Physics 170 Week 9 Lecture 2
|
|
- Gervase Melton
- 6 years ago
- Views:
Transcription
1 Physics 170 Week 9 Lecture 2 gordonws/170 Physics 170 Week 9 Lecture 2 1
2 Textbook Chapter 1: Section 1.6 Physics 170 Week 9 Lecture 2 2
3 Learning Goals: We will solve an example problem using tangential-normal coordinates. We will formulate Newton s second law in a cylindrical coordinate system. We shall study an example where Newton s law in cylindrical coordinates is used to solve a problem. You should also understand how to relate the orientations of the tangential-normal coordinate system and the cylindrical coordinate system for the case where the motion is entirely in a single plane. We will solve an example, where one needs to know how to do this. Physics 170 Week 9 Lecture 2
4 Example: The smooth block B, having a mass of 0.2 kg, is attached to the vertex A of the right circular cone using a light cord. The cone is rotating at a constant angular rate about the z axis such that the block attains a speed of 0.5 m/s. At this speed, determine the tension in the cord and the reaction which the cone exerts on the block. Neglect the size of the block. Physics 170 Week 9 Lecture 2 4
5 Free Body Diagram Physics 170 Week 9 Lecture 2 5
6 Find a Mathematical description of the force vectors Gravity: W = mg û b Normal reaction force N = N ) 4 5 ûn + 5 ûb Tension in cord: T = T ) 5 ûn ûb Physics 170 Week 9 Lecture 2 6
7 Find a Mathematical description of the acceleration vector The acceleration vector is at) = vt) û t + v2 ρ ûn and we know that v = 0, ρ = mm) and v = 0.5 m/s. a = 0.5 m/s)2 ûn m) Physics 170 Week 9 Lecture 2 7
8 Impose Newton s Second Law where F = W + N + T = mg û b +N F = m a 4 5 ûn + ) 5 ûb +T Remember that we computed the acceleration and found a = Putting this together, we have 0.2 kg) +N 0.5 m/s)2 ûn m) 0.5 m/s) m) ûn = 0.2 kg)9.81 m/s 2 ) û b 4 5 ûn + 5 ûb ) + T 5 ûn ûb 5 ûn + 4 ) 5 ûb ) Physics 170 Week 9 Lecture 2 8
9 Impose Newton s Second Law cont d We arrived at the formula 0.2 kg) +N 0.5 m/s) m) ûn = 0.2 kg)9.81 m/s 2 ) û b ) 4 5 ûn + 5 ûb + T 5 ûn + 4 ) 5 ûb First, consider the û b components. This gives an equation 5 N T = 0.2)9.81) N Then, consider the û n components. They give the formula 4 5 N + 5 T = 0.2 kg) 0.5) ) N Physics 170 Week 9 Lecture 2 9
10 Impose Newton s Second Law cont d Imposing Newton s law has resulted in the two equations 5 N T = 0.2)9.81) N 4 5 N + 5 Or, in matrix form [ ] [ /5 4/5 N 4/5 /5 T T = 0.2 kg) 0.5) ) N ] = [ 0.2)9.81) 0.2)0.5) ) Which we solve by multiplying by the inverse matrix to get [ ] [ ] [ ] N /5 4/5 0.2)9.81) = 0.2)0.5) T 4/5 /5 2 N ) ] N Physics 170 Week 9 Lecture 2 10
11 We have arrived at the equation [ ] [ N /5 4/5 = T 4/5 /5 Doing matrix multiplication gives ] [ 0.2)9.81) 0.2)0.5) ) ] N N = 5 0.2)9.81) 4 5 T = )9.81) )0.5) ) N 0.2)0.5) ) N N = N, T = 1.82 N Physics 170 Week 9 Lecture 2 11
12 Review of cylindrical coordinates Position: rt) = rt) û r + zt) ˆk Velocity: vt) = ṙt) û t + rt) θt) û θ + żt) ˆk Acceleration: at) = rt) rt) θt) ) 2 û r + 2ṙt) θt) ) + rt) θt) û θ + zt) ˆk Physics 170 Week 9 Lecture 2 12
13 The Cartesian cylindrical dictionary If you know the Cartesian vector r = x î + y ĵ + z ˆk, we can identify the components of the same vector in the cylindrical coordinate system using r = x 2 + y 2, θ = arctan y x, z = z Alternatively, if we know the cylindrical coordinates of a point r, θ, z) we can identify the Cartesian coordinates x = r cos θ, y = r sin θ, z = z Also, the unit vectors are related by û r = cos θ î + sin θ ĵ, û θ = sin θ î + cos θ ĵ Physics 170 Week 9 Lecture 2 1
14 Equation of Motion in Cylindrical Coordinates The equation of motion F = m a is an equation for vectors. When the vectors are written in cylindrical coordinates F = F r û r + F θ û θ + F z ˆk, a = ar û r + a θ û θ + a z ˆk the equation of motion is an equation for each of the three components F r = ma r, F θ = ma θ, F z = ma z If there is more than one force acting on the particle, F is the resultant, that is, the vector sum of the forces. Physics 170 Week 9 Lecture 2 14
15 Example: The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, θ = 0.7t) rad, and z = 0.5t) m, where t is in seconds. Determine the components of force F r, F θ, and F z which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy. Physics 170 Week 9 Lecture 2 15
