Chapter 16. Plane Motion Of Rigid Bodies: Forces And .

2y ago
19 Views
2 Downloads
3.45 MB
64 Pages
Last View : 2d ago
Last Download : 2m ago
Upload by : Ellie Forte
Transcription

Chapter 16. Plane Motion of Rigid Bodies:Forces and AccelerationsIntroductionEquations of Motion of a Rigid BodyAngular Momentum of a Rigid Body in Plane MotionPlane Motion of a Rigid Body: d’Alembert’s PrincipleAxioms of the Mechanics of Rigid BodiesSolution of Problems Involving the Motion of a Rigid BodySystems of Rigid BodiesConstrained Plane Motion

Rigid Body KineticsThe forces and moments applied to a robotic arm control the resulting kinematics, andtherefore the end position and forces of the actuator at the end of the robot arm.

Introduction In this chapter and in Chapters 17 and 18, we will be concerned with thekinetics of rigid bodies, i.e., relations between the forces acting on a rigidbody, the shape and mass of the body, and the motion produced. Results of this chapter will be restricted to: plane motion of rigid bodies, and rigid bodies consisting of plane slabs or bodies which are symmetricalwith respect to the reference plane. Our approach will be to consider rigid bodies as made of large numbers ofparticles and to use the results of Chapter 14 for the motion of systems ofparticles. F ma and M G HG

16.1 Kinematics of Rigid Body16.1A Equations of Motion for a Rigid Body Consider a rigid body acted upon byseveral external forces. Assume that the body is made of alarge number of particles. For the motion of the mass center Gof the body with respect to theNewtonian frame Oxyz, F maFor the motion of the body with respect to M G HGthe centroidal frame Gx’y’z’ ,

andmaHG. System of external forces is equipollent to the system consisting of16.1B Angular Momentum of a Rigid Body in Plane Motion Angular momentum of the slab may be computed byn H G (ri′ vi′Δmi )i 1n [ri′ (ω ri′)Δmi ]i 1 ( ω ri′ 2 Δmi Iω) After differentiation, H G I ω I α

Results are also valid for plane motion of bodies which are symmetrical withrespect to the reference plane. Results are not valid for asymmetrical bodies or three-dimensional motion. Consider a rigid slab in plane motion.16.1C Plane Motion of a Rigid Body Motion of a rigid body in plane motion is completelydefined by the resultant and moment resultant about G ofthe external forces. Fx ma x Fy ma y M G Iα The external forces and the collective effective forces ofthe slab particles are equipollent (reduce to the sameresultant and moment resultant) and equivalent (have thesame effect on the body).

d’Alembert’s Principle:The external forces acting on a rigid body are equivalent tothe effective forces of the various particles forming the body.*The most general motion of a rigid body that is symmetrical withrespect to the reference plane can be replaced by the sum of atranslation and a centroidal rotation.16.1D A Remark on the Axioms of the Mechanics of Rigid Bodies . The forces F and F ′ act at different points on arigid body but have the same magnitude, direction,and line of action. The forces produce the same moment about anypoint and are therefore, equipollent external forces. This proves the principle of transmissibility whereasit was previously stated as an axiom.

16.1E Problems Involving the Motion of a Rigid Body The fundamental relation between the forcesacting on a rigid body in plane motion and theacceleration of its mass center and the angularacceleration of the body is illustrated in a freebody-diagram equation. The techniques for solving problems of staticequilibrium may be applied to solve problems ofplane motion by utilizing d’Alembert’s principle, orprinciple of dynamic equilibrium These techniques may also be applied to problems involvingplane motion of connected rigid bodies by drawing a free-bodydiagram equation for each body and solving the correspondingequations of motion simultaneously.

