Dynamics 6

2. Diagrams. Always draw the complete free-body diagram of the body to be analyzed. Assign a convenient inertial coordinate system and label all known and unknown quantities. The kinetic diagram should also be constructed so as to clarify the equivalence between the applied forces and the resulting dynamic response.

3. Equations of Motion. Apply the three equations of motion from Eqs. 6/1, being consistent with the algebraic signs in relation to the choice of reference axes. Equation 6/2 or 6/3 may be employed as an alternative to the second of Eqs. 6/1. Combine these relations with the results from any needed kinematic analysis. Count the number of unknowns and be certain that there are an equal number of independent equations available. For a solvable rigid-body problem in plane motion, there can be no more than the five scalar unknowns which can be determined from the three scalar equations of motion, obtained from Eqs. 6/1, and the two scalar component relations which come from the relative-acceleration equation.

A virtual displacement is any assumed and arbitrary displacement, linear or angular, away from the natural or actual position. For a system of connected bodies, the virtual displacements must be consistent with the constraints of the system. For example, when one end of a link is hinged about a fixed pivot, the virtual displacement of the other end must be normal to the line joining the two ends. Such requirements for displacements consistent with the constraints are purely kinematic and provide what are known as the equations of constraint.

Although we have simplified this expression in Art. 6/2 in the course of deriving the moment equation of motion, we will pursue this same expression again for sake of emphasis by using the rigid body in plane motion represented in Fig. 6/13. The relative velocity becomes i, where the angular velocity of the body is k. The unit vector k is directed into the paper for the sense of shown. Because i, and are at right angles to one another, the magnitude of is i, and the magnitude of i is Thus, we may write HG where is the mass moment of inertia of the body about its mass center

Angular momentum is defined as the moment of linear momentum. In Art. 4/4 we expressed the angular momentum about the mass center of any prescribed system of mass as HG Σi mivi, which is merely the vector sum of the moments about G of the linear momenta of all particles. We showed in Art. 4/4 that this vector sum could also be written as HG Σi where is the velocity of mi with respect to G.

Any attempt to extend this simplified theory of impact utilizing a coefficient of restitution for the noncentral impact of rigid bodies of varying shape is a gross oversimplification which has little practical value. For this reason, we do not include such an exercise in this book, even though such a theory is easily developed and appears in certain references. We can and do, however, make full use of the principles of conservation of linear and angular momentum when they are applicable in discussing impact and other interactions of rigid bodies

As previously mentioned, the translational term will be expressed by its x-y, n-t, or r- components once the appropriate inertial reference system is designated. The equivalence depicted in Fig. 6/4 is basic to our understanding of the kinetics of plane motion and will be employed frequently in the solution of problems. Representation of the resultants and will help ensure that the force and moment sums determined from the free-body diagram are equated to their proper resultants.

Because the angular-momentum vector is always normal to the plane of motion, vector notation is generally unnecessary, and we may write the angular momentum about the mass center as the scalar (6/13) This angular momentum appears in the moment-angular-momentum relation, Eq. 4/9, which in scalar notation for plane motion, along with its integrated form, is (6/14) In words, the first of Eqs. 6/14 states that the sum of the moments about the mass center of all forces acting on the body equals the time rate of change of angular momentum about the mass center. The integrated form of Eq. 6/14 states that the initial angular momentum about the mass center G plus the external angular impulse about G equals the final angular momentum about G. The sense for positive rotati

Before carrying out this extension, you should review the definitions and concepts of work, kinetic energy, gravitational and elastic potential energy, conservative forces, and power treated in Arts. 3/6 and 3/7 because we will apply them to rigid-body problems. You should also review Arts. 4/3 and 4/4 on the kinetics of systems of particles, in which we extended the principles of Arts. 3/6 and 3/7 to encompass any general system of mass particles, which includes rigid bodies.

Chapter 6 is organized in the same three sections in which we treated the kinetics of particles in Chapter 3. Section A relates the forces and moments to the instantaneous linear and angular accelerations. Section B treats the solution of problems by the method of work and energy. Section C covers the methods of impulse and momentum. Virtually all of the basic concepts and approaches covered in these three sections were treated in Chapter 3 on particle kinetics. This repetition will help you with the topics of Chapter 6, provided you understand 412 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 412 the kinematics of rigid-body plane motion. In each of the three sections, we will treat three types of motion: translation, fixed-axis rotation, and general plane motion.