16 Strategy for solving: We are given the position as time-dependent cylindrical coordinates. From the position, we will find the velocity and the acceleration by taking time derivatives. Once we know the acceleration, we will use Newton s second law, F = m a in cylindrical coordinates to find the components of F. Physics 170 Week 9 Lecture 2 16
17 Solution: We are given the time dependent cylindrical coordinates: rt) = 1.5 m, θt) = 0.7)t rad, zt) = 0.50)t m where t is in seconds. From these, we find ṙt) = 0, θt) = 0.7) rad, żt) = 0.50) m rt) = 0, θt) = 0, zt) = 0 Physics 170 Week 9 Lecture 2 17
18 Solution cont d: We have found r, θ, z) = 1.5 m, 0.7)t rad, 0.50)t m), ṙ, θ, ż) = 0, 0.7) rad/s, 0.50) m/s), r, θ, z) = 0, 0, 0). We want to plug these into at) = rt) rt) θt) 2) û r + 2ṙt) θt) ) + rt) θt) û θ + zt) ˆk = 1.5 m)0.7 rad/s) 2 û r Physics 170 Week 9 Lecture 2 18
19 Solution cont d: We have found that the boy has acceleration. at) = 1.5 m)0.7 rad/s) 2 û r Apply Newton s second law F = m a to get the net force on the boy F = 40 kg)1.5 m)0.7 rad/s) 2 û r = 29.4 û r N Physics 170 Week 9 Lecture 2 19
20 Solution cont d: We have found that the net force on the boy is F = 40 kg)1.5 m)0.7 rad/s) 2 û r = 29.4 û r N This must be the resultant of the reaction of the slide F s and gravity W = mg ˆk: F = Fs mg ˆk. We find F s = 29.4 û r + 40 kg)9.81 m/s 2 ) ˆk N and F sr = 29.4 N, F sθ = 0, F sz = 92 N. Physics 170 Week 9 Lecture 2 20
21 Example: The 0.5 lb particle is guided along the circular path using the slotted arm guide. If the arm has an angular velocity θ = 4 rad/s and an angular acceleration θ = 8 rad/s 2 at the instant when θ = 0 degrees, determine the force of the guide on the particle. Motion occurs in the horizontal plane. Physics 170 Week 9 Lecture 2 21
22 Strategy for solving: We are given the angular speed and acceleration. We need to find rt) in terms of θt) in order to find r, ṙ and r at the instant in question. We will find rt) by first using geometry to find fθ). Then, we will use the result and information about θ, θ and θ to find r, ṙ and r. Physics 170 Week 9 Lecture 2 22
23 Strategy for solving cont d: We then use above information to write down the acceleration in cylindrical coordinates. Newton s second law tells us that the sum of all forces acting on the particle are equal to the mass times the acceleration. We can therefore multiply the acceleration that we have found by the mass to get the total force acting on the particle. Physics 170 Week 9 Lecture 2 2
24 Strategy for solving cont d: The reaction force of the guide is normal to the trajectory, and is easiest to describe in tangential-normal coordinates where forces are F = N û n + F û θ with F the force of the guide. Since motion occurs in the horizontal plane, we neglect gravity. We then equate the reaction and friction force to the total force that we found using Newton s second law. Physics 170 Week 9 Lecture 2 24
25 Strategy for solving cont d: In order to proceed, We shall have to find a relationship between the tangential-normal unit vectors û t and û n and the cylindrical unit vectors û r and û θ in order to solve this problem. Once we have the relation between unit vectors, we can solve for the normal reaction force in terms of known quantities. Physics 170 Week 9 Lecture 2 25
26 Solution: Let us find rθ). From the diagram we see that rθ) = 1.0 ft) cos θ Physics 170 Week 9 Lecture 2 26
27 Solution cont d: Using rθ) = 1.0 ft) cos θ, ṙθ) = 1.0 ft) θ sin θ ) rθ) = 1.0 ft) θ sin θ + θ2 cos θ) At the instant in question, θ = π/6 rad, θ = 4 rad/s, θ = 8 rad/s 2, r = 1.0ft) 2 = 2 ft, ṙ = 1.0)4)1 ft/s = 2 ft/s 2 r = 1.0) ) = ) ft/s 2 Physics 170 Week 9 Lecture 2 27
28 Solution cont d: The acceleration in cylindrical coordinates is at) = rt) rt) θt) 2) û r + 2ṙt) θt) ) + rt) θt) where we have put the z-component to zero. Plugging in: at) = 4 8 ) 2 42 ) û r + 2 2)4) + ) 2 8) û θ û θ ft/s 2 Physics 170 Week 9 Lecture 2 28
29 Solution cont d: The acceleration is at) = 4 8 at) = ) 2 42 ) 4 16 ) û r + û r + 2 2)4) ) ) 2 8) û θ ft/s 2 û θ ft/s 2 Physics 170 Week 9 Lecture 2 29
