A j = 0.1 cm 2 10 cm 10 cm 10 cm. W j Wj. W j W j. W j. 10 cm 10 cm 10 cm. r i
|
|
- Preston Terry
- 5 years ago
- Views:
Transcription
1 ME 131B Fluid Mechanics Solutions to Week Eight Problem Session: Angular Momentum Principle (3/2/98) 1. In control volume analysis, all governing principles share the same common structure: storage = inow outow + production The main dierence between dierent physical principles is in the production term, P. For the following principles, what is this production term equal to? (a) conservation of mass Mass can neither be created nor destroyed. P mass = (b) conservation of linear momentum According to Newton's law of motion, we can change the momentum of a system by apply an external force on it. P lin:mom: = X ~ F There are two main types of forces: { Surface force It is present along the control surfaces of your selected control volume. The best way to identify the surface forces is to trace along the entire control surface and ask yourself the question \What force does my control volume experience along this surface? Some examples of surface force are pressure force (normal direction) and friction (tangential direction). { Body force It is present due to the contents inside the selected control volume under the inuence of the surrounding force eld. Some examples of body force are gravitational force and electrostatic force. (c) conservation of angular momentum If we draw the analogy between linear momentum in translational motion and angular momentum in rotational motion, external force will be analogous to external torque. Hence, P ang:mom: = X ~ 1
2 Since the angular momentum equation is derived from the linear momentum equation, all the external forces, ~ F, (both surface and body) in the linear momentum equation are capable of generating torque, ~r ~ F, on the same control volume as long as the line of action of the force does not pass through the center of rotation. (d) conservation of energy We can change the total energy of a system by adding heat () or doing work (W ) on the system P energy = in + W in 2. The Reynolds Transport Theorem is the core basis in control volume analysis. It serves as a bridge between the control mass and the control volume approach. We can state it as:! dn ( dv ) + V ~ d A ~ system Most physical laws are Lagrangian in nature, i.e. they are derived for a system with a xed amount of substance (control mass approach). However, this approach is not easy to follow for a uid system simply because a uid can be deformed continuously as it moves around in space. The Reynolds Transport Theorem relates the rate of change of an extensive property N of a control mass system! dn dt system with the rate of change of the same property in a ( dv ) + {z } storage ~ V d ~ A {z } outow - inow There are two major components in the above equation: { The rst one is the storage term which accounts for the rate of increase in property N within the control volume. { The second one is the net outow term which accounts for the loss of property N due to the uid motion in and out of the control volume. Only with the Reynolds Transport Theorem, we can then relate the physical laws to what we measure in a xed region in space (control volume approach). 2
3 For the following principles, what are the quantities N and? The quantity N is an extensive property of the system whereas the quantity is its intensive counterpart. (a) conservation of mass N = M (total mass of system), = 1 (b) conservation of linear momentum N = ~ M (total linear momentum of system), = ~ V (c) conservation of angular momentum N = ~ A (total angular momentum of system), = ~r ~ V (d) conservation of energy N = E (total energy of system), = u + j~ V j g z Remarks: With the results of uestion 1 and 2, we can summarize all the conservation laws in the ( dv ) + ~ V d ~ A = P 3. What is the main criterion in choosing a suitable control volume in problem solving? We should put the control surfaces at places where { we know how the ow behaves, for example, ( ~ V ; P ) { we want to know something about, for example, frictional shear, exit pressure. 4. From what physical principle is the angular momentum equation derived? The angular momentum equation is derived by taking the cross product between the position vector, ~r, and the linear momentum equation. Hence, its main physics comes from Newton's law of motion. The main dierence is that the linear momentum equation governs the translational motion while the the angular momentum equation governs the rotational motion of the system. 3