For a translating body, then, our general equations for plane motion, Eqs. 6/1, may be written (6/6) For rectilinear translation, illustrated in Fig. 6/8a, if the x-axis is chosen in the direction of the acceleration, then the two scalar force equations become ΣFx and ΣFy 0. For curvilinear translation, Fig. 6/8b, if we use n-t coordinates, the two scalar force equations become ΣFn and ΣFt In both cases, ΣMG 0. We may also employ the alternative moment equation, Eq. 6/2, with the aid of the kinetic diagram. For rectilinear translation we see that ΣMP and ΣMA 0. For curvilinear translation the kinetic diagram permits us to write ΣMA in the clockwise sense and ΣMB in the counterclockwise sense. Thus, we have complete freedom to choose a convenient moment center.

For our purpose in this chapter, a body which can be approximated as a thin slab with its motion confined to the plane of the slab will be considered to be in plane motion. The plane of motion will contain the mass center, and all forces which act on the body will be projected onto the plane of motion. A body which has appreciable dimensions normal to the plane of motion but is symmetrical about that plane of motion through the mass center may be treated as having plane motion. These idealizations clearly fit a very large category of rigid-body motions.

From the kinetic diagram in Fig. 6/9c, we may obtain Eq. 6/4 very easily by evaluating the moment of the resultants about O, which becomes ΣMO Application of the parallel-axis theorem for mass moments of inertia, IO gives ΣMO (IO IO. For the common case of rotation of a rigid body about a fixed axis through its mass center G, clearly, 0, and therefore ΣF 0. The resultant of the applied forces then is the couple We may combine the resultant-force component and resultant couple by moving to a parallel position through point Q on line OG, Fig. 6/10, located by Using the parallel-axis theorem and IO gives q Point Q is called the center of percussion and has the unique property that the resultant of all forces applied to the body must pass through it. It follows that the sum of the moments of all forces about the center of percussion is always zero, ΣMQ 0.

Gravitational potential energy Vg and elastic potential energy Ve were covered in detail in Art. 3/7. Recall that the symbol U (rather than U) is used to denote the work done by all forces except the weight and elastic forces, which are accounted for in the potential-energy terms. The work-energy relation, Eq. 3/15a, was introduced in Art. 3/6 for particle motion and was generalized in Art. 4/3 to include the motion of a general system of particles. This equation [4/2] applies to any mechanical system. For application to the motion of a single rigid body, the terms T1 and T2 must include the effects of translaton and rotation as given by Eqs. 6/7, 6/8, 6/9, or 6/10, and U1-2 is the work done by all external forces. On the other hand, if we choose to express the effects of weight and springs by means of potential energy rather than work, we may rewrite the above equation as [4/3a] where the prime denotes the work done by all forces other than weight and spring forces. When applied to an interconnected system of rigid bodies, Eq. 4/3a includes the effect of stored elastic energy in the connections, as well as that of gravitational potential energy for the various members. The term includes the work of all forces external to the system (other than gravitational forces), as well as the negative work of internal friction forces, if any. The terms T1 and T2 are the initial and final kinetic energies of all moving parts over the interval of motion in question. When the work-energy principle is applied to a single rigid body, either a free-body diagram or an active-force diagram should be used. In the case of an interconnected system of rigid bodies, an active-force diagram of the entire system should be drawn in order to isolate the system and disclose all forces which do work on the system. Diagrams should U 1-2 T1 V1 U 1-2 T2 V2 T1 U1-2 T2 1 2 I2, 1 22Σmii 2 Σ1 2mi (i )2 ˙i T 1 2 IC2 Article 6/6 Work-Energy Relations 461 c06.qxd 2/10/12 2:13 PM Page 461 also be drawn to disclose the initial and final positions of the system for the given interval of motion. The work-energy equation provides a direct relationship between the forces which do work and the corresponding changes in the motion of a mechanical system. However, if there is appreciable internal mechanical friction, then the system must be dismembered in order to disclose the kinetic-friction forces and account for the negative work which they do. When the system is dismembered, however, one of the primary advantages of the work-energy approach is automatically lost. The work-energy method is most useful for analyzing conservative systems of interconnected bodies, where energy loss due to the negative work of friction forces is negligible.