30 Solution cont d: Newton s second law is F = m a Inserting the mass previous page, Nû n + F û θ = slug) and the acceleration from the 4 16 ) û r ) û θ lb To calculate N, we take the dot-product of both sides with û r. We need to know û r û n. Physics 170 Week 9 Lecture 2 0
31 Theoretical Interlude When the motion is confined to a plane, that is, when the xy-plane, equivalently rθ-plane, is the osculating plane, it might be useful to know the relative orientations of the tangential-normal and the spherical polar coordinates: To find it, we write the velocity in tangential-normal and cylindrical coordinates v = v û t, v = ṙ û r + r θ û θ These are the same vector. The dot product with û t is û t û r = cos ψ = ṙ ṙ 2 + r 2 θ, tan ψ = r θ 2 ṙ = r dθ dr Physics 170 Week 9 Lecture 2 1
32 Theoretical Interlude cont d Summary: For motion in the xy=rθ-plane: [ ] [ ût û r û t û θ cos ψ sin ψ = û n û r û n û θ sin ψ cos ψ ] cos ψ = ṙ ṙ 2 + r 2 θ 2, sin ψ = tan ψ = r dθ dr = r dr/dθ r θ ṙ 2 + r 2 θ 2 Physics 170 Week 9 Lecture 2 2
33 Solution cont d: Nû n + F û θ = ) û r We take the dot-product of both sides with û n to get ) Nû r û n = N sin ψ) = ) ) û θ lb Physics 170 Week 9 Lecture 2
34 Solution cont d: We have found N sin ψ) = ) Now tan ψ = r θ/ṙ =, sin ψ = 2, cos ψ = 1 2 ) 2 N = lb N = Physics 170 Week 9 Lecture 2 4
35 Solution cont d: Nû n + F û θ = ) û r We take the dot-product of both sides with û t to get F û t û θ = ) û t û r ) F = cot ψ) ) û θ lb ) û t û θ ) lb Physics 170 Week 9 Lecture 2 5
36 Solution cont d: F = F = F = 0.14 lb 4 16 ) cot ψ) ) ) ) lb lb Physics 170 Week 9 Lecture 2 6
37 For the next lecture, please read Textbook Chapter 14: Section Physics 170 Week 9 Lecture 2 7
Physics 170 Week 10, Lecture 1
Physics 170 Week 10, Lecture 1 http://www.phas.ubc.ca/ gordonws/170 Physics 170 Week 10, Lecture 1 1 Textbook Chapter 14: Section 14.1-3 Physics 170 Week 10, Lecture 1 2 Learning Goals: We will define
More informationEQUATIONS OF MOTION: CYLINDRICAL COORDINATES (Section 13.6)
EQUATIONS OF MOTION: CYLINDRICAL COORDINATES (Section 13.6) Today s Objectives: Students will be able to analyze the kinetics of a particle using cylindrical coordinates. APPLICATIONS The forces acting
More informationME 230 Kinematics and Dynamics
ME 230 Kinematics and Dynamics Wei-Chih Wang Department of Mechanical Engineering University of Washington Lecture 6: Particle Kinetics Kinetics of a particle (Chapter 13) - 13.4-13.6 Chapter 13: Objectives
More informationEQUATIONS OF MOTION: CYLINDRICAL COORDINATES
Today s Objectives: Students will be able to: 1. Analyze the kinetics of a particle using cylindrical coordinates. EQUATIONS OF MOTION: CYLINDRICAL COORDINATES In-Class Activities: Check Homework Reading
More informationPhysics 170 Week 5, Lecture 2
Physics 170 Week 5, Lecture 2 http://www.phas.ubc.ca/ gordonws/170 Physics 170 Week 5 Lecture 2 1 Textbook Chapter 5:Section 5.5-5.7 Physics 170 Week 5 Lecture 2 2 Learning Goals: Review the condition
More informationNEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES
NEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES Objectives: Students will be able to: 1. Write the equation of motion for an accelerating body. 2. Draw the
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5
1 / 40 CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa 2 / 40 EQUATIONS OF MOTION:RECTANGULAR COORDINATES
More information3 Space curvilinear motion, motion in non-inertial frames
3 Space curvilinear motion, motion in non-inertial frames 3.1 In-class problem A rocket of initial mass m i is fired vertically up from earth and accelerates until its fuel is exhausted. The residual mass
More informationEQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5)
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5) Today s Objectives: Students will be able to apply the equation of motion using normal and tangential coordinates. APPLICATIONS Race
More informationLecture 10. Example: Friction and Motion
Lecture 10 Goals: Exploit Newton s 3 rd Law in problems with friction Employ Newton s Laws in 2D problems with circular motion Assignment: HW5, (Chapter 7, due 2/24, Wednesday) For Tuesday: Finish reading