4 5. Choose the best answer in the following question: Pressure is always directed into the control volume of interest. Pressure is a compressive force. Hence, it is always directed into the system of interest. 6. The angular momentum principle can be expressed in the following two forms: Form 1: Form @t ~r ~ F s + ~r ~ V ~r ~ F s + (~r ~g) ( dv ) + ~ T shaft ( dv ) + ~r ~ V (~r ~g) ( dv ) + ~ T shaft ~ V d ~ A ~r h 2~! V ~ + ~! (~! ~r) + ~! _ i ~r ( dv ) ~r V ~ ( dv ) + ~r V ~ V ~ d A ~ (a) What is the main dierence between the above two forms? Form 1 is derived in an inertial frame. Form 2 is derived in a rotating (non-inertial) frame. We need to make sure that the velocity vector, ~ V, is consistent with the corresponding choice of reference frame when we invoke the angular momentum principle. As long as we use the two forms in a consistent manner, they should give identical results. (b) Give a verbal description to each term in the equations. ~r F ~ s is the torque generated by surface force, F ~ s. R (~r ~g)( dv ) is the torque generated by gravitational force. R ~r (2~! V ~ ) ( dv ) is the torque generated by Coriolis force. R ~r [~! (~! ~r)] ( dv ) is the torque generated by centripetal force. R ~r ( ~! _ ~r) ( dv ) is the \ctitious torque due to angular acceleration of the rotating reference (~r V ~ ) ( dv ) is the rate of increase in angular momentum within the control volume. R (~r V ~ ) ( V ~ d A) ~ is the net outow of angular momentum caused by uid motion in and out of the control volume. 4
5 7. A total water discharge of 2 cm 3 is issued from a sprinkler as shown in the following gure: A j =.1 cm 2 1 cm 1 cm 1 cm W j Wj W j o 4 W j W j W j 1 cm 1 cm 1 cm A o 4 Assume that the jet speed is the same from all the holes. We rst choose a control volume to include the entire sprinkler arm as indicated above. Let us solve this angular momentum problem using an inertial reference frame and see how the analysis works. The corresponding angular momentum equation is X ~r V ~ ( dv ) + ~r V ~ V ~ d A ~ We then examine every individual term in the above equation: { The sources of external torque in this problem come from shaft torque, ~ shaft frictional torque, ~ f { The storage term is zero because we are dealing with a steady problem. { The net angular momentum outow term can be evaluated by rst considering the jet out of one hole only: ω W j r i For this single jet case, r i ω ~r V ~ V ~ d A ~ = (r Vt ) ( W j A j ) ~e k = (r V t ) ~e k 6 where V t is the tangential velocity component measured in an inertial frame. 5
6 { Consider the relative motion equation ~ V = ~ U + ~ W, we can resolve it in the tangential direction as V t = U t + W t = r i! + W j sin ) r V t = r 2 i! + r i W j sin = r 2 i! + r i sin { Hence, the angular momentum outow from this hole is equal to r i sin r 2 i!! 6 { The total angular momentum outow can be obtained by summing the contribution from all six holes together ~r V ~ V ~ d A ~ = 2 3X i = 1 r i ~e k! sin r 2 i 6 A! # ~e k j 6 where r 1 = 1 cm, r 2 = 2 cm, r 3 = 3 cm. The entire angular momentum equation can then be simplied to shaft + f = 3 # (r 1 + r 2 + r 3 ) sin! r r r 2 3 This general equation of motion forms the common basis for the following special cases of interests. (a) Static Case: Determine the torque that must be applied to the sprinkler arms to hold them from rotating. In this static case, we have {! = { f = The restraining torque is shaft = 2 18 A j! (r 1 + r 2 + r 3 ) sin (b) Frictionless Case: Determine the angular speed if the arms are free to rotate and there is no friction. In this frictionless case, we have 6
7 { shaft = { f = The angular speed of the sprinkler arm is! = r 1 + r 2 + r 3 r r r 2 3 sin (c) Frictional Case: Determine the angular speed if there is a constant frictional torque of 1 N-m resisting rotation of the arms. In this frictional case, we have { shaft = The angular speed of the sprinkler arm is! = 1 r r r 2 3 (r 1 + r 2 + r 3 ) sin which reduces to the results in Part (b) for f =. # 3 f (d) Relative Motion: Determine the absolute velocity of the uid leaving Hole A in Part (b) and (c). In both cases, we can apply the relative motion equation V ~ = U ~ + W ~ to analyze the velocity components. { Radial component: { Tangential component: V r = U r + W r = + W j cos = cos V t = U t + W t = r 3! + W j sin = sin r 3! The absolute velocity of the uid is given by V = = = q vu u t vu u t V 2 t + Vr 2! 2 sin r 3! +! 2 cos! 2 + (r 3!) 2 3 A j r 3! sin 7
8 (e) Sketch the corresponding velocity vector diagrams for Part (d). W = V r = W r V V t = U W t U = r 3 ω 8. Refer to the schematic below, a \wye joint splits a pipe ow into two equal amounts, =2, which exit at a distance R from the x-axis. The system rotates about the x axis at a rate. /2 Τ, Ω R >> D pipe R x /2 (a) Inertial Frame Analysis: Apply the angular momentum principle in an inertial frame to i. determine the torque required to turn the pipe (constant speed). ii. determine the additional torque which is required to generate an angular acceleration _ on the existing system (constant acceleration). The angular momentum equation in an inertial frame is X ~r ~ V ( dv ) + ~r ~ V ~ V d ~ A Let us rst consider a control volume which includes the upper tube only: 8
