DOCSLIB.ORG

Formulas for Motorized Linear Motion Systems

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Formulas for Motorized Linear Motion Systems FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS Haydon Kerk Motion Solutions : 203 756 7441 Pittman Motors : 267 933 2105 www.HaydonKerkPittman.com 2 Symbols andSYMBOLS Units AND UNITS Symbol Description Units Symbol Description Units a linear acceleration m/s2 P power W 2 a angular acceleration rad/s Pavg average power W 2 ain angular acceleration rad/s PCV power required at constant velocity W angular acceleration 2 power required at system input aout rad/s Pin W α temperature coefficient /°C Ploss dissipated power W DV viscous damping factor Nm s/rad Pmax maximum power W F force N Pmax(f) maximum power (final“hot”) W Fa linear force required during acceleration (FJ +Ff +Fg) N Pmax(i) maximum power (initial “cold”) W Ff linear force required to overcome friction N Pout output power W Fg linear force required to overcome gravity N PPK peak power W FJ linear force required to overcome load inertia N Rm motor regulation RPM/Nm 2 2 g gravitational constant (9.8 m/s ) m/s Rmt motor terminal resistance Ω I current A Rmt(f) motor terminal resistance (final“hot”) Ω ILR locked rotor current A Rmt(i) motor terminal resistance (initial “cold”) Ω IO no-load current A Rth thermal resistance °C/W IPK peak current A r radius m IRMS RMS current A S linear distance m J inertia Kg-m2 t time s inertia of belt Kg-m2 torque Nm JB T inertia of the gear box Kg-m2 T torque required to overcome load inertia Nm JGB a reflected inertia at system input Kg-m2 T torque required at motor shaft during acceleration Jin a (motor) Nm 2 continuous rated motor torque Jout reflected inertia at system output Kg-m TC Nm inertia of pully 1 Kg-m2 coulomb friction torque Nm JP1 TCF inertia of pully 2 Kg-m2 reverse torque required for deceleration JP2 Td Nm lead screw inertia Kg-m2 drag / preload torque Nm JS TD voltage constant torque required to overcome friction KE V/rad/s Tf Nm T torque required to overcome gravity Nm K(f) KE or KT (final“hot”) Nm/A or V/(rad/s) g torque required at system input K(i) KE or KT (initial “cold”) Nm/A or V/(rad/s) Tin Nm motor constant locked rotor torque Km Nm/ w TLR Nm KT torque constant Nm/A Tout torque required at system output Nm L screw lead m/rev TRMS RMS torque required over the total duty cycle Nm m mass Kg TRMS(motor) RMS torque required at the motor shaft Nm N gear ratio n/a μ coefficient of friction n/a η efficiency n/a VT motor terminal voltage V n speed RPM Vbus DC drive bus voltage V no load speed nO RPM v linear velocity m/s peak speed final linear velocity m/s nPK RPM vf θ load orientation (horizontal = 0°, vertical = 90°) degrees vi initial linear velocity m/s °C peak linear velocity m/s Θa ambient temperature vPK Θf motor temperature (final“hot”) °C W energy J angular velocity Θi motor temperature (initial “cold”) °C ω rad/s Θm motor temperature °C ωin angular velocity at system input rad/s Θr motor temperature rise °C ωout angular velocity at system output rad/s Θrated rated motor temperature °C ωo no load angular velocity rad/s ωPK peak angular velocity rad/s 1 www.HaydonKerkPittman.com Haydon Kerk Motion Solutions: 203 756 7441/Pittman Motors: 267 933 2105 MOTION SYSTEM 3 Motion System FormulasFORMULAS Any motion system should be broken down into its individual components. Power Controller / Mechanical Lead Screw / Supply Drive Motor Transmission Support Rails + + F x S Watts t = Out (V)(I) I I Watts (V)( ) Watts (V)( ) Watts (T)(ω) Watts (T)(ω) Watts Watts Lost Watts Lost Watts Lost Watts Lost Watts Lost All analysis begins with the linear motion profile of the output and the force required to move the load. Motion Profiles 1/3 1/3 1/3 Trapazoidal Motion Profile vpk - v a a t 1 t 2 t 3 t dwell Time t move Total Cycle Time Optimized for minimum power t 1 = t 2= t 3 3S S is the total move distance vPK = t is the total move time 2t Area under profile curve represents distance moved www.HaydonKerkPittman.com 2 Haydon Kerk Motion Solutions: 203 756 7441/Pittman Motors: 267 933 2105 MOTION SYSTEM 4 Motion System FormulasFORMULAS Triangular Motion Profiles vPK a -a Optimized for minimum acceleration slope v t 1 = t 2 t 1 t 2 Time t dwell t 2S move = Total Cycle Time vPK t is the total move distance S Move Profile Peak Velocity (VPK) Acceleration (a ) t is the total move time 1/3 1/3 1/3 Area under profile curve represents distance moved Triangular Complex Motion Profile v2 2 a - v1 a 3 v a 1 t 1 t 2 t 3 t 4 t 5 t dwell Time t move Total Cycle Time Linear Motion Formulas Linear to Rotary Formulas through a Lead Screw vf - vi (1) s = –––– t 2 2 π a 2 a = = rad/s (2) vf = vi + a t L 1 2 (3) s = t – at 2 vi + 2 πv ω = = rad/s 2 2 L (4) 2 as = vf - vi ( v - vi) 60 v f = Δv = = RPM (5) a = n ( tf - ti ) Δt L 3 known quantities are needed to solve for the other 2. Each segment calculated individually. 