More informationCircular Motion Dynamics
Circular Motion Dynamics 8.01 W04D2 Today s Reading Assignment: MIT 8.01 Course Notes Chapter 9 Circular Motion Dynamics Sections 9.1-9.2 Announcements Problem Set 3 due Week 5 Tuesday at 9 pm in box outside
More informationLecture 5. Dynamics. Forces: Newton s First and Second
Lecture 5 Dynamics. Forces: Newton s First and Second What is a force? It s a pull or a push: F F Force is a quantitative description of the interaction between two physical bodies that causes them to
More informationparticle p = m v F ext = d P = M d v cm dt
Lecture 11: Momentum and Collisions; Introduction to Rotation 1 REVIEW: (Chapter 8) LINEAR MOMENTUM and COLLISIONS The first new physical quantity introduced in Chapter 8 is Linear Momentum Linear Momentum
More informationPHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009
PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively.
More informationPhysics 2211 M Quiz #2 Solutions Summer 2017
Physics 2211 M Quiz #2 Solutions Summer 2017 I. (16 points) A block with mass m = 10.0 kg is on a plane inclined θ = 30.0 to the horizontal, as shown. A balloon is attached to the block to exert a constant
More information1 Summary of Chapter 2
General Astronomy (9:61) Fall 01 Lecture 7 Notes, September 10, 01 1 Summary of Chapter There are a number of items from Chapter that you should be sure to understand. 1.1 Terminology A number of technical
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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. Physics 8.01 Fall Problem Set 2: Applications of Newton s Second Law Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall 2012 Problem 1 Problem Set 2: Applications of Newton s Second Law Solutions (a) The static friction force f s can have a magnitude
More informationChapter 9- Static Equilibrium
Chapter 9- Static Equilibrium Changes in Office-hours The following changes will take place until the end of the semester Office-hours: - Monday, 12:00-13:00h - Wednesday, 14:00-15:00h - Friday, 13:00-14:00h
More information1 Kinematics 1. 2 Particle Dynamics Planar Dynamics 42
Dynamics: 4600 03 Example Problems Contents 1 Kinematics 1 Particle Dynamics 13 3 Planar Dynamics 4 1 Kinematics Problem 1: 0 pts. The two blocks shown to the right are constrained the move in orthogonal
More informationPractice Midterm Exam 1. Instructions. You have 60 minutes. No calculators allowed. Show all your work in order to receive full credit.
MATH202X-F01/UX1 Spring 2015 Practice Midterm Exam 1 Name: Answer Key Instructions You have 60 minutes No calculators allowed Show all your work in order to receive full credit 1 Consider the points P
More informationInstructions: (62 points) Answer the following questions. SHOW ALL OF YOUR WORK. A B = A x B x + A y B y + A z B z = ( 1) + ( 1) ( 4) = 5
AP Physics C Fall, 2016 Work-Energy Mock Exam Name: Answer Key Mr. Leonard Instructions: (62 points) Answer the following questions. SHOW ALL OF YOUR WORK. (12 pts ) 1. Consider the vectors A = 2 î + 3
More informationLecture D10 - Angular Impulse and Momentum
J. Peraire 6.07 Dynamics Fall 2004 Version.2 Lecture D0 - Angular Impulse and Momentum In addition to the equations of linear impulse and momentum considered in the previous lecture, there is a parallel
More informationDynamics Kinetics of a particle Section 4: TJW Force-mass-acceleration: Example 1
Section 4: TJW Force-mass-acceleration: Example 1 The beam and attached hoisting mechanism have a combined mass of 1200 kg with center of mass at G. If the inertial acceleration a of a point P on the hoisting
More informationProblem 1 Problem 2 Problem 3 Problem 4 Total
Name Section THE PENNSYLVANIA STATE UNIVERSITY Department of Engineering Science and Mechanics Engineering Mechanics 12 Final Exam May 5, 2003 8:00 9:50 am (110 minutes) Problem 1 Problem 2 Problem 3 Problem
More informationINTRODUCTION. The three general approaches to the solution of kinetics. a) Direct application of Newton s law (called the forcemass-acceleration
INTRODUCTION According to Newton s law, a particle will accelerate when it is subjected to unbalanced force. Kinetics is the study of the relations between unbalanced forces and resulting changes in motion.