9 1 /2 d r Τ, Ω r R y x z According to the coordinates system chosen, we can express the position and velocity vectors as ~r = r cos + sin ~j cos + sin ~j + r sin ~ k ~V = 2 A We can then perform the cross product evaluation as follows: ~r V ~ = r cos + sin ~j cos + sin ~j + r sin ~ k 2 A r = 2 A sin cos ~ i ~j + r2 sin cos + r 2 A sin cos ~ j + r2 sin 2 = r 2 sin sin cos ~j ~ j ~ k ~ i ~ k The total angular momentum stored in the upper tube can be obtained by integrating the above expression along the entire upper tube length ~r V 1 ~ R = sin h i ( dv ) = r 2 sin sin cos ~j ( A dr) = A R3 sin cos ~j 2 The storage term can then be obtained by take the time derivative of the above expression. (Notice that only the angular speed is a function of ~r 1 ~ A R 3 _ ( dv ) = sin cos ~j 2 The angular momentum outow term can be obtained by evaluating the ~r V ~ expression at r = R = sin 1 ~r ~ V ~ V d ~ A = R 2 sin sin cos ~j # 2 9
10 These procedures conclude the analysis of the upper tube. We can then proceed to analyze the lower tube by following the same procedures. Τ, Ω y r R z x d r 2 /2 The main dierences will be in the expressions of position and velocity vectors ~r = r cos sin ~j ~V = cos sin ~j r sin 2 A ~ k After algebraic manipulations, we ~r 2 ~ A R 3 _ ( dv ) = sin + cos ~j 2 and ~r V 2 ~ V ~ d A ~ R 2 = sin To obtain the global conservation equation for the entire system, we need to sum up the results from the upper and the ~r ~ V ( dv ) + ~r ~ V sin + cos ~j # 2 ~ V d ~ A = R A R3 _ From the angular momentum equation in an inertial frame, we deduce that the applied torque is ~ = R A R3 _ The rst term corresponds to the torque required to turn the pipe at constant speed,, while the second term corresponds to the additional torque required to produce an angular acceleration, _, on the existing system. Hence, ~ = R 2 ~ _ = 2 A _ R3!! 1
11 (b) Rotating Frame Analysis: Repeat the analysis in Part (a) in a rotating frame. The angular momentum equation in a rotating frame is ~ ~r h 2~! ~ V + ~! (~! ~r) + _ ~! ~r i ( dv ) ~r V ~ ( dv ) + ~r V ~ V ~ d A Let us rst consider a control volume which includes the upper tube only: 1 /2 d r Τ, Ω r R y x z According to the coordinates system chosen, we can express the position, velocity and angular velocity vectors as ~r = r cos + sin ~j ~V = cos + sin ~j 2 A ~! = Since ~r is collinear to ~ V, their cross product is equal to zero ~r ~ V = ~ Hence, the storage term and the net outow term both equal zero and do not contribute to the angular momentum balance in the rotating frame. Let us evaluate each cross product term in the \ctitious torque carefully: 2 ~! ~ V = 2 2 A = A sin ~ i ~j = A sin ~ k cos + sin ~j ~! ~r = h i r cos + sin ~j ~ i ~j = r sin = r sin ~ k 11
12 These combine to give and ~! (~! ~r) = r sin ~ k ~ i k ~ = r 2 sin = r 2 sin ~j _~! ~r = _ h i r cos + sin ~j ~ i ~j = r _ sin = r _ sin ~ k 2 ~! ~ V + ~! (~! ~r) + _ ~! ~r = sin ~r h 2 ~! V ~ + ~! (~! ~r) + ~! _ i ~r = r sin 2 A A + r _ ~ k r 2 sin ~j + r _ r sin cos A + r _ ~j r 2 2 sin cos ~ k The combined \ctitious torque can be obtained by integrating the above expression along the entire upper tube length i ( dv ) = 1 ~r h 2 ~! ~ V + ~! (~! ~r) + _ ~! ~r R = sin = ( A) R 2 r sin 2 A i r 2 2 sin cos ~ k ( A dr) 2 A + R _! # R 3 2 cos ~ k 2 + r _ r sin cos R 2 cos sin A + r _ ~j 2 A + R _! ~j These procedures conclude the analysis of the upper tube. We can then proceed to analyze the lower tube by following the same procedures. 12
13 Τ, Ω y r R z x d r 2 /2 The main dierences will be in the expressions of position and velocity vectors ~r = r cos sin ~j ~V = cos sin ~j 2 A After algebraic manipulations, we obtain = ( A) ~r h 2 ~! V ~ + ~! (~! ~r) + ~! _ ~r 2 R 2 2 A + R _ # ~ k + R3 2 cos 2! + R2 cos sin i ( dv ) 2 A + R _! ~j To obtain the global conservation equation for the entire system, we need to sum up the results from the upper and the lower tubes. = ( A) ~r h 2 ~! ~ V + ~! (~! ~r) + _ ~! ~r i ( dv ) 2 R 2 2 A + R _! # From the angular momentum equation in a rotating frame, we deduce that the applied torque is ~ = = ( A) ~r h 2 ~! V ~ + ~! (~! ~r) + ~! _ i ~r ( dv ) 2 R 2 2 A + R _! # The rst term corresponds to the torque required to turn the pipe at constant speed,, while the second term corresponds to the additional torque required to produce an angular acceleration, _, on the existing system. Hence, ~ = R 2 ~ _ = 2 A _ R3 13
14 This example demonstrates that we can obtain the same results by choosing either the inertial or rotating reference frame to analyze angular momentum problems as long as the velocity vector is consistent with the corresponding chosen reference frame. In this problem, { the production, the storage and the net outow terms are all active in the inertial frame analysis; { the production term and the \ctitious torque are both active but the storage and the net outow terms are both zero in the rotating frame analysis. 14