3 www.HaydonKerkPittman.com Haydon Kerk Motion Solutions: 203 756 7441/Pittman Motors: 267 933 2105 MOTION SYSTEM 5 Motion System FormulasFORMULAS Inertia, Acceleration, Velocity, and Required Torque Lead Screw System 2 L 1 J 2 Jin = m X η + S Kg-m 2 π 2 2 πa rad/sec in = a L 2 π v rad/sec ωin = Jin mout L ain a out Τin= Τa+Τf +Τg+ΤD N-m N-m ωin vout Τa= Jin ain Tin Fout cosØmgμL Τf = N-m 2 π η sinØmgL Τg = N-m 2 π η ΤD = drag/preload N-m refer to manufacturer’s data Belt and Pulley System Jout 1 2 Jin = X + JP1+ JP2+ JB Kg-m N 2 η 2 = rad/sec ain (aout)(N) ωin = ωout N rad/sec Jin Jout ( )( ) Τout 1 ain aout Τ = X N-m in N η ωin ωout For a 1st approximation analysis, JP1, JP2, and JB can be disregarded. For higher precision, pulley and Tin Tout belt inertias should be included, as well as the reflected inertia of these components. www.HaydonKerkPittman.com 4 Haydon Kerk Motion Solutions: 203 756 7441/Pittman Motors: 267 933 2105 MOTION SYSTEM 6 Motion System FormulasFORMULAS Gearbox System Jout 1 2 J = X J Kg-m in 2 η + GB N 2 = rad/sec ain (aout)(N) ωin = ωout N rad/sec Jin Jout ( )( ) Τout 1 ain aout Τ = X ΤD N-m in N η + ωin ωout Drag torque can be significant (TD) depending on the Tin Tout viscosity of the lubricant. For a first approximation analysis, this can be left out. Gearbox inertia may be found in manufacturer’s data sheets, or calculated using gear dimensions and materials / mass. Mechanical Power and RMS Torque FS Linear Power = t = watts Rotary Power = Tω = watts 2 2 2 2 ΤRMS = Τ1 t 1 + Τ2 t2 + Τ3 t 3 ...+ Τn t n t 1 + t 2 + t 3 ... + t n + tdwell 5 www.HaydonKerkPittman.com Haydon Kerk Motion Solutions: 203 756 7441/Pittman Motors: 267 933 2105 MOTION SYSTEM 7 Motion System FormulasFORMULAS DC Motor Formulas Motor No Load Speed VT – (IO x Rmt) no = 9.5493 x KE Assuming IO is very small, it can be ignored for a quick approximation of motor no-load speed. If more accurate results are needed, IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv). As the no-load speed increases with increased terminal voltage on a given motor, IO will also increase based on the motor’s Dv value found on the manufacturer’s motor data sheet. Τ Viscous Damping – – – Torque (Dv) Breakaway Torque – – – – – – – – – – – – – Coulomb Friction – – – ο ω Motor Locked Rotor Current / Locked Rotor Torque VT ILR = Rmt TLR = ILR x KT VT = x KT Rmt Motor Regulation – Theoretical using motor constants Rmt Rm = 9.5493 x KE x KT – Regulation calculated using test data nO Rm = TLR www.HaydonKerkPittman.com 6 Haydon Kerk Motion Solutions: 203 756 7441/Pittman Motors: 267 933 2105 MOTION SYSTEM 8 Motion System FormulasFORMULAS Motor Power Relationships – Watts lost due to winding resistance 2 Ploss = I x Rmt – Output power at any point on the motor curve Pout = ω x T – Motor maximum output power Pmax = 0.25 x ωo x TLR – Motor maximum output power (Theoretical) 2 VT Pmax = 0.25 x Rmt Temperature Effects on Motor Constants – Change in terminal resistance Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf – Θi) ] αconductor = 0.0040/°C (copper) In the case of a mechanically commutated motor with graphite brushes, this analysis will result in a slight error because of the negative temperature coefficient of carbon. For estimation purposes, this can be ignored, but for a more rigorous analysis, the behavior of the brush material must be taken into account. This is not an issue when evaluating a brushless dc motor. – Change in torque constant and voltage constant ( KT = KE ) K(f) = K(i) x [ 1 + αmagnet (Θf – Θi) ] Magnetic Material αmagnet (/°C) Θmax (°C) Ceramic -0.0020/°C 300°C SmCo -0.0004/°C 300°C AlNiCo -0.0002/°C 540°C NdFeB -0.0012/°C 150°C The magnetic material values above represent average figures for particular material classes and estimating performance.