Free Body Diagrams and Kinetic DiagramsThe free body diagram is the same as you have done in statics and in Ch. 13; we willadd the kinetic diagram in our dynamic analysis.Ex :1.Isolate the body of interest (free body)2.Draw your axis system (Cartesian, polar, path)3.Add in applied forces (e.g., weight)4. Replace supports with forces (e.g., tension force)5.Draw appropriate dimensions (angles and distances)yxInclude yourpositive zaxis direction

Put the inertial terms for the body of interest on the kinetic diagram.Ex 2:1.Isolate the body of interest (free body)2.Draw in the mass times acceleration of the particle; if unknown, do this in the positivedirection according to your chosen axes. For rigid bodies, also include the rotationalterm, I G a.ΣF maΣM G Iα

Ex 3:Draw the FBD and KD for the bar AB of mass m .A known force P is applied at the bottom of thebar.

yCyAL/2CIαxmgB3. Applied forcesma yP4. Replace supports with forces5. Dimensions6. Kinetic diagramG maxGL/22. AxesxCr1. Isolate body

Ex 4: A drum of 100 mm radius is attachedto a disk of 200 mm radius. Thecombined drum and disk had acombined mass of 5 kg.A cord isattached as shown, and a force ofmagnitude P 25 N is applied.Thecoefficients of static and kineticfriction between the wheel and groundare m s 0.25 and m k 0.20,respectively.for the wheel.Draw the FBD and KD

1.Isolate body2.Axes3.Applied forces4. Replace supports with forces5.Dimensions6.Kinetic diagram

ma y100mmIα 200 mmmaxWFNyx

Ex 5:The ladder AB slides down the wall as shown.The wall and floor are both rough. Draw theFBD and KD for the ladder.

1.Isolate body 2. Axes3.Applied forces5.NBθFBWyNADimensionsma yIα FA4. Replace supports with forcesxmax6. Kinetic diagram

Sample Problem 16.1At a forward speed of 10 m/s, the truck brakeswere applied, causing the wheels to stoprotating.It was observed that the truck toskidded to a stop in 7 m.Determinethemagnitudeofthenormalreaction and the friction force at each wheel asthe truck skidded to a stop.STRATEGY: Calculate the acceleration during the skidding stop by assuming uniform acceleration. Draw the free-body-diagram equation expressing the equivalence of the external andeffective forces. Apply the three corresponding scalar equations to solve for the unknown normalwheel forces at the front and rear and the coefficient of friction between the wheelsand road surface.

MODELING and ANALYSIS: Calculate the acceleration during the skidding stop byassuming uniform acceleration.a 7.14m ms2Draw a free-body-diagram equation expressing theequivalence of the external and inertial terms. Apply the corresponding scalar equations. Fy (Fy )eff Fx (Fx )effN A NB W 0FA FBma N A NB k m a kWa 7.140.728 kg 9.81

MAApply the corresponding scalar equations. ( M A )effN A W N B 0.341WN rear1 N2 A12 Frear µ k N rearN front1 N2 V F front µ k N front( 0.341W )N rear 0.1705W( 0.728)( 0.1705W )Frear 0.1241W( 0.659W )N front 0.3295W( 0.728)( 0.3295W )F front 0.240W12

Sample Problem 16.3The thin plate of mass 8 kg is held in place asshown.Neglecting the mass of the links, determineimmediately after the wire has been cut (a) theacceleration of the plate, and (b) the force in eachlink.STRATEGY: Note that after the wire is cut, all particles of the plate move along parallel circular paths ofradius 150 mm.The plate is in curvilinear translation. Draw the free-body-diagram equation expressing the equivalence of the external and effectiveforces. Resolve into scalar component equations parallel and perpendicular to the path of the masscenter. Solve the component equations and the moment equation for the unknown acceleration and link

forces.MODELING and ANALYSIS:*Note that after the wire is cut, all particles of theplate move along parallel circular paths of radius 150mm. The plate is in curvilinear translation.*Draw the free-body-diagram equation expressingthe equivalence of the external and effective forces.*Resolve the diagram equation into componentsparallel and perpendicular to the path of the masscenter. Ft (Ft )effW cos 30 mamg cos 30

()a 9.81m/s 2 cos 30 a 8.50 m s 2 o60Solve the component equations and the momentequation for the unknown acceleration and link forces. M G ( M G )eff(FAE sin 30 )(250 mm ) (FAE cos 30 )(100 mm )(FDF sin 30 )(250 mm ) (FDF cos 30 )(100 mm ) 038.4 FAE 211.6 FDF 0FDF 0.1815 FAE Fn (Fn )effFAE FDF W sin 30 0FAE 0.1815 FAE W sin 30 0(FAE 0.619(8 kg ) 9.81m s 2)FAE 47.9 N T