If a set of virtual displacements satisfying the equations of constraint and therefore consistent with the constraints is assumed for a mechanical system, the proper relationship between the coordinates which specify the configuration of the system will be determined by applying the work-energy relationship of Eq. 6/11, expressed in terms of virtual changes. Thus, (6/11a) It is customary to use the differential symbol d to refer to differential changes in the real displacements, whereas the symbol is used to signify virtual changes, that is, differential changes which are assumed rather than real.

If there are more than three remaining unknowns in a system, however, the three independent scalar equations of motion, when applied to the system, are not sufficient to solve the problem. In this case, more advanced methods such as virtual work (Art. 6/7) or Lagrange's equations (not discussed in this book*) could be employed, or else the system could be dismembered and each part analyzed separately with the resulting equations solved simultaneously.

If we had wished to eliminate reference to F2 and F3, for example, by choosing their intersection as the reference point, then P would lie on the opposite side of the vector, and the clockwise moment of ma ma ΣMP I mad mad ma H˙ G I. ma a ΣMP H˙ G ma ma I ma ma I 416 Chapter 6 Plane Kinetics of Rigid Bodies a - G P d ma - P G α I - α - ρ ≡ F1 F2 F3 Free-Body Diagram Kinetic Diagram Figure 6/5 c06.qxd 2/10/12 2:13 PM Page 416 about P would be a negative term in the equation. Equation 6/2 is easily remembered as it is merely an expression of the familiar principle of moments, where the sum of the moments about P equals the combined moment about P of their sum, expressed by the resultant couple ΣMG and the resultant force ΣF In Art. 4/4 we also developed an alternative moment equation about P, Eq. 4/13, which is [4/13] For rigid-body plane motion, if P is chosen as a point fixed to the body, then in scalar form becomes IP, where IP is the mass moment of inertia about an axis through P and is the angular acceleration of the body. So we may write the equation as (6/3) where the acceleration of P is aP and the position vector from P to G is When 0, point P becomes the mass center G, and Eq. 6/3 reduces to the scalar form ΣMG previously derived. When point P becomes a point O fixed in an inertial reference system and attached to the body (or body extended), then aP 0, and Eq. 6/3 in scalar form reduces to (6/4) Equation 6/4 then applies to the rotation of a rigid body about a nonaccelerating point O fixed to the body and is the two-dimensional simplification of Eq. 4/7.

Impact phenomena involve a fairly complex interrelationship of energy and momentum transfer, energy dissipation, elastic and plastic deformation, relative impact velocity, and body geometry. In Art. 3/12 we treated the impact of bodies modeled as particles and considered only the case of central impact, where the contact forces of impact passed through the mass centers of the bodies, as would always happen with colliding smooth spheres, for example. To relate the conditions after impact to those before impact required the introduction of the so-called coefficient of restitution e or impact coefficient, which compares the relative separation velocity with the relative approach velocity measured along the direction of the contact forces. Although in the classical theory of impact, e was considered a constant for given materials, more modern investigations show that e is highly dependent on geometry and impact velocity as well as on materials. At best, even for spheres and rods under direct central and longitudinal impact, the coefficient of restitution is a complex and variable factor of limited use.

In Art. 4/4 of Chapter 4 on systems of particles, we developed a general equation for moments about an arbitrary point P, Eq. 4/11, which is [4/11] where is the vector from P to the mass center G and is the mass-center acceleration. As we have shown earlier in this article, for a rigid body in plane motion becomes Also, the cross product is simply the moment of magnitude of about P. Therefore, for the two-dimensional body illustrated in Fig. 6/5 with its free-body diagram and kinetic diagram, we may rewrite Eq. 4/11 simply as (6/2) Clearly, all three terms are positive in the counterclockwise sense for the example shown, and the choice of P eliminates reference to F1 and F3.