More informationDynamics 4600:203 Homework 09 Due: April 04, 2008 Name:
Dynamics 4600:03 Homework 09 Due: April 04, 008 Name: Please denote your answers clearly, i.e., box in, star, etc., and write neatly. There are no points for small, messy, unreadable work... please use
More informationChapter 5. The Laws of Motion
Chapter 5 The Laws of Motion The Laws of Motion The description of an object in There was no consideration of what might influence that motion. Two main factors need to be addressed to answer questions
More informationy(t) = y 0 t! 1 2 gt 2. With y(t final ) = 0, we can solve this for v 0 : v 0 A ĵ. With A! ĵ =!2 and A! = (2) 2 + (!
1. The angle between the vector! A = 3î! 2 ĵ! 5 ˆk and the positive y axis, in degrees, is closest to: A) 19 B) 71 C) 90 D) 109 E) 161 The dot product between the vector! A = 3î! 2 ĵ! 5 ˆk and the unit
More informationUNIT-07. Newton s Three Laws of Motion
1. Learning Objectives: UNIT-07 Newton s Three Laws of Motion 1. Understand the three laws of motion, their proper areas of applicability and especially the difference between the statements of the first
More informationChapter 17 Two Dimensional Rotational Dynamics
Chapter 17 Two Dimensional Rotational Dynamics 17.1 Introduction... 1 17.2 Vector Product (Cross Product)... 2 17.2.1 Right-hand Rule for the Direction of Vector Product... 3 17.2.2 Properties of the Vector
More informationChapter 5. The Laws of Motion
Chapter 5 The Laws of Motion Sir Isaac Newton 1642 1727 Formulated basic laws of mechanics Discovered Law of Universal Gravitation Invented form of calculus Many observations dealing with light and optics
More informationSolutionbank M1 Edexcel AS and A Level Modular Mathematics
Page of Solutionbank M Exercise A, Question A particle P of mass 0. kg is moving along a straight horizontal line with constant speed m s. Another particle Q of mass 0.8 kg is moving in the same direction
More informationPHYS-2010: General Physics I Course Lecture Notes Section V
PHYS-2010: General Physics I Course Lecture Notes Section V Dr. Donald G. Luttermoser East Tennessee State University Edition 2.5 Abstract These class notes are designed for use of the instructor and students
More informationPhysics 12. Unit 5 Circular Motion and Gravitation Part 1
Physics 12 Unit 5 Circular Motion and Gravitation Part 1 1. Nonlinear motions According to the Newton s first law, an object remains its tendency of motion as long as there is no external force acting
More informationEQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES Today s Objectives: Students will be able to: 1. Apply the equation of motion using normal and tangential coordinates. In-Class Activities: Check
More informationPhysics UCSB TR 2:00-3:15 lecture Final Exam Wednesday 3/17/2010
Physics @ UCSB TR :00-3:5 lecture Final Eam Wednesday 3/7/00 Print your last name: Print your first name: Print your perm no.: INSTRUCTIONS: DO NOT START THE EXAM until you are given instructions to do
More informationPhysics 111. Lecture 22 (Walker: ) Torque Rotational Dynamics Static Equilibrium Oct. 28, 2009
Physics 111 Lecture 22 (Walker: 11.1-3) Torque Rotational Dynamics Static Equilibrium Oct. 28, 2009 Lecture 22 1/26 Torque (τ) We define a quantity called torque which is a measure of twisting effort.