FDF 0.1815 ( 47.9 N )FDF 8.70 N CREFLECT and THINK: If AE and DF had been cables rather than links,the answers you just determined indicate that DFwould have gone slack (i.e., you can’t push on arope), since the analysis showed that it would bein compression. Therefore, the plate would not beundergoing curvilinear translation, but it wouldhave been undergoing general plane motion. It is important to note that that there is always more than one way to solve problemslike this, since you can choose to take moments about any point you wish. In thiscase, you took them about G, but you could have also chosen to take them about Aor D.

Sample Problem 16.4A pulley of mass 6 kg and having a radius of gyration of 200mm is connected to two blocks as shown.Assuming no axle friction, determine the angular accelerationof the pulley and the acceleration of each block.STRATEGY: Determine the direction of rotation by evaluating the netmoment on the pulley due to the two blocks. Relate the acceleration of the blocks to the angularacceleration of the pulley.* Draw the free-body-diagram equation expressing theequivalence of the external and effective forces on thecomplete pulley plus blocks system. Solve the corresponding moment equation for the pulley angular acceleration.

MODELING and ANALYSIS: Determine the direction of rotation by evaluating the netmoment on the pulley due to the two blocks.Rotation is counterclockwise.

Note: Relate the acceleration of the blocks to the angular acceleration of the pulley. Draw the free-body-diagram equation expressing theequivalence of the external and effective forces on thecomplete pulley and blocks system. Solve the corresponding moment equation for thepulley angular acceleration. MG ( M G )eff 2.41rad s 2

Then,a A 0.603m s 2aA 0.603 maB 0.362m s20.362 maB s2s2

REFLECT and THINK: You could also solve this problem by considering thepulley and each block as separate systems, but you would havemore resulting equations. You would have to use this approachif you wanted to know the forces in the cables.

Sample Problem 16.5A cord is wrapped around a homogeneous disk of mass 15 kg.The cord is pulled upwards with a force T 180 N.Determine: (a) the acceleration of the center of the disk, (b)the angular acceleration of the disk, and (c) the accelerationof the cord.STRATEGY: Draw the free-body-diagram equation expressing the equivalence of the external and effectiveforces on the disk. Solve the three corresponding scalar equilibrium equations for the horizontal, vertical, andangular accelerations of the disk. Determine the acceleration of the cord by evaluating the tangential acceleration of the point Aon the disk.

MODELING and ANALYSIS: Drawthefree-body-diagramequationexpressing the equivalence of the external andeffective forces on the disk. Solve the three scalar equilibrium equations. Fx (Fx )effax 00 ma x Fy (Fy )effT W ma y()T W 180 N - (15 kg ) 9.81m s 2ay m15 kga y 2.19 m s 2 M G (M G )eff

Tr I α α (12 mr 2 )α2T2(180 N ) (15 kg )(0.5 m )mrα 48.0 rad s 2Determineacceleration of the cord by evaluating the tangentialacceleration of the point A on the disk. acord ( aA )t a ( aA G )t 2.19 m s 2 ( 0.5 m ) ( 48 rad s 2 )REFLECT and THINK: The angular acceleration is clockwise, as we would expect. Asimilar analysis would apply in many practical situations,such as pulling wire off a spool or paper off a roll. In suchcases, you would need to be sure that the tension pullingon the disk is not larger than the tensile strength of the material.acord 26.2 m s 2

Sample Problem 16.6A uniform sphere of mass m and radius r is projected alonga rough horizontal surface with a linear velocity v 0 .Thecoefficient of kinetic friction between the sphere and thesurface isµk .Determine: (a) the time t 1 at which the sphere will startrolling without sliding, and (b) the linear and angularvelocities of the sphere at time t 1 .STRATEGY: Draw the free-body-diagram equation expressing the equivalence of the external and effectiveforces on the sphere. Solve the three corresponding scalar equilibrium equations for the normal reaction from thesurface and the linear and angular accelerations of the sphere. Apply the kinematic relations for uniformly accelerated motion to determine the time at whichthe tangential velocity of the sphere at the surface is zero, i.e., when the sphere stops sliding.