Load more

Recommended publications
  • Rotational Motion (The Dynamics of a Rigid Body)
    University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body) Henry Semat City College of New York Robert Katz University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/physicskatz Part of the Physics Commons Semat, Henry and Katz, Robert, "Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body)" (1958). Robert Katz Publications. 141. https://digitalcommons.unl.edu/physicskatz/141 This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 11 Rotational Motion (The Dynamics of a Rigid Body) 11-1 Motion about a Fixed Axis The motion of the flywheel of an engine and of a pulley on its axle are examples of an important type of motion of a rigid body, that of the motion of rotation about a fixed axis. Consider the motion of a uniform disk rotat­ ing about a fixed axis passing through its center of gravity C perpendicular to the face of the disk, as shown in Figure 11-1. The motion of this disk may be de­ scribed in terms of the motions of each of its individual particles, but a better way to describe the motion is in terms of the angle through which the disk rotates.
    [Show full text]
  • Kinematics Study of Motion
    Kinematics Study of motion Kinematics is the branch of physics that describes the motion of objects, but it is not interested in its causes. Itziar Izurieta (2018 october) Index: 1. What is motion? ............................................................................................ 1 1.1. Relativity of motion ................................................................................................................................ 1 1.2.Frame of reference: Cartesian coordinate system ....................................................................................................................................................................... 1 1.3. Position and trajectory .......................................................................................................................... 2 1.4.Travelled distance and displacement ....................................................................................................................................................................... 3 2. Quantities of motion: Speed and velocity .............................................. 4 2.1. Average and instantaneous speed ............................................................ 4 2.2. Average and instantaneous velocity ........................................................ 7 3. Uniform linear motion ................................................................................. 9 3.1. Distance-time graph .................................................................................. 10 3.2. Velocity-time
    [Show full text]
  • Introduction to Robotics Lecture Note 5: Velocity of a Rigid Body
    ECE5463: Introduction to Robotics Lecture Note 5: Velocity of a Rigid Body Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018 Lecture 5 (ECE5463 Sp18) Wei Zhang(OSU) 1 / 24 Outline Introduction • Rotational Velocity • Change of Reference Frame for Twist (Adjoint Map) • Rigid Body Velocity • Outline Lecture 5 (ECE5463 Sp18) Wei Zhang(OSU) 2 / 24 Introduction For a moving particle with coordinate p(t) 3 at time t, its (linear) velocity • R is simply p_(t) 2 A moving rigid body consists of infinitely many particles, all of which may • have different velocities. What is the velocity of the rigid body? Let T (t) represent the configuration of a moving rigid body at time t.A • point p on the rigid body with (homogeneous) coordinate p~b(t) and p~s(t) in body and space frames: p~ (t) p~ ; p~ (t) = T (t)~p b ≡ b s b Introduction Lecture 5 (ECE5463 Sp18) Wei Zhang(OSU) 3 / 24 Introduction Velocity of p is d p~ (t) = T_ (t)p • dt s b T_ (t) is not a good representation of the velocity of rigid body • - There can be 12 nonzero entries for T_ . - May change over time even when the body is under a constant velocity motion (constant rotation + constant linear motion) Our goal is to find effective ways to represent the rigid body velocity. • Introduction Lecture 5 (ECE5463 Sp18) Wei Zhang(OSU) 4 / 24 Outline Introduction • Rotational Velocity • Change of Reference Frame for Twist (Adjoint Map) • Rigid Body Velocity • Rotational Velocity Lecture 5 (ECE5463 Sp18) Wei Zhang(OSU) 5 / 24 Illustrating
    [Show full text]
  • Linear and Angular Velocity Examples