In Art. 4/5, we expressed the principles of conservation of momentum for a general mass system by Eqs. 4/15 and 4/16. These principles are applicable to either a single rigid body or a system of interconnected rigid bodies. Thus, if ΣF 0 for a given interval of time, then [4/15] which says that the linear-momentum vector undergoes no change in the absence of a resultant linear impulse. For the system of interconnected rigid bodies, there may be linear-momentum changes of individual parts of the system during the interval, but there will be no resultant momentum change for the system as a whole if there is no resultant linear impulse

In Arts. 4/2 and 4/4 we derived the force and moment vector equations of motion for a general system of mass. We now apply these results by starting, first, with a general rigid body in three dimensions. The force equation, Eq. 4/1, [4/1] tells us that the resultant ΣF of the external forces acting on the body equals the mass m of the body times the acceleration of its mass center G. The moment equation taken about the mass center, Eq. 4/9, [4/9] shows that the resultant moment about the mass center of the external forces on the body equals the time rate of change of the angular momentum of the body about the mass center

In Chapter 3 we found that two force equations of motion were required to define the motion of a particle whose motion is confined to a plane. For the plane motion of a rigid body, an additional equation is needed to specify the state of rotation of the body. Thus, two force equations and one moment equation or their equivalent are required to determine the state of rigid-body plane motion

In Eq. 6/11 the differential motions are differential changes in the real or actual displacements which occur. For a mechanical system which assumes a steady-state configuration during constant acceleration, we often find it convenient to introduce the concept of virtual work. The concepts of virtual work and virtual displacement were introduced and used to establish equilibrium configurations for static systems of interconnected bodies (see Chapter 7 of Vol. 1 Statics).

In addition to using the work-energy equation to determine the velocities due to the action of forces acting over finite displacements, we may also use the equation to establish the instantaneous accelerations of the members of a system of interconnected bodies as a result of the active forces applied. We may also modify the equation to determine the configuration of such a system when it undergoes a constant acceleration.

In general, dynamics problems which involve physical constraints to motion require a kinematic analysis relating linear to angular acceleration before the force and moment equations of motion can be solved. It is for this reason that an understanding of the principles and methods of Chapter 5 is so vital to the work of Chapter 6.

In our study of the kinetics of particles in Arts. 3/6 and 3/7, we developed the principles of work and energy and applied them to the motion of a particle and to selected cases of connected particles. We found that these principles were especially useful in describing motion which resulted from the cumulative effect of forces acting through distances. Furthermore, when the forces were conservative, we were able to determine velocity changes by analyzing the energy conditions at the beginning and end of the motion interval. For finite displacements, the work-energy method eliminates the necessity for determining the acceleration and integrating it over the interval to obtain the velocity change. These same advantages are realized when we extend the workenergy principles to describe rigid-body motion.

In the case of an interconnected system, the system center of mass is generally inconvenient to use. As was illustrated previously in Articles 3/9 and 3/10 in the chapter on particle motion, the use of momentum principles greatly facilitates (HO)1 (HO)2 or (HG)1 (HG)2 G1 G2 Article 6/8 Impulse-Momentum Equations 489 F1 Ga Gb F6 F4 F5 F3 Ga = mav _ a Gb = mbv _ b F2 a b (HG)a = I _ a a ω (HG)b = I _ b b ω O Figure 6/15 c06.qxd 2/10/12 2:13 PM Page 489 the analysis of situations where forces and couples act for very short periods of time.

In the kinetics of rigid bodies which have angular motion, we must introduce a property of the body which accounts for the radial distribution of its mass with respect to a particular axis of rotation normal to the plane of motion. This property is known as the mass moment of inertia of the body, and it is essential that we be able to calculate this property in order to solve rotational problems. We assume that you are familiar with the calculation of mass moments of inertia. Appendix B treats this topic for those who need instruction or review.