MODELING and ANALYSIS: Draw the free-body-diagram equation expressing theequivalence of external and effective forces on thesphere. Solve the three scalar equilibrium equations. Fy (Fy )effN W mgN W 0 Fx (Fx )eff F ma µ k mg M G (M G )effFr I α(µ k mg )r (2 mr 23NOTE:a µk g)αα 5 µk g2 rAs long as the sphere both rotates and slides, its linear and angular motionsare uniformly accelerated.

Apply the kinematic relations for uniformly accelerated motion todetermine the time at which the tangential velocity of the sphereat the surface is zero, i.e., when the sphere stops sliding.v v 0 a t v 0 ( µ k g )t5µ gω ω 0 αt 0 k t 2 r a µk g5 µk gα 2 rAt the instant t 1 when the sphere stops sliding,v1 rω1 5 µk g v0 µk gt1 r t12r 5 µk g 5 µk g 2 v0 tω1 1 2 r 2 r 7 µk g 5 v v1 r ω1 r 0 7 r t1 2 v07 µk gω1 5 v07 rv1 75 v0

16.2 Constrained Plane MotionThe reactions of the automobile crankshaft bearingsdepend on the mass, mass moment of inertia, andthe kinematics of the crankshaft.The forces one the wind turbine blades are alsodependent on mass, mass moment of inertia,and kinematics.

Most engineering applications involve rigid bodies which aremoving under given constraints, e.g., cranks, connecting rods,and non-slippingwheels. Constrained plane motion:motions with definite relationsbetween the components of acceleration of the mass center andthe angular acceleration of the body. Solution of a problem involving constrained plane motionbegins with a kinematic analysis. e.g., given q, w, and a, find P , N A , and N B .- kinematic analysis yields- application of d’Alembert’s principle yields P , N A ,and N B .

Noncentroidal Rotation Noncentroidal rotation:motion of a body is constrained to rotateabout a fixed axis that does not pass through its mass center. Kinematic relation between the motion of the mass center G and themotion of the body about G, (16.7)at r αan r ω 2Fig.16.14 The kinematic relations are used to eliminatefrom equations derived from d’Alembert’s principle orfrom the method of dynamic equilibrium.Fig.16.15

Rolling Motion For a balanced disk constrained to roll withoutsliding,x rθ a rα Rolling, no sliding:F µs Na rαRolling, sliding impending:F µs Na rαRotating and sliding:F µk N a, rαindependentFor the geometric center of an unbalanced disk,aO rαThe acceleration of the mass center, aG aO aG O aO aG O aG Otn() ()

Sample Problem 16.7The portion AOB of the mechanism is actuated by gearD and at the instant shown has a clockwise angularvelocity of 8 rad/s and a counterclockwise angularacceleration of 40 rad/s2.Determine: a) tangential force exerted by gear D, and b)components of the reaction at shaft O.STRATEGY: Draw the free-body-equation for AOB, expressing the equivalence of the external andeffective forces. Evaluate the external forces due to the weights of gear E and arm OB and the effectiveforces associated with the angular velocity and acceleration. Solve the three scalar equations from the free-body-equation for the tangential forceat A and the horizontal and vertical components of reaction at shaft O.

MODELING and ANALYSIS: Draw the free-body-equation for AOB. Evaluate the external forces due to the weights ofgear E and arm OB and the effective forces.()WE (4 kg ) 9.81m s 2 39.2 N()WOB (3 kg ) 9.81m s 2 29.4 N2I Eα m E kE α( 4kg )( 0.085 m )2( 40 rad s )2 1.156 N mmOB ( aOB )t m OB ( r α )mE 4 kgk E 85 mmmOB 3 kgω 8 rad/s( 3kg )( 0.200 m ) ( 40 rads2 ) 24.0 N2m m OB ( aOB ) nOB ( r ω ) 38.4 Nα 40 rad s 221 I OBα mLαOB12() 1.600 N m112( 3kg )( 0.200 m )(8 rad s )( 3kg )( 0.400 m )22( 40 rad s )2