    Linear and Angular Velocity Examples Example 1 Determine the angular displacement in radians of 6.5 revolutions. Round to the nearest tenth. Each revolution equals 2 radians. For 6.5 revolutions, the number of radians is 6.5 2 or 13 . 13 radians equals about 40.8 radians. Example 2 Determine the angular velocity if 4.8 revolutions are completed in 4 seconds. Round to the nearest tenth. The angular displacement is 4.8 2 or 9.6 radians. = t 9.6 = = 9.6 , t = 4 4 7.539822369 Use a calculator. The angular velocity is about 7.5 radians per second. Example 3 AMUSEMENT PARK Jack climbs on a horse that is 12 feet from the center of a merry-go-round. 1 The merry-go-round makes 3 rotations per minute. Determine Jack’s angular velocity in radians 4 per second. Round to the nearest hundredth. 1 The merry-go-round makes 3 or 3.25 revolutions per minute. Convert revolutions per minute to radians 4 per second. 3.25 revolutions 1 minute 2 radians 0.3403392041 radian per second 1 minute 60 seconds 1 revolution Jack has an angular velocity of about 0.34 radian per second. Example 4 Determine the linear velocity of a point rotating at an angular velocity of 12 radians per second at a distance of 8 centimeters from the center of the rotating object. Round to the nearest tenth. v = r v = (8)(12 ) r = 8, = 12 v 301.5928947 Use a calculator. The linear velocity is about 301.6 centimeters per second.
    [Show full text]
  • Guide for the Use of the International System of Units (SI)
    Guide for the Use of the International System of Units (SI) m kg s cd SI mol K A NIST Special Publication 811 2008 Edition Ambler Thompson and Barry N. Taylor NIST Special Publication 811 2008 Edition Guide for the Use of the International System of Units (SI) Ambler Thompson Technology Services and Barry N. Taylor Physics Laboratory National Institute of Standards and Technology Gaithersburg, MD 20899 (Supersedes NIST Special Publication 811, 1995 Edition, April 1995) March 2008 U.S. Department of Commerce Carlos M. Gutierrez, Secretary National Institute of Standards and Technology James M. Turner, Acting Director National Institute of Standards and Technology Special Publication 811, 2008 Edition (Supersedes NIST Special Publication 811, April 1995 Edition) Natl. Inst. Stand. Technol. Spec. Publ. 811, 2008 Ed., 85 pages (March 2008; 2nd printing November 2008) CODEN: NSPUE3 Note on 2nd printing: This 2nd printing dated November 2008 of NIST SP811 corrects a number of minor typographical errors present in the 1st printing dated March 2008. Guide for the Use of the International System of Units (SI) Preface The International System of Units, universally abbreviated SI (from the French Le Système International d’Unités), is the modern metric system of measurement. Long the dominant measurement system used in science, the SI is becoming the dominant measurement system used in international commerce. The Omnibus Trade and Competitiveness Act of August 1988 [Public Law (PL) 100-418] changed the name of the National Bureau of Standards (NBS) to the National Institute of Standards and Technology (NIST) and gave to NIST the added task of helping U.S.
    [Show full text]
  • Unit 1: Motion
    Macomb Intermediate School District High School Science Power Standards Document Physics The Michigan High School Science Content Expectations establish what every student is expected to know and be able to do by the end of high school. They also outline the parameters for receiving high school credit as dictated by state law. To aid teachers and administrators in meeting these expectations the Macomb ISD has undertaken the task of identifying those content expectations which can be considered power standards. The critical characteristics1 for selecting a power standard are: • Endurance – knowledge and skills of value beyond a single test date. • Leverage - knowledge and skills that will be of value in multiple disciplines. • Readiness - knowledge and skills necessary for the next level of learning. The selection of power standards is not intended to relieve teachers of the responsibility for teaching all content expectations. Rather, it gives the school district a common focus and acts as a safety net of standards that all students must learn prior to leaving their current level. The following document utilizes the unit design including the big ideas and real world contexts, as developed in the science companion documents for the Michigan High School Science Content Expectations. 1 Dr. Douglas Reeves, Center for Performance Assessment Unit 1: Motion Big Ideas The motion of an object may be described using a) motion diagrams, b) data, c) graphs, and d) mathematical functions. Conceptual Understandings A comparison can be made of the motion of a person attempting to walk at a constant velocity down a sidewalk to the motion of a person attempting to walk in a straight line with a constant acceleration.