Chapter 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 informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 15 Conservation Equations in Fluid Flow Part III Good afternoon. I welcome you all
More informationIn this lecture... Centrifugal compressors Thermodynamics of centrifugal compressors Components of a centrifugal compressor
Lect- 3 In this lecture... Centrifugal compressors Thermodynamics of centrifugal compressors Components of a centrifugal compressor Centrifugal compressors Centrifugal compressors were used in the first
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 informationChapter 11 Rotational Dynamics and Static Equilibrium. Copyright 2010 Pearson Education, Inc.
Chapter 11 Rotational Dynamics and Static Equilibrium Units of Chapter 11 Torque Torque and Angular Acceleration Zero Torque and Static Equilibrium Center of Mass and Balance Dynamic Applications of Torque
More informationLecture Outline Chapter 11. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.
Lecture Outline Chapter 11 Physics, 4 th Edition James S. Walker Chapter 11 Rotational Dynamics and Static Equilibrium Units of Chapter 11 Torque Torque and Angular Acceleration Zero Torque and Static
More informationChapter 12. Static Equilibrium and Elasticity
Chapter 12 Static Equilibrium and Elasticity Static Equilibrium Equilibrium implies that the object moves with both constant velocity and constant angular velocity relative to an observer in an inertial
More informationTutorial Materials for ME 131B Fluid Mechanics (Compressible Flow & Turbomachinery) Calvin Lui Department of Mechanical Engineering Stanford University Stanford, CA 94305 March 1998 Acknowledgments This
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 informationChapter 4 Dynamics: Newton s Laws of Motion
Chapter 4 Dynamics: Newton s Laws of Motion Units of Chapter 4 Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal
More information2. FLUID-FLOW EQUATIONS SPRING 2019
2. FLUID-FLOW EQUATIONS SPRING 2019 2.1 Introduction 2.2 Conservative differential equations 2.3 Non-conservative differential equations 2.4 Non-dimensionalisation Summary Examples 2.1 Introduction Fluid
More informationChapter 6: Momentum Analysis
6-1 Introduction 6-2Newton s Law and Conservation of Momentum 6-3 Choosing a Control Volume 6-4 Forces Acting on a Control Volume 6-5Linear Momentum Equation 6-6 Angular Momentum 6-7 The Second Law of
More informationV (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t)
IV. DIFFERENTIAL RELATIONS FOR A FLUID PARTICLE This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common
More informationDynamics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Dynamics Semester 1, / 18
Dynamics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2016-17 B. Bona (DAUIN) Dynamics Semester 1, 2016-17 1 / 18 Dynamics Dynamics studies the relations between the 3D space generalized forces
More informationChapter 5. Sound Waves and Vortices. 5.1 Sound waves
Chapter 5 Sound Waves and Vortices In this chapter we explore a set of characteristic solutions to the uid equations with the goal of familiarizing the reader with typical behaviors in uid dynamics. Sound
More information10.52 Mechanics of Fluids Spring 2006 Problem Set 3
10.52 Mechanics of Fluids Spring 2006 Problem Set 3 Problem 1 Mass transfer studies involving the transport of a solute from a gas to a liquid often involve the use of a laminar jet of liquid. The situation
More informationAP Physics C Mechanics Objectives
AP Physics C Mechanics Objectives I. KINEMATICS A. Motion in One Dimension 1. The relationships among position, velocity and acceleration a. Given a graph of position vs. time, identify or sketch a graph
More informationObjectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation
Objectives Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Conservation of Mass Conservation of Mass Mass, like energy, is a conserved
More informationRotation. Kinematics Rigid Bodies Kinetic Energy. Torque Rolling. featuring moments of Inertia
Rotation Kinematics Rigid Bodies Kinetic Energy featuring moments of Inertia Torque Rolling Angular Motion We think about rotation in the same basic way we do about linear motion How far does it go? How
More information4 Mechanics of Fluids (I)
1. The x and y components of velocity for a two-dimensional flow are u = 3.0 ft/s and v = 9.0x ft/s where x is in feet. Determine the equation for the streamlines and graph representative streamlines in
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 informationNewton's second law of motion