In words, the first of Eqs. 6/12 and 6/12a states that the resultant force equals the time rate of change of momentum. The integrated form of Eqs. 6/12 and 6/12a states that the initial linear momentum plus the linear impulse acting on the body equals the final linear momentum. As in the force-mass-acceleration formulation, the force summations in Eqs. 6/12 and 6/12a must include all forces acting externally on the body considered. We emphasize again, therefore, that in the use of the impulse-momentum equations, it is essential to construct the complete impulse-momentum diagrams so as to disclose all external impulses. In contrast to the method of work and energy, all forces exert impulses, whether they do work or not. Angular Momentum Angular momentum is defined as th

It is instructive to use an alternative approach to derive the moment equation by referring directly to the forces which act on the representative particle of mass mi, as shown in Fig. 6/3. The acceleration of mi equals the vector sum of and the relative terms i2 and i, where the mass center G is used as the reference point. It follows that the resultant of all forces on mi has the components mii2 , and mii in the directions shown. The sum of the moments of these force components about G in the sense of becomes Similar moment expressions exist for all particles in the body, and the sum of these moments about G for the resultant forces acting on all particles may be written as But the origin of coordinates is taken at the mass center, so that Σmixi 0 and Σmi yi 0. Thus, the moment sum becomes as before. The contribution to ΣMG of the forces internal to the body is, of course, zero since they occur in pairs of equal and opposite forces of action and reaction between interacting particles. Thus, ΣMG, as before, represents the sum of moments about the mass center G of only the external forces acting on the body, as disclosed by the free-body diagram. We note that the force component mii2 has no moment about G and conclude, therefore, that the angular velocity has no influence on the moment equation about the mass center

Keep in mind the following considerations when solving planemotion problems. Choice of Coordinate System. The force equation of Eq. 6/1 should be expressed in whatever coordinate system most readily describes the acceleration of the mass center. You should consider rectangular, normal-tangential, and polar coordinates. Choice of Moment Equation. In Art. 6/2 we also showed, with the aid of Fig. 6/5, the application of the alternative relation for moments about any point P, Eq. 6/2. This figure and this equation are also repeated here for easy reference. [6/2] In some instances, it may be more convenient to use the alternative moment relation of Eq. 6/3 when moments are taken about a point P whose acceleration is known. Note also that the equation for moments about a ΣMP I mad c06.qxd 2/10/12 2:13 PM Page 443 nonaccelerating point O on the body, Eq. 6/4, constitutes still another alternative moment relation and at times may be used to advantage. Constrained versus Unconstrained Motion. In working a problem in general plane motion, we first observe whether the motion is unconstrained or constrained, as illustrated in the examples of Fig. 6/6. If the motion is constrained, we must account for the kinematic relationship between the linear and the angular accelerations and incorporate it into our force and moment equations of motion. If the motion is unconstrained, the accelerations can be determined independently of one another by direct application of the three motion equations, Eqs. 6/1. Number of Unknowns. In order for a rigid-body problem to be solvable, the number of unknowns cannot exceed the number of independent equations available to describe them, and a check on the sufficiency of the relationships should always be made. At the most, for plane motion we have three scalar equations of motion and two scalar components of the vector relative-acceleration equation for constrained motion. Thus, we can handle as many as five unknowns for each rigid body. Identification of the Body or System. We emphasize the importance of clearly choosing the body to be isolated and representing this isolation by a correct free-body diagram. Only after this vital step has been completed can we properly evaluate the equivalence between the external forces and their resultants. Kinematics. Of equal importance in the analysis of plane motion is a clear understanding of the kinematics involved. Very often, the difficulties experienced at this point have to do with kinematics, and a thorough review of the relative-acceleration relations for plane motion will be most helpful. Consistency of Assumptions. In formulating the solution to a problem, we recognize that the directions of certain forces or accelerations may not be known at the outset, so that it may be necessary to make initial assumptions whose validity will be proved or disproved when the solution is carried out. It is essential, however, that all assumptions made be consistent with the principle of action and reaction 444 Chapter 6 Plane Kinetics of Rigid Bodies a - G P d ma - P G α I - α - ρ ≡ F1 F2 F3 Free-Body Diagram Kinetic Diagram Figure 6/5, repeated c06.qxd 2/10/12 2:13 PM Page 444 Article 6/5 General Plane Motion 445 and with any kinematic requirements, which are also called conditions of constraint. Thus, for example, if a wheel is rolling on a horizontal surface, its center is constrained to move on a horizontal line. Furthermore, if the unknown linear acceleration a of the center of the wheel is assumed positive to the right, then the unknown angular acceleration will be positive in a clockwise sense in order that a r, if we assume the wheel does not slip. Also, we note that, for a wheel which rolls without slipping, the static friction force between the wheel and its supporting surface is generally less than its maximum value, so that F sN. But if the wheel slips as it rolls, a r, and a kinetic friction force is generated which is given by F kN. It may be necessary to test the validity of either assumption, slipping or no slipping, in a given problem. The difference between the coefficients of static and kinetic friction, s and k, is sometimes ignored, in which case, is used for either or both coefficients.