    [Show full text]
  • TECHNICAL Nbte D-626
    NASA TN D-626 LIBRARY COPY NOV 16 1960 SPACE FLIGHT LANGLEY FIELD, VIRGINIA t TECHNICAL NbTE D-626 SATELLITE ATTDUDE CONTROL USING A COMBINATION OF INERTIA WHEELS AND A BAR MAGNET By James J. Adams and Roy F. Brissenden Langley Research Center Langley Field, Va. NATIONAL AERONAUTICS AND SPACE ADMINISTRATION WASHINGTON November 1960 . NATIONAL AERONAUTICS AND SPACE ADMINISTRATION TECHNICAL NOTE D-626 SATELLITE ATTITUDE CONTROL USING A COMBINATION OF INERTIA WHEELS AND A BAR MAGNET By James J. Adams and Roy F. Brissenden SUMMARY The use of inertia wheels to control the attitude of a satellite has currently aroused much interest. The stability of such a system has been studied in this investigation. A single-degree-of-freedom analysis indicates that a response with suitable dynamic characteristics and pre- cise control can be achieved by commanding the angular velocity of the inertia wheel with an error signal that is the sum of the attitude error, the attitude rate, and the integral of the attitude error. A digital computer was used to study the three-degree-of-freedom response to step displacements, and the results indicate that the cross-coupling effects of inertia coupling and precession coupling had no effect on system sta- bility. A study was also made of the use of a bar magnet to supplement the inertia wheels by providing a means of removing any momentum intro- duced into the system by disturbances such as aerodynamic torques. A study of a case with large aerodynamic torques, with a typical orbit, indicated that the magnet was a suitable device for supplying the essen- tial trimming force.
    [Show full text]
  • Newton Euler Equations of Motion Examples
    Newton Euler Equations Of Motion Examples Alto and onymous Antonino often interloping some obligations excursively or outstrikes sunward. Pasteboard and Sarmatia Kincaid never flits his redwood! Potatory and larboard Leighton never roller-skating otherwhile when Trip notarizes his counterproofs. Velocity thus resulting in the tumbling motion of rigid bodies. Equations of motion Euler-Lagrange Newton-Euler Equations of motion. Motion of examples of experiments that a random walker uses cookies. Forces by each other two examples of example are second kind, we will refer to specify any parameter in. 213 Translational and Rotational Equations of Motion. Robotics Lecture Dynamics. Independence from a thorough description and angular velocity as expected or tofollowa userdefined behaviour does it only be loaded geometry in an appropriate cuts in. An interface to derive a particular instance: divide and author provides a positive moment is to express to output side can be run at all previous step. The analysis of rotational motions which make necessary to decide whether rotations are. For xddot and whatnot in which a very much easier in which together or arena where to use them in two backwards operation complies with respect to rotations. Which influence of examples are true, is due to independent coordinates. On sameor adjacent joints at each moment equation is also be more specific white ellipses represent rotations are unconditionally stable, for motion break down direction. Unit quaternions or Euler parameters are known to be well suited for the. The angular momentum and time and runnable python code. The example will be run physics examples are models can be symbolic generator runs faster rotation kinetic energy.
    [Show full text]
  • Rotation: Moment of Inertia and Torque
    Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. This tutorial discusses the dynamics of an object rotating about a fixed axis and introduces the concepts of torque and moment of inertia. These concepts allows us to get a better understanding of why pushing a door towards its hinges is not very a very effective way to make it open, why using a longer wrench makes it easier to loosen a tight bolt, etc. This module begins by looking at the kinetic energy of rotation and by defining a quantity known as the moment of inertia which is the rotational analog of mass. Then it proceeds to discuss the quantity called torque which is the rotational analog of force and is the physical quantity that is required to changed an object's state of rotational motion. Moment of Inertia Kinetic Energy of Rotation Consider a rigid object rotating about a fixed axis at a certain angular velocity. Since every particle in the object is moving, every particle has kinetic energy. To find the total kinetic energy related to the rotation of the body, the sum of the kinetic energy of every particle due to the rotational motion is taken. The total kinetic energy can be expressed as ..