OpenStax-CNX module: m14042 1 Newton's second law of motion Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 2.0 Abstract Second law of
More informationDynamics. Dynamics of mechanical particle and particle systems (many body systems)
Dynamics Dynamics of mechanical particle and particle systems (many body systems) Newton`s first law: If no net force acts on a body, it will move on a straight line at constant velocity or will stay at
More informationThe... of a particle is defined as its change in position in some time interval.
Distance is the. of a path followed by a particle. Distance is a quantity. The... of a particle is defined as its change in position in some time interval. Displacement is a.. quantity. The... of a particle
More informationIntroduction & Basic Concepts of Thermodynamics
Introduction & Basic Concepts of Thermodynamics Reading Problems 2-1 2-8 2-53, 2-67, 2-85, 2-96 Introduction to Thermal Sciences Thermodynamics Conservation of mass Conservation of energy Second law of
More informationg (z) = 1 (1 + z/a) = 1
1.4.2 Gravitational Force g is the gravitational force. It always points towards the center of mass, and it is proportional to the inverse square of the distance above the center of mass: g (z) = GM (a
More informationFluid Mechanics II. Newton s second law applied to a control volume
Fluid Mechanics II Stead flow momentum equation Newton s second law applied to a control volume Fluids, either in a static or dnamic motion state, impose forces on immersed bodies and confining boundaries.
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 informationWork - kinetic energy theorem for rotational motion *
OpenStax-CNX module: m14307 1 Work - kinetic energy theorem for rotational motion * Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 2.0
More informationViscous Fluids. Amanda Meier. December 14th, 2011
Viscous Fluids Amanda Meier December 14th, 2011 Abstract Fluids are represented by continuous media described by mass density, velocity and pressure. An Eulerian description of uids focuses on the transport
More information1 2 Models, Theories, and Laws 1.5 Distinguish between models, theories, and laws 2.1 State the origin of significant figures in measurement
Textbook Correlation Textbook Correlation Physics 1115/2015 Chapter 1 Introduction, Measurement, Estimating 1.1 Describe thoughts of Aristotle vs. Galileo in describing motion 1 1 Nature of Science 1.2
More informationManipulator Dynamics 2. Instructor: Jacob Rosen Advanced Robotic - MAE 263D - Department of Mechanical & Aerospace Engineering - UCLA
Manipulator Dynamics 2 Forward Dynamics Problem Given: Joint torques and links geometry, mass, inertia, friction Compute: Angular acceleration of the links (solve differential equations) Solution Dynamic
More informationPhysics 106b/196b Problem Set 9 Due Jan 19, 2007
Physics 06b/96b Problem Set 9 Due Jan 9, 2007 Version 3: January 8, 2007 This problem set focuses on dynamics in rotating coordinate systems (Section 5.2), with some additional early material on dynamics
More informationP 1 P * 1 T P * 1 T 1 T * 1. s 1 P 1
ME 131B Fluid Mechanics Solutions to Week Three Problem Session: Isentropic Flow II (1/26/98) 1. From an energy view point, (a) a nozzle is a device that converts static enthalpy into kinetic energy. (b)
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 informationPhysics 351 Wednesday, February 28, 2018
Physics 351 Wednesday, February 28, 2018 HW6 due Friday. For HW help, Bill is in DRL 3N6 Wed 4 7pm. Grace is in DRL 2C2 Thu 5:30 8:30pm. To get the most benefit from the homework, first work through every
More informationPhysics A - PHY 2048C
Physics A - PHY 2048C and 11/15/2017 My Office Hours: Thursday 2:00-3:00 PM 212 Keen Building Warm-up Questions 1 Did you read Chapter 12 in the textbook on? 2 Must an object be rotating to have a moment
More informationFluid Mechanics. Spring 2009
Instructor: Dr. Yang-Cheng Shih Department of Energy and Refrigerating Air-Conditioning Engineering National Taipei University of Technology Spring 2009 Chapter 1 Introduction 1-1 General Remarks 1-2 Scope
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 informationPhysics Kinematics, Projectile Motion, Free-Body Diagrams, and Rotational Motion
Physics Kinematics, Projectile Motion, Free-Body Diagrams, and Rotational Motion Kinematics and Projectile Motion Problem Solving Steps 1. Read and Re-Read the whole problem carefully before trying to
More informationIntroduction to Fluid Machines, and Compressible Flow Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Introduction to Fluid Machines, and Compressible Flow Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 09 Introduction to Reaction Type of Hydraulic
More informationShow all work in answering the following questions. Partial credit may be given for problems involving calculations.