Part a of the figure shows the relevant free-body diagram. Part b of the figure shows the equivalent force-couple system with the resultant force applied through G. Part c of the figure is a kinetic diagram, which represents the resulting dynamic effects as specified by Eqs. 4/1 and 4/9. The equivalence between the free-body diagram and the kinetic diagram enables us to clearly visualize and easily remember the separate translational and rotational effects of the forces applied to a rigid body. We will express this equivalence mathematically as we apply these results to the treatment of rigid-body plane motion

Recall from our study of statics that a general system of forces acting on a rigid body may be replaced by a resultant force applied at a chosen point and a corresponding couple. By replacing the external forces by their equivalent force-couple system in which the resultant force acts through the mass center, we may visualize the action of the forces and the corresponding dynamic response of the body with the aid of Fig. 6/1.

Rigid-body translation in plane motion was described in Art. 5/1 and illustrated in Figs. 5/1a and 5/1b, where we saw that every line in a translating body remains parallel to its original position at all times. In rectilinear translation all points move in straight lines, whereas in curvilinear translation all points move on congruent curved paths. In either case, there is no angular motion of the translating body, so that both and are zero. Therefore, from the moment relation of Eqs. 6/1, we see that all reference to the moment of inertia is eliminated for a translating body.

Rotation of a rigid body about a fixed axis O was described in Art. 5/2 and illustrated in Fig. 5/1c. For this motion, we saw that all points in the body describe circles about the rotation axis, and all lines of the body in the plane of motion have the same angular velocity and angular acceleration .

Similarly, if the resultant moment about a given fixed point O or about the mass center is zero during a particular interval of time for a single rigid body or for a system of interconnected rigid bodies, then [4/16] which says that the angular momentum either about the fixed point or about the mass center undergoes no change in the absence of a corresponding resultant angular impulse. Again, in the case of the interconnected system, there may be angular-momentum changes of individual components during the interval, but there will be no resultant angularmomentum change for the system as a whole if there is no resultant angular impulse about the fixed point or the mass center. Either of Eqs. 4/16 may hold without the other

Successful application of kinetics requires that you isolate the body or system to be analyzed. The isolation technique was illustrated and used in Chapter 3 for particle kinetics and will be employed consistently in the present chapter. For problems involving the instantaneous relationships among force, mass, and acceleration, the body or system should be explicitly defined by isolating it with its free-body diagram. When the principles of work and energy are employed, an activeforce diagram which shows only those external forces which do work on the system may be used in lieu of the free-body diagram. The impulsemomentum diagram should be constructed when impulse-momentum methods are used. No solution of a problem should be attempted without first defining the complete external boundary of the body or system and identifying all external forces which act on it

The acceleration components of the mass center for circular motion are most easily expressed in n-t coordinates, so we have an and at as shown in Fig. 6/9a for rotation of the rigid body about the fixed axis through O. Part b of the figure represents the free-body diagram, and the equivalent kinetic diagram in part c of the figure shows the force resultant in terms of its n- and t-components and the resultant couple Our general equations for plane motion, Eqs. 6/1, are directly applicable and are repeated here. [6/1] Thus, the two scalar components of the force equation become ΣFn and ΣFt In applying the moment equation about G, we must account for the moment of the force applied to the body at O, so this force must not be omitted from the free-body diagram. For fixed-axis rotation, it is generally useful to apply a moment equation directly about the rotation axis O. We derived this equation previously as Eq. 6/4, which is repeated here.

