Chapter 18 KINETICS OF RIGID BODIES IN THREE DIMENSIONS. The two fundamental equations for the motion of a system of particles .

Similar documents
Dynamics and control of mechanical systems

SOLUTION (y 2 + z 2 )dm + (x 2 + y 2 )dm. (x 2 + z 2 )dm + I xx + I yy + I zz = = 2. (x 2 + y 2 + z 2 )dm

Moving Reference Frame Kinematics Homework

Equilibrium of Rigid Bodies

Solution 11. Kinetics of rigid body(newton s Second Law)

Infinitesimal Rotations

Contents. Dynamics and control of mechanical systems. Focuses on

KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS

10. The dimensional formula for c) 6% d) 7%

Rigid bodies - general theory

I xx + I yy + I zz = (y 2 + z 2 )dm + (x 2 + y 2 )dm. (x 2 + z 2 )dm + (x 2 + y 2 + z 2 )dm = 2

Lecture 38: Equations of Rigid-Body Motion

Force Couple Systems = Replacement of a Force with an Equivalent Force and Moment (Moving a Force to Another Point)

Lecture 41: Highlights

Problem 1. Mathematics of rotations

Plane Motion of Rigid Bodies: Momentum Methods

Lecture 38: Equations of Rigid-Body Motion

1/30. Rigid Body Rotations. Dave Frank

PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)

Review before final exam. Guide how to identify type of the problem

Classical Mechanics III (8.09) Fall 2014 Assignment 3

PHY4116 From Newton to Einstein. Mid-Term Test, 10a.m. Wed. 16 th Nov Duration: 50 minutes. There are 25 marks in Section A and 25 in Section B.

CH. 1 FUNDAMENTAL PRINCIPLES OF MECHANICS

Integration of the equation of motion with respect to time rather than displacement leads to the equations of impulse and momentum.

LAWS OF GYROSCOPES / CARDANIC GYROSCOPE

2. Supports which resist forces in two directions. Fig Hinge. Rough Surface. Fig Rocker. Roller. Frictionless Surface

27. Euler s Equations

Pan Pearl River Delta Physics Olympiad 2005

Fluid Mechanics II. Newton s second law applied to a control volume

Chapter 11. Angular Momentum

Advanced Dynamics. - Lecture 1 Constraints. Paolo Tiso Spring Semester 2017 ETH Zürich

ME 321: FLUID MECHANICS-I

Kinetic Energy of Rolling

StaticS Fourteenth edition in si units

ENGI 4430 Advanced Calculus for Engineering Faculty of Engineering and Applied Science Problem Set 3 Solutions [Multiple Integration; Lines of Force]

Torque and Rotation Lecture 7

COIN TOSS MODELING RĂZVAN C. STEFAN, TIBERIUS O. CHECHE

Note on Mathematical Development of Plate Theories

Angular Momentum. Objectives CONSERVATION OF ANGULAR MOMENTUM

Moving Reference Frame Kinematics Homework

Lecture 9 Kinetics of rigid bodies: Impulse and Momentum

Laws of gyroscopes / cardanic gyroscope

Lecture 37: Principal Axes, Translations, and Eulerian Angles

Rigid Body Dynamics, SG2150 Solutions to Exam,

Mechanics: Scalars and Vectors

Video 2.1a Vijay Kumar and Ani Hsieh

PHY 5246: Theoretical Dynamics, Fall Assignment # 10, Solutions. (1.a) N = a. we see that a m ar a = 0 and so N = 0. ω 3 ω 2 = 0 ω 2 + I 1 I 3

Mechanics Departmental Exam Last updated November 2013

APPLIED MECHANICS I Resultant of Concurrent Forces Consider a body acted upon by co-planar forces as shown in Fig 1.1(a).