More informationChapter 8. Centripetal Force and The Law of Gravity
Chapter 8 Centripetal Force and The Law of Gravity Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration
More informationPhysics 170 Lecture 19. Chapter 12 - Kinematics Sections 8-10
Phys 170 Lecture 0 1 Physics 170 Lecture 19 Chapter 1 - Kinematics Sections 8-10 Velocity & Acceleration in Polar / Cylinical Coordinates Pulley Problems Phys 170 Lecture 0 Polar Coordinates Polar coordinates
More informationPhys101 First Major-111 Zero Version Monday, October 17, 2011 Page: 1
Monday, October 17, 011 Page: 1 Q1. 1 b The speed-time relation of a moving particle is given by: v = at +, where v is the speed, t t + c is the time and a, b, c are constants. The dimensional formulae
More informationPhysics 8 Friday, October 21, 2011
Physics 8 Friday, October 21, 2011 Bill and Zoey are away next week at Medical Imaging Conference in warm, sunny Valencia, Spain. Simon Hastings (simonhas@sas.upenn.edu) will run the class meetings on
More informationDr. Gundersen Phy 205DJ Test 2 22 March 2010
Signature: Idnumber: Name: Do only four out of the five problems. The first problem consists of five multiple choice questions. If you do more only your FIRST four answered problems will be graded. Clearly
More informationEQUATIONS OF MOTION: RECTANGULAR COORDINATES
EQUATIONS OF MOTION: RECTANGULAR COORDINATES Today s Objectives: Students will be able to: 1. Apply Newton s second law to determine forces and accelerations for particles in rectilinear motion. In-Class
More informationPHYSICS 221, FALL 2010 EXAM #1 Solutions WEDNESDAY, SEPTEMBER 29, 2010
PHYSICS 1, FALL 010 EXAM 1 Solutions WEDNESDAY, SEPTEMBER 9, 010 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
More informationRotation. Rotational Variables
Rotation Rigid Bodies Rotation variables Constant angular acceleration Rotational KE Rotational Inertia Rotational Variables Rotation of a rigid body About a fixed rotation axis. Rigid Body an object that
More informationChapter 5. The Laws of Motion
Chapter 5 The Laws of Motion The Laws of Motion The description of an object in motion included its position, velocity, and acceleration. There was no consideration of what might influence that motion.
More information= v 0 x. / t = 1.75m / s 2.25s = 0.778m / s 2 nd law taking left as positive. net. F x ! F
Multiple choice Problem 1 A 5.-N bos sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the bos sliding to the right at 1.75 m/s
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 informationN - W = 0. + F = m a ; N = W. Fs = 0.7W r. Ans. r = 9.32 m
91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 479 R1 1. The ball is thrown horizontally with a speed of 8 m>s. Find the equation of the path, y = f(x), and then determine the ball s velocity and the normal
More informationEngineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Kinematics
Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Kinematics Module 10 - Lecture 24 Kinematics of a particle moving on a curve Today,
More informationNewton s first and second laws
Lecture 2 Newton s first and second laws Pre-reading: KJF 4.1 to 4.7 Please log in to Socrative, room HMJPHYS1002 Recall Forces are either contact Pushes / Pulls Tension in rope Friction Normal force (virtually
More informationPotential Energy & Conservation of Energy
PHYS 101 Previous Exam Problems CHAPTER 8 Potential Energy & Conservation of Energy Potential energy Conservation of energy conservative forces Conservation of energy friction Conservation of energy external
More informationTorque. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Physics 6A Torque is what causes angular acceleration (just like a force causes linear acceleration) Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque
More informationPHY321 Homework Set 10
PHY321 Homework Set 10 1. [5 pts] A small block of mass m slides without friction down a wedge-shaped block of mass M and of opening angle α. Thetriangular block itself slides along a horizontal floor,
More informationDynamics II Motion in a Plane. Review Problems
Dynamics II Motion in a Plane Review Problems Problem 1 A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 8.0 N thrust
More informationPhysics 1 Second Midterm Exam (AM) 2/25/2010
Physics Second Midterm Eam (AM) /5/00. (This problem is worth 40 points.) A roller coaster car of m travels around a vertical loop of radius R. There is no friction and no air resistance. At the top of
More informationCHAPTER 10 ROTATION OF A RIGID OBJECT ABOUT A FIXED AXIS WEN-BIN JIAN ( 簡紋濱 ) DEPARTMENT OF ELECTROPHYSICS NATIONAL CHIAO TUNG UNIVERSITY
CHAPTER 10 ROTATION OF A RIGID OBJECT ABOUT A FIXED AXIS WEN-BIN JIAN ( 簡紋濱 ) DEPARTMENT OF ELECTROPHYSICS NATIONAL CHIAO TUNG UNIVERSITY OUTLINE 1. Angular Position, Velocity, and Acceleration 2. Rotational
More informationPLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION
PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION I. Moment of Inertia: Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate.