Physics 3210, Spring 2018 Final Exam Name: Signature: UID: Please read the following before continuing: Show all work in answering the following questions. Partial credit may be given for problems involving
More informationCEE 3310 Control Volume Analysis, Oct. 7, D Steady State Head Form of the Energy Equation P. P 2g + z h f + h p h s.
CEE 3310 Control Volume Analysis, Oct. 7, 2015 81 3.21 Review 1-D Steady State Head Form of the Energy Equation ( ) ( ) 2g + z = 2g + z h f + h p h s out where h f is the friction head loss (which combines
More informationAppendix W. Dynamic Models. W.2 4 Complex Mechanical Systems. Translational and Rotational Systems W.2.1
Appendix W Dynamic Models W.2 4 Complex Mechanical Systems W.2.1 Translational and Rotational Systems In some cases, mechanical systems contain both translational and rotational portions. The procedure
More information2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity
2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity Energy 7 4 Kinematics Free fall Collisions 3 5 Dynamics
More informationCOURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour. Basic Equations in fluid Dynamics
COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour Basic Equations in fluid Dynamics Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET 1 Description of Fluid
More information= o + t = ot + ½ t 2 = o + 2
Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the
More informationChapter 9 TORQUE & Rotational Kinematics
Chapter 9 TORQUE & Rotational Kinematics This motionless person is in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. This car is in dynamic equilibrium
More information3.8 The First Law of Thermodynamics and the Energy Equation
CEE 3310 Control Volume Analysis, Sep 30, 2011 65 Review Conservation of angular momentum 1-D form ( r F )ext = [ˆ ] ( r v)d + ( r v) out ṁ out ( r v) in ṁ in t CV 3.8 The First Law of Thermodynamics and
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 informationEngineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati
Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Module No. - 01 Basics of Statics Lecture No. - 01 Fundamental of Engineering Mechanics
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 informationMOMENTUM PRINCIPLE. Review: Last time, we derived the Reynolds Transport Theorem: Chapter 6. where B is any extensive property (proportional to mass),
Chapter 6 MOMENTUM PRINCIPLE Review: Last time, we derived the Reynolds Transport Theorem: where B is any extensive property (proportional to mass), and b is the corresponding intensive property (B / m
More informationPhysics 121, April 3, Equilibrium and Simple Harmonic Motion. Physics 121. April 3, Physics 121. April 3, Course Information
Physics 121, April 3, 2008. Equilibrium and Simple Harmonic Motion. Physics 121. April 3, 2008. Course Information Topics to be discussed today: Requirements for Equilibrium (a brief review) Stress and
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 PowerPoints. Chapter 8 Physics: Principles with Applications, 6 th edition Giancoli
Lecture PowerPoints Chapter 8 Physics: Principles with Applications, 6 th edition Giancoli 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the
More informationLesson 6 Review of fundamentals: Fluid flow
Lesson 6 Review of fundamentals: Fluid flow The specific objective of this lesson is to conduct a brief review of the fundamentals of fluid flow and present: A general equation for conservation of mass
More informationAxis Balanced Forces Centripetal force. Change in velocity Circular Motion Circular orbit Collision. Conservation of Energy
When something changes its velocity The rate of change of velocity of a moving object. Can result from a change in speed and/or a change in direction On surface of earth, value is 9.8 ms-²; increases nearer
More informationPhysics for Scientists and Engineers. Chapter 5 Force and Motion
Physics for Scientists and Engineers Chapter 5 Force and Motion Spring, 2008 Ho Jung Paik Force Forces are what cause any change in the velocity of an object The net force is the vector sum of all the
More informationChapter Four fluid flow mass, energy, Bernoulli and momentum
4-1Conservation of Mass Principle Consider a control volume of arbitrary shape, as shown in Fig (4-1). Figure (4-1): the differential control volume and differential control volume (Total mass entering
More informationRutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 19. Home Page. Title Page. Page 1 of 36.
Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 19 Page 1 of 36 12. Equilibrium and Elasticity How do objects behave under applied external forces? Under
More information4. Find the average velocities and average accelerations of a particle moving in 1-D given its position at various times.
PHYSICS 201: TEST 1 STUDY SHEET 1. Convert a quantity from one set of units to another set of units. 2. Convert a 2-D vector from rectangular form (components) to polar form (magnitude and angle), or from
More informationPLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)
PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when
More information8.012 Physics I: Classical Mechanics Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 8.012 Physics I: Classical Mechanics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE
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 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 informationVideo 2.1a Vijay Kumar and Ani Hsieh
Video 2.1a Vijay Kumar and Ani Hsieh Robo3x-1.3 1 Introduction to Lagrangian Mechanics Vijay Kumar and Ani Hsieh University of Pennsylvania Robo3x-1.3 2 Analytical Mechanics Aristotle Galileo Bernoulli
More informationGame Physics. Game and Media Technology Master Program - Utrecht University. Dr. Nicolas Pronost
Game and Media Technology Master Program - Utrecht University Dr. Nicolas Pronost Essential physics for game developers Introduction The primary issues Let s move virtual objects Kinematics: description
More informationChapter 6: Momentum Analysis of Flow Systems
Chapter 6: Momentum Analysis of Flow Systems Introduction Fluid flow problems can be analyzed using one of three basic approaches: differential, experimental, and integral (or control volume). In Chap.
More informationLecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation
Lecture 4 Differential Analysis of Fluid Flow Navier-Stockes equation Newton second law and conservation of momentum & momentum-of-momentum A jet of fluid deflected by an object puts a force on the object.
More informationENGR 292 Fluids and Thermodynamics
ENGR 292 Fluids and Thermodynamics Scott Li, Ph.D., P.Eng. Mechanical Engineering Technology Camosun College Timeline Last week, Reading Break Feb.21: Thermodynamics 1 Feb.24: Midterm Review (Fluid Statics
More informationCEE 3310 Control Volume Analysis, Oct. 10, = dt. sys
CEE 3310 Control Volume Analysis, Oct. 10, 2018 77 3.16 Review First Law of Thermodynamics ( ) de = dt Q Ẇ sys Sign convention: Work done by the surroundings on the system < 0, example, a pump! Work done
More informationDEVIL PHYSICS BADDEST CLASS ON CAMPUS IB PHYSICS
DEVIL PHYSICS BADDEST CLASS ON CAMPUS IB PHYSICS OPTION B-1A: ROTATIONAL DYNAMICS Essential Idea: The basic laws of mechanics have an extension when equivalent principles are applied to rotation. Actual
More informationPh1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004
Ph1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004 Problem 1 (10 points) - The Delivery A crate of mass M, which contains an expensive piece of scientific equipment, is being delivered to Caltech.
More informationwhere = rate of change of total energy of the system, = rate of heat added to the system, = rate of work done by the system
The Energy Equation for Control Volumes Recall, the First Law of Thermodynamics: where = rate of change of total energy of the system, = rate of heat added to the system, = rate of work done by the system
More informationFluid Dynamics Exercises and questions for the course
Fluid Dynamics Exercises and questions for the course January 15, 2014 A two dimensional flow field characterised by the following velocity components in polar coordinates is called a free vortex: u r
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 informationConservation of Angular Momentum
10 March 2017 Conservation of ngular Momentum Lecture 23 In the last class, we discussed about the conservation of angular momentum principle. Using RTT, the angular momentum principle was given as DHo
More informationProject TOUCAN. A Study of a Two-Can System. Prof. R.G. Longoria Update Fall ME 144L Prof. R.G. Longoria Dynamic Systems and Controls Laboratory
Project TOUCAN A Study of a Two-Can System Prof. R.G. Longoria Update Fall 2009 Laboratory Goals Gain familiarity with building models that reflect reality. Show how a model can be used to guide physical
More informationOscillatory Motion SHM
Chapter 15 Oscillatory Motion SHM Dr. Armen Kocharian Periodic Motion Periodic motion is motion of an object that regularly repeats The object returns to a given position after a fixed time interval A
More information20 Torque & Circular Motion
Chapter 0 Torque & Circular Motion 0 Torque & Circular Motion The mistake that crops up in the application of Newton s nd Law for Rotational Motion involves the replacement of the sum of the torques about
More informationIntroduction to Aerodynamics. Dr. Guven Aerospace Engineer (P.hD)
Introduction to Aerodynamics Dr. Guven Aerospace Engineer (P.hD) Aerodynamic Forces All aerodynamic forces are generated wither through pressure distribution or a shear stress distribution on a body. The
More informationLecture PowerPoints. Chapter 4 Physics: Principles with Applications, 6 th edition Giancoli
Lecture PowerPoints Chapter 4 Physics: Principles with Applications, 6 th edition Giancoli 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the
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 informationArtificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik. Robot Dynamics. Dr.-Ing. John Nassour J.
Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik Robot Dynamics Dr.-Ing. John Nassour 25.1.218 J.Nassour 1 Introduction Dynamics concerns the motion of bodies Includes Kinematics
More informationPhysics for Scientists and Engineers 4th Edition, 2017
A Correlation of Physics for Scientists and Engineers 4th Edition, 2017 To the AP Physics C: Mechanics Course Descriptions AP is a trademark registered and/or owned by the College Board, which was not
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 informationRotational Kinematics and Dynamics. UCVTS AIT Physics
Rotational Kinematics and Dynamics UCVTS AIT Physics Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position,
More informationGeneral Definition of Torque, final. Lever Arm. General Definition of Torque 7/29/2010. Units of Chapter 10
Units of Chapter 10 Determining Moments of Inertia Rotational Kinetic Energy Rotational Plus Translational Motion; Rolling Why Does a Rolling Sphere Slow Down? General Definition of Torque, final Taking
More informationChapter 3. Shallow Water Equations and the Ocean. 3.1 Derivation of shallow water equations
Chapter 3 Shallow Water Equations and the Ocean Over most of the globe the ocean has a rather distinctive vertical structure, with an upper layer ranging from 20 m to 200 m in thickness, consisting of
More informationCLASS Fourth Units (Second part)
CLASS Fourth Units (Second part) Energy analysis of closed systems Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. MOVING BOUNDARY WORK Moving boundary work (P
More informationMechanics. In the Science Program, Mechanics contributes to the following program goals described in the Exit Profile:
Mechanics Objectives: 00UR Discipline: Physics Ponderation: 3-2-3 Course Code: 203-NYA-05 Prerequisite: Sec. V Physics 534, Mathematics 536 (or equivalent) Course Credit: 2 2/3 Corequisite: 00UP (Calculus
More informationvector H. If O is the point about which moments are desired, the angular moment about O is given:
The angular momentum A control volume analysis can be applied to the angular momentum, by letting B equal to angularmomentum vector H. If O is the point about which moments are desired, the angular moment
More informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 11 Kinematics of Fluid Part - II Good afternoon, I welcome you all to this session
More informationChapter 5 Newton s Laws of Motion
Chapter 5 Newton s Laws of Motion Newtonian Mechanics Mass Mass is an intrinsic characteristic of a body The mass of a body is the characteristic that relates a force on the body to the resulting acceleration.
More informationd v 2 v = d v d t i n where "in" and "rot" denote the inertial (absolute) and rotating frames. Equation of motion F =
Governing equations of fluid dynamics under the influence of Earth rotation (Navier-Stokes Equations in rotating frame) Recap: From kinematic consideration, d v i n d t i n = d v rot d t r o t 2 v rot
More informationCenter of Gravity. The location of the center of gravity is defined by: n mgx. APSC 111 Review Page 7
Center of Gravity We have said that for rigid bodies, all of the forces act at the centre of mass. This is a normally a very good approximation, but strictly speaking, the forces act at the centre of gravity,
More informationSymmetries 2 - Rotations in Space
Symmetries 2 - Rotations in Space This symmetry is about the isotropy of space, i.e. space is the same in all orientations. Thus, if we continuously rotated an entire system in space, we expect the system
More informationRotational Dynamics continued
Chapter 9 Rotational Dynamics continued 9.1 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 informationp = mv L = Iω L =! r x p or, if we use translational parameters:
ANGULAR MOMENTUM Torque is the rotational counterpart to force. So whereas when a net force is applied, a body accelerates, when a net torque is applied, a body angularly accelerates. Angular momentum
More information