III. Work and Energy

3-D Kinetics of Rigid Bodies

Chapter 5 Equilibrium of a Rigid Body Objectives

Chapter 8 Rotational Equilibrium and Rotational Dynamics Force vs. Torque Forces cause accelerations Torques cause angular accelerations Force and

Physics 106a, Caltech 4 December, Lecture 18: Examples on Rigid Body Dynamics. Rotating rectangle. Heavy symmetric top

Physics 312, Winter 2007, Practice Final

UNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 20, 2011 Professor A. Dolovich

AS3010: Introduction to Space Technology

6. 3D Kinematics DE2-EA 2.1: M4DE. Dr Connor Myant

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS AP PHYSICS

5. Plane Kinetics of Rigid Bodies

PHY 5246: Theoretical Dynamics, Fall Assignment # 9, Solutions. y CM (θ = 0) = 2 ρ m

PROBLEM Area of Problem I = 471,040 mm xy 2(471,040) 252,757 1,752,789 = or θ m = and = (1,002,773 ± 885,665) mm

Classical Mechanics. Luis Anchordoqui

Physics 121, March 27, Angular Momentum, Torque, and Precession. Department of Physics and Astronomy, University of Rochester

SOLUTION Determine the moment of inertia for the shaded area about the x axis. I x = y 2 da = 2 y 2 (xdy) = 2 y y dy

INSTRUCTIONS TO CANDIDATES:

13. Rigid Body Dynamics II

PHYS 1401 Chapter 11 Solutions

Example: Inverted pendulum on cart

Physical Dynamics (PHY-304)

9 Kinetics of 3D rigid bodies - rotating frames

Mechanics Physics 151

AP Pd 3 Rotational Dynamics.notebook. May 08, 2014

DYNAMICS OF SERIAL ROBOTIC MANIPULATORS

5.3 Rigid Bodies in Three-Dimensional Force Systems

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (Section 15.1)

Solution: (a) (b) (N) F X =0: A X =0 (N) F Y =0: A Y + B Y (54)(9.81) 36(9.81)=0

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: Plane Motion of Rigid Bodies: Energy and Momentum Methods. Seventh Edition CHAPTER

Problem Solving Session 11 Three Dimensional Rotation and Gyroscopes Solutions

The first change comes in how we associate operators with classical observables. In one dimension, we had. p p ˆ

ENGI 4430 Multiple Integration Cartesian Double Integrals Page 3-01

r CM = ir im i i m i m i v i (2) P = i

8.012 Physics I: Classical Mechanics Fall 2008

PHYSICS PART II SECTION- I. Straight objective Type

Chapter 16 Mechanical Waves

DESIGN OF A FUZZY PID CONTROLLER FOR APPLICATION IN SATELLITE ATTITUDE CONTROL SYSTEM. Willer Gomes. Evandro Marconi Rocco

16. Rotational Dynamics

Physics 351 Monday, April 3, 2017

Lecture 35: The Inertia Tensor

Properties of surfaces II: Second moment of area

An example of Lagrangian for a non-holonomic system

Distributed Forces: Moments of Inertia

Rotational Kinematics and Dynamics. UCVTS AIT Physics

x y plane is the plane in which the stresses act, yy xy xy Figure 3.5.1: non-zero stress components acting in the x y plane

- Beams are structural member supporting lateral loadings, i.e., these applied perpendicular to the axes.

12. Rigid Body Dynamics I

Space frames. b) R z φ z. R x. Figure 1 Sign convention: a) Displacements; b) Reactions

Aircraft Structures Structural & Loading Discontinuities

Equilibrium at a Point

Lecture 6: Distributed Forces Part 2 Second Moment of Area

Transcription:

hapter 18 KINETIS F RIID DIES IN THREE DIMENSINS The to fundamental equations for the motion of a sstem of particles ΣF = ma ΣM = H H provide the foundation for three dimensional analsis, just as the do in the case of plane motion of rigid bodies The computation of the angular momentum H and its derivative H, hoever, are no considerabl more involved The rectangular components of the angular momentum H of a rigid bod ma be epressed in terms of the components of its angular velocit and of its centroidal moments and products of inertia: H = +I ω - I ω - I ω H = -I ω + I ω - I ω H = -I ω - I ω + I ω If principal aes of inertia are used, these relations reduce to H H = I ω H = I ω H = I ω

H The sstem of the momenta of the particles forming a rigid bod ma be reduced to the vector mv attached at and the couple H nce these are determined, the angular momentum H of the bod about an given point ma be obtained b riting H = r mv + H In general, the angular momentum H and the angular velocit do not have the same direction The ill, hoever, have the same direction if is directed along one of the principal aes of inertia of the bod H r mv In the particular case of a rigid bod constrained to rotate about a fied point, the components of the angular momentum H of the bod about ma be obtained directl from the components of its angular velocit and from its moments and products of inertia ith respect to aes through H H = +I ω - I ω - I ω H = -I ω + I ω - I ω H = -I ω - I ω + I ω