The angular momentum about the mass center for the general system was expressed in Eq. 4/8a as HG Σi where i is the position vector relative to G of the representative particle of mass mi. For our rigid body, the velocity of mi relative to G is i, which has a magnitude i and lies in the plane of motion normal to i. The product i is then a vector normal to the x-y plane in the sense of , and its magnitude is Thus, the magnitude of HG becomes HG The summation, which may also be written as 2 dm, is defined as the mass moment of inertia of the body about the z-axis through G. (See Appendix B for a discussion of the calculation of mass moments of inertia.)

The bar in Fig. 6/6b, on the other hand, undergoes a constrained motion, where the vertical and horizontal guides for the ends of the bar impose a kinematic relationship between the acceleration components of the mass center and the angular acceleration of the bar. Thus, it is necessary to determine this kinematic relationship from the principles established in Chapter 5 and to combine it with the force and moment equations of motion before a solution can be carried out

The concept of power was discussed in Art. 3/6, which treated workenergy for particle motion. Recall that power is the time rate at which work is performed. For a force F acting on a rigid body in plane motion, the power developed by that force at a given instant is given by Eq. 3/16 and is the rate at which the force is doing work. The power is given by where dr and v are, respectively, the differential displacement and the velocity of the point of application of the force. Similarly, for a couple M acting on the body, the power developed by the couple at a given instant is the rate at which it is doing work, and is given by where d and are, respectively, the differential angular displacement and the angular velocity of the body. If the senses of M and are the same, the power is positive and energy is supplied to the body. Conversely, if M and have opposite senses, the power is negative and energy is removed from the body. If the force F and the couple M act simultaneously, the total instantaneous power is We may also express power by evaluating the rate at which the total mechanical energy of a rigid body or a system of rigid bodies is changing. The work-energy relation, Eq. 4/3, for an infinitesimal displacement is where dU is the work of the active forces and couples applied to the body or to the system of bodies. Excluded from dU are the work of dU dT dV P F v M P dU dt M d dt M P dU dt F dr dt F v 462 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 462 gravitational forces and that of spring forces, which are accounted for in the dV term. Dividing by dt gives the total power of the active forces and couples as Thus, we see that the power developed by the active forces and couples equals the rate of change of the total mechanical energy of the body or system of bodies. We note from Eq. 6/9 that, for a given body, the first term may be written where R is the resultant of all forces acting on the body and is the resultant moment about the mass center G of all forces. The dot product accounts for the case of curvilinear motion of the mass center, where and are not in the same direction.

The dynamics of a rigid body in general plane motion combines translation and rotation. In Art. 6/2 we represented such a body in Fig. 6/4 with its free-body diagram and its kinetic diagram, which discloses the dynamic resultants of the applied forces. Figure 6/4 and Eqs. 6/1, which apply to general plane motion, are repeated here for convenient reference. [6/1] Direct application of these equations expresses the equivalence between the externally applied forces, as disclosed by the free-body diagram, and their force and moment resultants, as represented by the kinetic diagram.

The equations of impulse and momentum may also be used for a system of interconnected rigid bodies since the momentum principles are applicable to any general system of constant mass. Figure 6/15 shows the combined free-body diagram and momentum diagram for two interconnected bodies a and b. Equations 4/6 and 4/7, which are ΣF and ΣMO O where O is a fixed reference point, may be written for each member of the system and added. The sums are (6/18) In integrated form for a finite time interval, these expressions become (6/19) t2 t1 ΣF dt (G)system t2 t1 ΣMO dt (HO)system ΣMO (H˙ O)a (H˙ O)b ... ΣF G˙ a G˙ b ... H˙ G˙ ΣMO H˙O and (HO)1 t2 t1 ΣMO dt (HO)2 HO IO I mr 2 (I mr 2). v r r HO I mvd mv I, 488 Chapter 6 Plane Kinetics of Rigid Bodies O O G HG = I _ ω HG = I _ ω ω G = mv _ d v _ G (b) (a) ω G = mv _ v _ r _ Figure 6/14 c06.qxd 2/10/12 2:13 PM Page 488 We note that the equal and opposite actions and reactions in the connections are internal to the system and cancel one another so they are not involved in the force and moment summations. Also, point O is one fixed reference point for the entire system.