    [Show full text]
  • Lecture 24 Angular Momentum
    LECTURE 24 ANGULAR MOMENTUM Instructor: Kazumi Tolich Lecture 24 2 ¨ Reading chapter 11-6 ¤ Angular momentum n Angular momentum about an axis n Newton’s 2nd law for rotational motion Angular momentum of an rotating object 3 ¨ An object with a moment of inertia of � about an axis rotates with an angular speed of � about the same axis has an angular momentum, �, given by � = �� ¤ This is analogous to linear momentum: � = �� Angular momentum in general 4 ¨ Angular momentum of a point particle about an axis is defined by � � = �� sin � = ��� sin � = �-� = ��. � �- ¤ �⃗: position vector for the particle from the axis. ¤ �: linear momentum of the particle: � = �� �⃗ ¤ � is moment arm, or sometimes called “perpendicular . Axis distance.” �. Quiz: 1 5 ¨ A particle is traveling in straight line path as shown in Case A and Case B. In which case(s) does the blue particle have non-zero angular momentum about the axis indicated by the red cross? A. Only Case A Case A B. Only Case B C. Neither D. Both Case B Quiz: 24-1 answer 6 ¨ Only Case A ¨ For a particle to have angular momentum about an axis, it does not have to be Case A moving in a circle. ¨ The particle can be moving in a straight path. Case B ¨ For it to have a non-zero angular momentum, its line of path is displaced from the axis about which the angular momentum is calculated. ¨ An object moving in a straight line that does not go through the axis of rotation has an angular position that changes with time. So, this object has an angular momentum.
    [Show full text]
  • Rotational Motion and Angular Momentum 317
    CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 317 10 ROTATIONAL MOTION AND ANGULAR MOMENTUM Figure 10.1 The mention of a tornado conjures up images of raw destructive power. Tornadoes blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing winds as high as 500 km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Learning Objectives 10.1. Angular Acceleration • Describe uniform circular motion. • Explain non-uniform circular motion. • Calculate angular acceleration of an object. • Observe the link between linear and angular acceleration. 10.2. Kinematics of Rotational Motion • Observe the kinematics of rotational motion. • Derive rotational kinematic equations. • Evaluate problem solving strategies for rotational kinematics. 10.3. Dynamics of Rotational Motion: Rotational Inertia • Understand the relationship between force, mass and acceleration. • Study the turning effect of force. • Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. 10.4. Rotational Kinetic Energy: Work and Energy Revisited • Derive the equation for rotational work. • Calculate rotational kinetic energy. • Demonstrate the Law of Conservation of Energy. 10.5. Angular Momentum and Its Conservation • Understand the analogy between angular momentum and linear momentum. • Observe the relationship between torque and angular momentum. • Apply the law of conservation of angular momentum. 10.6. Collisions of Extended Bodies in Two Dimensions • Observe collisions of extended bodies in two dimensions. • Examine collision at the point of percussion.
    [Show full text]
  • Shape Rotational Inertia, I I = M1r + M2r + ... (Point Masses) I = ∫ R Dm
    Chs. 10 - Rotation & Angular Motion Dan Finkenstadt, fi[email protected] October 26, 2015 Shape Rotational Inertia, I I one aspect of shape is the center of mass I How difficult is it to spin an object about (c.o.m.) any axis? I another is how mass is distributed about the I Formulas: (distance r from axis) c.o.m. 2 2 I = m1r1 + m2r2 + ::: I an important parameter describing shape is called, (point masses) 1. Rotational Inertia (in our book) Z 2. Moment of Inertia (in most other books) I = r2 dm (solid object) 3.2 nd Moment of Mass (in engineering, math books) Solution: 8 kg(2 m)2 + 5 kg(8 m2) + 6 kg(2 m)2 = 96 kg · m2 P 2 Rotational Inertia: I = miri Two you should memorize: Let's calculate one of these: I mass M concentrated at radius R I = MR2 I Disk of mass M, radius R, about center 1 I = MR2 solid disk 2 Parallel Axis Theorem Active Learning Exercise Problem: Rotational Inertia of Point We can calculate the rotational inertia about an Masses axis parallel to the c.o.m. axis Four point masses are located at the corners of a square with sides 2.0 m, 2 I = Icom + m × shift as shown. The 1.0 kg mass is at the origin. What is the rotational inertia of the four-mass system about an I Where M is the total mass of the body axis passing through the mass at the origin and pointing out of the screen I And shift is the distance between the axes (page)? I REMEMBER: The axes must be parallel! What is so useful about I? Rotation I radians! It connects up linear quantities to angular I right-hand rule quantities, e.g., kinetic energy angular speed v ! = rad r s I v = r! 2 2 I ac = r! 1 2 1 2v 1 2 mv ! mr ! I! ang.
    [Show full text]