Solve the three scalar equations derived from thefree-body-equation for the tangential force at A andthe horizontal and vertical components of reaction atO. M O ( M O )effF (0.120m ) I Eα mOB (aOB )t (0.200m ) I OBα 1.156 N m (24.0 N )(0.200m ) 1.600 N mF 63.0 NFI Eα 1.156 N mmOB (aOB )t 24.0 N 63.0 N Fx ( Fx )effRx mOB (aOB )t 24.0 NRx 24.0 NRx 24.0 N

Fy ( Fy )effmOB (aOB )n 38.4 NI OBα 1.600 N mR y F WE WOB mOB (aOB )R y 63.0 N 39.2 N 29.4 N 38.4 NR y 24.0 NREFLECT and THINK: When you drew your kinetic diagram, you put yourinertia terms at the center of mass for the gear and for therod. Alternatively, you could have found the center of massfor the system and put the vectors I AOB α, m AOB a x andm AOB a y on the diagram. Finally, you could have found an overall I O for thecombined gear and rod and used Eq. 16.8 to solve for force F.

Sample Problem 16.9A sphere of weight W is released with no initial velocity and rollswithout slipping on the incline.Determine: a) the minimum value of the coefficient of friction, b)the velocity of G after the sphere has rolled 3 m and c) the velocityof G if the sphere were to move 3 m down a frictionless incline.STRATEGY: Draw the free-body-equation for the sphere, expressing the equivalence of the external and effectiveforces. With the linear and angular accelerations related, solve the three scalar equations derived from thefree-body-equation for the angular acceleration and the normal and tangential reactions at C. Calculate the friction coefficient required for the indicated tangential reaction at C. Calculate the velocity after 3 m of uniformly accelerated motion. Assuming no friction, calculate the linear acceleration down the incline and the corresponding velocityafter 3 m.

MODELING and ANALYSIS: Draw the free-body-equation for the sphere, expressingthe equivalence of the external and effective forces. With the linear and angular accelerations related, solvethe three scalar equations derived from thefree-body-equationangular acceleration andreactions at C.forthethe normal and tangential M C (M C )eff(W sin θ )r (ma )r I

kinetics of rigid bodies, i.e., relations between the forces acting on a rigid body, the shape and mass of the body, and the motion produced. Results of this chapter will be restricted to: plane motion of rigid bodies, and rigid bodies consisting of plane slabs or bodies which are symmetrical with respect to the reference plane.

Related Documents:

Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .

TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26

both translational motion and rotational motion. This combination is called general plane motion. F x m (a G) x F y m (a G) y M G I G a Using an x-y inertial coordinate system, the scalar equations of motions about the center of mass, G, may be written as: EQUATIONS OF MOTION: GENERAL PLANE MOTION (Section 17.5)

DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .

Plane Kinematics of Rigid Bodies Motion Relative to Rotating Axes Consider plane motion of two particles A and B (moving independently of each other) in fixed X-Y plane. Observing motion of point A from a moving reference frame x-y (origin attached to B) that rotates with ω the vector is normal to the plane of the motion

Brief Contents CHAPTER 1 Representing Motion 2 CHAPTER 2 Motion in One Dimension 30 CHAPTER 3 Vectors and Motion in Two Dimensions 67 CHAPTER 4 Forces and Newton’s Laws of Motion 102 CHAPTER 5 Applying Newton’s Laws 131 CHAPTER 6 Circular Motion, Orbits, and Gravity 166 CHAPTER 7 Rotational Motion 200

Fig. 9.5: (a) the downward motion; (b) the upward motion of a marble on an inclined plane; and (c) on a double inclined plane. Newton further studied Galileo’s ideas on force and motion and presented three fundamental laws that govern the motion of objects. These three laws are known as Newton’s laws of motion. The first law of motion is .

Edexcel IGCSE Accounting Pg 10 1.3 The accounting equation learn a simple Statement oflist examples of Refer to Sec Syllabus assets and Financial Position(horizontal/T The Principles of liabilities style) which explains the basic Double-Entry accounting equationclassification of . items as assets calculate the value of assets, Go For Accounting Pg and liabilities liabilities and capital using .