More informationCircular Motion Kinematics 8.01 W03D1
Circular Motion Kinematics 8.01 W03D1 Announcements Open up the Daily Concept Questions page on the MITx 8.01x Webpage. Problem Set 2 due Tue Week 3 at 9 pm Week 3 Prepset due Friday Week 3 at 8:30 am
More informationKinetics of Particles
Kinetics of Particles A- Force, Mass, and Acceleration Newton s Second Law of Motion: Kinetics is a branch of dynamics that deals with the relationship between the change in motion of a body and the forces
More informationChapter 18 Static Equilibrium
Chapter 8 Static Equilibrium Chapter 8 Static Equilibrium... 8. Introduction Static Equilibrium... 8. Lever Law... 3 Example 8. Lever Law... 5 8.3 Generalized Lever Law... 6 8.4 Worked Examples... 8 Example
More informationTwo-Dimensional Rotational Dynamics
Two-Dimensional Rotational Dynamics 8.01 W09D2 W09D2 Reading Assignment: MIT 8.01 Course Notes: Chapter 17 Two Dimensional Rotational Dynamics Sections 17.1-17.5 Chapter 18 Static Equilibrium Sections
More informationDYNAMICS ME HOMEWORK PROBLEM SETS
DYNAMICS ME 34010 HOMEWORK PROBLEM SETS Mahmoud M. Safadi 1, M.B. Rubin 2 1 safadi@technion.ac.il, 2 mbrubin@technion.ac.il Faculty of Mechanical Engineering Technion Israel Institute of Technology Spring
More informationRotational Kinetic Energy
Lecture 17, Chapter 10: Rotational Energy and Angular Momentum 1 Rotational Kinetic Energy Consider a rigid body rotating with an angular velocity ω about an axis. Clearly every point in the rigid body
More informationPhysics 8 Monday, October 9, 2017
Physics 8 Monday, October 9, 2017 Pick up a HW #5 handout if you didn t already get one on Wednesday. It s due this Friday, 10/13. It contains some Ch9 (work) problems, some Ch10 (motion in a plane) problems,
More information= constant of gravitation is G = N m 2 kg 2. Your goal is to find the radius of the orbit of a geostationary satellite.
Problem 1 Earth and a Geostationary Satellite (10 points) The earth is spinning about its axis with a period of 3 hours 56 minutes and 4 seconds. The equatorial radius of the earth is 6.38 10 6 m. The
More informationForces Part 1: Newton s Laws
Forces Part 1: Newton s Laws Last modified: 13/12/2017 Forces Introduction Inertia & Newton s First Law Mass & Momentum Change in Momentum & Force Newton s Second Law Example 1 Newton s Third Law Common
More informationPHYSICS I. Lecture 1. Charudatt Kadolkar. Jul-Nov IIT Guwahati
PHYSICS I Lecture 1 Charudatt Kadolkar IIT Guwahati Jul-Nov 2014 Section 1 Introduction to the Course Syllabus Topics Classical Mechanics: Kinetic Energy rest mass energy Syllabus Topics Classical Mechanics:
More informationPhysics 2514 Lecture 13
Physics 2514 Lecture 13 P. Gutierrez Department of Physics & Astronomy University of Oklahoma Physics 2514 p. 1/18 Goals We will discuss some examples that involve equilibrium. We then move on to a discussion
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 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 informationLecture Outline Chapter 6. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.
Lecture Outline Chapter 6 Physics, 4 th Edition James S. Walker Chapter 6 Applications of Newton s Laws Units of Chapter 6 Frictional Forces Strings and Springs Translational Equilibrium Connected Objects
More informationPhysics 6010, Fall Relevant Sections in Text: Introduction
Physics 6010, Fall 2016 Introduction. Configuration space. Equations of Motion. Velocity Phase Space. Relevant Sections in Text: 1.1 1.4 Introduction This course principally deals with the variational
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 informationDynamics ( 동역학 ) Ch.3 Kinetic of Particles (3.1)
Dynamics ( 동역학 ) Ch.3 Kinetic of Particles (3.1) Introduction This chapter exclusively deals with the Newton s second Law Newton s second law: - A particle will have an acceleration proportional to the
More informationCircular Motion Kinematics
Circular Motion Kinematics 8.01 W04D1 Today s Reading Assignment: MIT 8.01 Course Notes Chapter 6 Circular Motion Sections 6.1-6.2 Announcements Math Review Week 4 Tuesday 9-11 pm in 26-152. Next Reading
More informationFigure 1 Answer: = m
Q1. Figure 1 shows a solid cylindrical steel rod of length =.0 m and diameter D =.0 cm. What will be increase in its length when m = 80 kg block is attached to its bottom end? (Young's modulus of steel
More informationFirst Year Physics: Prelims CP1. Classical Mechanics: Prof. Neville Harnew. Problem Set III : Projectiles, rocket motion and motion in E & B fields
HT017 First Year Physics: Prelims CP1 Classical Mechanics: Prof Neville Harnew Problem Set III : Projectiles, rocket motion and motion in E & B fields Questions 1-10 are standard examples Questions 11-1
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003 NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 14 pages. Make sure none are missing 2. There is
More information= m. 30 m. The angle that the tangent at B makes with the x axis is f = tan-1
1 11. When the roller coaster is at B, it has a speed of 5 m>s, which is increasing at at = 3 m>s. Determine the magnitude of the acceleration of the roller coaster at this instant and the direction angle
More informationF1.9AB2 1. r 2 θ2 + sin 2 α. and. p θ = mr 2 θ. p2 θ. (d) In light of the information in part (c) above, we can express the Hamiltonian in the form
F1.9AB2 1 Question 1 (20 Marks) A cone of semi-angle α has its axis vertical and vertex downwards, as in Figure 1 (overleaf). A point mass m slides without friction on the inside of the cone under the
More informationPhysics 111: Mechanics Lecture 5
Physics 111: Mechanics Lecture 5 Bin Chen NJIT Physics Department Forces of Friction: f q When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion.