The principle of impulse and momentum for a rigid bod in threedimensional motion is epressed b the same fundamental formula used for a rigid bod in plane motion Sst Momenta 1 + Sst Et Imp 1 2 = Sst Momenta 2 The initial and final sstem momenta should be represented as shon in the figure and computed from or H = +I ω - I ω - I ω H = -I ω + I ω - I ω H = -I ω - I ω + I ω H = I ω H = I ω H r mv H = I ω The kinetic energ of a rigid bod in three-dimensional motion ma be divided into to parts, one associated ith the motion of its mass center, and the other ith its motion about Using principal aes,,, e rite 1 2 1 2 2 2 2 T = mv 2 + (I ω + I ω + I ω ) here v = velocit of the mass center = angular velocit m = mass of rigid bod I, I, I = principal centroidal moments of inertia

In the case of a rigid bod constrained to rotate about a fied point, the kinetic energ ma be epressed as 1 2 2 2 T = 2 (I ω + I ω + I ω ) here,, and aes are the principal aes of inertia of the bod at The equations for kinetic energ make it possible to etend to the three-dimensional motion of a rigid bod the application of the principle of ork and energ and of the principle of conservation of energ The fundamental equations ΣF = ma ΣM H = H can be applied to the motion of a rigid bod in three dimensions We first recall that H represents the angular momentum of the bod relative to a centroidal frame of fied orientation and that H represents the rate of change of H ith respect to that frame s the bod rotates, its moments and products of inertia ith respect to change continuall It is therefore more convenient to use a frame rotating ith the bod to resolve into components and to compute the moments and products of inertia hich are used to determine H

H ΣF = ma ΣM = H H represents the rate of change of H ith respect to the frame of fied orientation, therefore H = (H ) + W H here H = angular momentum of the bod ith respect to the frame of fied orientation (H ) = rate of change of H ith respect to the rotating frame W = angular velocit of the rotating frame H ΣF = ma ΣM = H H = (H ) + W H Substituting H above into Σ M, ΣM = (H ) + W H If the rotating frame is attached to the bod, its angular velocit W is identical to the angular velocit of the bod Setting W =, using principal aes, and riting this equation in scalar form, e obtain Euler s equations of moton

H In the case of a rigid bod constrained to rotate about a fied point, an alternative method of solution ma be used, involving moments of the forces and the rate of change of the angular momentum about point ΣM = (H ) + W H here ΣM = sum of the moments about of the forces applied to the rigid bod H = angular momentum of the bod ith respect to the frame (H ) = rate of change of H ith respect to the rotating frame W = angular velocit of the rotating frame φ D ψ D When the motion of groscopes and other aismmetrical bodies are considered, the Eulerian angles φ,, and ψ are introduced to define the position of a groscope The time derivatives of these angles represent, respectivel, the rates of precession, nutation, and spin of the groscope The angular velocit is epressed in terms of these derivatives as = -φ sin i + j + (ψ + φ cos )k

φ D ψ D = -φ sin i + j + (ψ + φ cos )k The unit vectors are associated ith the frame attached to the inner gimbal of the groscope (figure to the right)and rotate, therfore, ith the angular velocit W = -φ sin i + j + φ cos k φk j ψk φ D ψ Denoting b I the moment of inertia of the groscope ith respect to its spin ais and b I its moment of inertia ith respect to a transverse ais through, e rite φk j ψk H = -I φ sin i + I j + I(ψ + φ cos )k W = -φ sin i + j + φ cos k ΣM = (H ) + W H Substituting for H and D leads to the differential equations defining the motion of the groscope into

φk ψk ΣM In the particular case of the stead precession of a groscope, the angle, the rate of precession φ, and the rate of spin ψ remain constant Such motion is possible onl if the moments of the eternal forces about satisf the relation ΣM = (Iω - I φ cos )φ sin j ie, if the eternal forces reduce to a couple of moment equal to the right-hand member of the equation above and applied about an ais perpendicular to the precession ais and to the spin ais