Challenge Problems Problem 1: A drum A of mass m and radius R is suspended from a drum B also of mass m and radius R, which is free to rotate about its axis. The suspension is
More informationRolling without slipping Angular Momentum Conservation of Angular Momentum. Physics 201: Lecture 19, Pg 1
Physics 131: Lecture Today s Agenda Rolling without slipping Angular Momentum Conservation o Angular Momentum Physics 01: Lecture 19, Pg 1 Rolling Without Slipping Rolling is a combination o rotation and
More informationRolling, Torque & Angular Momentum
PHYS 101 Previous Exam Problems CHAPTER 11 Rolling, Torque & Angular Momentum Rolling motion Torque Angular momentum Conservation of angular momentum 1. A uniform hoop (ring) is rolling smoothly from the
More informationSolution to Problem. Part A. x m. x o = 0, y o = 0, t = 0. Part B m m. range
PRACTICE PROBLEMS: Final Exam, December 4 Monday, GYM, 6 to 9 PM Problem A Physics Professor did a daredevil stunt in his spare time. In the figure below he tries to cross a river from a 53 ramp at an
More informationPhysics 123 Quizes and Examinations Spring 2002 Porter Johnson
Physics 123 Quizes and Examinations Spring 2002 Porter Johnson Physics can only be learned by thinking, writing, and worrying. -David Atkinson and Porter Johnson (2002) There is no royal road to geometry.
More informationAP Physics. Harmonic Motion. Multiple Choice. Test E
AP Physics Harmonic Motion Multiple Choice Test E A 0.10-Kg block is attached to a spring, initially unstretched, of force constant k = 40 N m as shown below. The block is released from rest at t = 0 sec.
More informationPC 1141 : AY 2012 /13
NUS Physics Society Past Year Paper Solutions PC 1141 : AY 2012 /13 Compiled by: NUS Physics Society Past Year Solution Team Yeo Zhen Yuan Ryan Goh Published on: November 17, 2015 1. An egg of mass 0.050
More informationAngular Momentum Conservation of Angular Momentum
Lecture 22 Chapter 12 Physics I Angular Momentum Conservation of Angular Momentum Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi IN THIS CHAPTER, you will continue discussing rotational
More informationPH1104/PH114S MECHANICS
PH04/PH4S MECHANICS SEMESTER I EXAMINATION 06-07 SOLUTION MULTIPLE-CHOICE QUESTIONS. (B) For freely falling bodies, the equation v = gh holds. v is proportional to h, therefore v v = h h = h h =.. (B).5i
More informationPhysics of Rotation. Physics 109, Introduction To Physics Fall 2017
Physics of Rotation Physics 109, Introduction To Physics Fall 017 Outline Next two lab periods Rolling without slipping Angular Momentum Comparison with Translation New Rotational Terms Rotational and
More informationClassical Mechanics Lecture 15
Classical Mechanics Lecture 5 Today s Concepts: a) Parallel Axis Theorem b) Torque & Angular Acceleration Mechanics Lecture 5, Slide Unit 4 Main Points Mechanics Lecture 4, Slide Unit 4 Main Points Mechanics
More informationTwo-Dimensional Rotational Kinematics
Two-Dimensional Rotational Kinematics Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing 2.
More informationPhysics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 20: Rotational Motion. Slide 20-1
Physics 1501 Fall 2008 Mechanics, Thermodynamics, Waves, Fluids Lecture 20: Rotational Motion Slide 20-1 Recap: center of mass, linear momentum A composite system behaves as though its mass is concentrated
More informationReview for 3 rd Midterm
Review for 3 rd Midterm Midterm is on 4/19 at 7:30pm in the same rooms as before You are allowed one double sided sheet of paper with any handwritten notes you like. The moment-of-inertia about the center-of-mass
More informationa +3bt 2 4ct 3) =6bt 12ct 2. dt 2. (a) The second hand of the smoothly running watch turns through 2π radians during 60 s. Thus,
1. a) Eq. 11-6 leads to ω d at + bt 3 ct 4) a +3bt 4ct 3. dt b) And Eq. 11-8 gives α d a +3bt 4ct 3) 6bt 1ct. dt. a) The second hand of the smoothly running watch turns through π radians during 60 s. Thus,
More informationCP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford TT 2017
CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford TT 2017 1 OUTLINE : CP1 REVISION LECTURE 3 : INTRODUCTION TO CLASSICAL MECHANICS 1. Angular velocity and
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Two men, Joel and Jerry, push against a wall. Jerry stops after 10 min, while Joel is
More informationFriction is always opposite to the direction of motion.
6. Forces and Motion-II Friction: The resistance between two surfaces when attempting to slide one object across the other. Friction is due to interactions at molecular level where rough edges bond together:
More informationPHY2053 General Physics I
PHY2053 General Physics I Section 584771 Prof. Douglas H. Laurence Final Exam May 3, 2018 Name: 1 Instructions: This final exam is a take home exam. It will be posted online sometime around noon of the
More informationYour Name: PHYSICS 101 MIDTERM. Please Circle your section 1 9 am Galbiati 2 10 am Wang 3 11 am Hasan 4 12:30 am Hasan 5 12:30 pm Olsen
1 Your Name: PHYSICS 101 MIDTERM October 27, 2005 2 hours Please Circle your section 1 9 am Galbiati 2 10 am Wang 3 11 am Hasan 4 12:30 am Hasan 5 12:30 pm Olsen Problem Score 1 /16 2 /16 3 /16 4 /18 5
More informationPhysics 110 Homework Solutions Week #5
Physics 110 Homework Solutions Week #5 Wednesday, October 7, 009 Chapter 5 5.1 C 5. A 5.8 B 5.34. A crate on a ramp a) x F N 15 F 30 o mg Along the x-axis we that F net = ma = Fcos15 mgsin30 = 500 cos15
More informationQuiz Number 4 PHYSICS April 17, 2009
Instructions Write your name, student ID and name of your TA instructor clearly on all sheets and fill your name and student ID on the bubble sheet. Solve all multiple choice questions. No penalty is given
More informationTorque/Rotational Energy Mock Exam. Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK.
AP Physics C Spring, 2017 Torque/Rotational Energy Mock Exam Name: Answer Key Mr. Leonard Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK. (22 pts ) 1. Two masses are attached
More informationUniversity of Alabama Department of Physics and Astronomy. PH 125 / LeClair Fall Exam III Solution
University of Alabama Department of Physics and Astronomy PH 5 / LeClair Fall 07 Exam III Solution. A child throws a ball with an initial speed of 8.00 m/s at an anle of 40.0 above the horizontal. The
More informationClass XI Chapter 7- System of Particles and Rotational Motion Physics
Page 178 Question 7.1: Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie
More informationAP Physics C: Rotation II. (Torque and Rotational Dynamics, Rolling Motion) Problems
AP Physics C: Rotation II (Torque and Rotational Dynamics, Rolling Motion) Problems 1980M3. A billiard ball has mass M, radius R, and moment of inertia about the center of mass I c = 2 MR²/5 The ball is
More informationPHYS 21: Assignment 4 Solutions
PHYS 1: Assignment 4 Solutions Due on Feb 4 013 Prof. Oh Katharine Hyatt 1 Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.43 We see immediately that there is friction in this problem,
More information