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 informationPlane Motion of Rigid Bodies: Forces and Accelerations
Plane Motion of Rigid Bodies: Forces and Accelerations Reference: Beer, Ferdinand P. et al, Vector Mechanics for Engineers : Dynamics, 8 th Edition, Mc GrawHill Hibbeler R.C., Engineering Mechanics: Dynamics,
More informationPhys101 Lecture 5 Dynamics: Newton s Laws of Motion
Phys101 Lecture 5 Dynamics: Newton s Laws of Motion Key points: Newton s second law is a vector equation Action and reaction are acting on different objects Free-Body Diagrams Ref: 4-1,2,3,4,5,6,7. Page
More informationDistance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is:
Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v =
More informationProjectile Motion and 2-D Dynamics
Projectile Motion and 2-D Dynamics Vector Notation Vectors vs. Scalars In Physics 11, you learned the difference between vectors and scalars. A vector is a quantity that includes both direction and magnitude
More informationPhysics 101 Lecture 11 Torque
Physics 101 Lecture 11 Torque Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com Force vs. Torque q Forces cause accelerations q What cause angular accelerations? q A door is free to rotate about an axis
More informationChapter 8. Rotational Kinematics
Chapter 8 Rotational Kinematics 8.3 The Equations of Rotational Kinematics 8.4 Angular Variables and Tangential Variables The relationship between the (tangential) arc length, s, at some radius, r, and
More informationPhysics Final Exam Formulas
INSTRUCTIONS: Write your NAME on the front of the blue exam booklet. The exam is closed book, and you may have only pens/pencils and a calculator (no stored equations or programs and no graphing). Show
More informationFigure 17.1 The center of mass of a thrown rigid rod follows a parabolic trajectory while the rod rotates about the center of mass.
17.1 Introduction A body is called a rigid body if the distance between any two points in the body does not change in time. Rigid bodies, unlike point masses, can have forces applied at different points
More informationChapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity
Chapter 8 Rotational Equilibrium and Rotational Dynamics 1. Torque 2. Torque and Equilibrium 3. Center of Mass and Center of Gravity 4. Torque and angular acceleration 5. Rotational Kinetic energy 6. Angular
More informationChapter 6: Vector Analysis
Chapter 6: Vector Analysis We use derivatives and various products of vectors in all areas of physics. For example, Newton s 2nd law is F = m d2 r. In electricity dt 2 and magnetism, we need surface and
More informationWork and Energy (Work Done by a Constant Force)
Lecture 11 Chapter 7 Physics I 10.16.2013 Work and Energy (Work Done by a Constant Force) Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi Lecture Capture: http://echo360.uml.edu/danylov2013/physics1fall.html
More informationPhysics 0174(CHS) Exam #1 Academic Year NAME
. Physics 0174(CHS) Exam #1 Academic Year 2015-2016 NAME This exam consists of 6 pages in addition to this page; please check to see that you have all of them. Be sure to show clearly how you arrive at
More informationPHYS 101 Previous Exam Problems. Force & Motion I
PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward
More informationPROBLEM 16.4 SOLUTION
PROBLEM 16.4 The motion of the.5-kg rod AB is guided b two small wheels which roll freel in horizontal slots. If a force P of magnitude 8 N is applied at B, determine (a) the acceleration of the rod, (b)
More informationExam 02: Chapters 16 19
NAME: Exam 02: Chapters 16 19 Instructions Solve six of the following problems to the best of your ability. You have two hours in which to complete this exam. Choose one problem from each chapter, then
More informationChapter 6. Circular Motion and Other Applications of Newton s Laws
Chapter 6 Circular Motion and Other Applications of Newton s Laws Circular Motion Two analysis models using Newton s Laws of Motion have been developed. The models have been applied to linear motion. Newton
More information