The kinetic relationships which form the basis for most of the analysis of rigid-body motion were developed in Chapter 4 for a general system of particles. Frequent reference will be made to these equations as they are further developed in Chapter 6 and applied specifically to the plane motion of rigid bodies. You should refer to Chapter 4 frequently as you study Chapter 6. Also, before proceeding make sure that you have a firm grasp of the calculation of velocities and accelerations as developed in Chapter 5 for rigid-body plane motion. Unless you can determine accelerations correctly from the principles of kinematics, you frequently will be unable to apply the force and moment principles of kinetics. Consequently, you should master the necessary kinematics, including the calculation of relative accelerations, before proceeding

The kinetics of rigid bodies treats the relationships between the external forces acting on a body and the corresponding translational and rotational motions of the body. In Chapter 5 we developed the kinematic relationships for the plane motion of rigid bodies, and we will use these relationships extensively in this present chapter, where the effects of forces on the two-dimensional motion of rigid bodies are examined

The motion of a rigid body may be unconstrained or constrained. The rocket moving in a vertical plane, Fig. 6/6a, is an example of unconstrained motion as there are no physical confinements to its motion. ΣMO IO I, . ΣMP IP maP (H˙ P)rel ΣMP (H˙ P)rel maP ma. I Article 6/2 General Equations of Motion 417 B G A F mg y x G (a) Unconstrained Motion (b) Constrained Motion T a - x a - x a - y a - y α α y x Figure 6/6 c06.qxd 2/10/12 2:13 PM Page 417 The two components and of the mass-center acceleration and the angular acceleration may be determined independently of one another by direct application of Eqs. 6/1.

The principles of impulse and momentum were developed and used in Articles 3/9 and 3/10 for the description of particle motion. In that treatment, we observed that those principles were of particular importance when the applied forces were expressible as functions of the time and when interactions between particles occurred during short periods of time, such as with impact. Similar advantages result when the impulse-momentum principles are applied to the motion of rigid bodies. In Art. 4/2 the impulse-momentum principles were extended to cover any defined system of mass particles without restriction as to the connections between the particles of the system. These extended relations all apply to the motion of a rigid body, which is merely a special case of a general system of mass. We will now apply these equations directly to rigid-body motion in two dimensions

The results embodied in our basic equations of motion for a rigid body in plane motion, Eqs. 6/1, are represented diagrammatically in Fig. 6/4, ΣMG Σmii 2 I mx my ΣMG Σmii 2 a sin Σmixi a cos Σmi yi MGi mii 2 (mia sin )xi (mia cos )yi mia, a Article 6/2 General Equations of Motion 415 y x mi mia - xi G or β ω α ω mi i ρ α mi 2 ρ ωi yi Figure 6/3 a - G ma - G α ≡ F1 F2 F3 I - α Free-Body Diagram Kinetic Diagram Figure 6/4 c06.qxd 2/10/12 2:13 PM Page 415 which is the two-dimensional counterpart of parts a and c of Fig. 6/1 for a general three-dimensional body. The free-body diagram discloses the forces and moments appearing on the left-hand side of our equations of motion. The kinetic diagram discloses the resulting dynamic response in terms of the translational term and the rotational term which appear on the right-hand side of Eqs. 6/1.

The term dU represents the total work done by all active nonpotential forces acting on the system under consideration during the infinitesimal displacement of the system. The work of potential forces is included in the dV-term. If we use the subscript i to denote a representative body of the interconnected system, the differential change in kinetic energy T for the entire system becomes where d and di are the respective changes in the magnitudes of the velocities and where the summation is taken over all bodies of the system. But for each body, d and where represents the infinitesimal linear displacement of the center of mass and where di represents the infinitesimal angular displacement of the body in the plane of motion. We note that d is identical to d where is the component of along the tangent to the curve described by the mass center of the body in question. Also i represents the angular acceleration of the representative body. Consequently, for the entire system d This change may also be written as d where Ri and are the resultant force and resultant couple acting on body i and where di dik. These last two equations merely show us that the differential change in kinetic energy equals the differential work done on the system by the resultant forces and resultant couples acting on all the bodies of the system. The term dV represents the differential change in the total gravitational potential energy Vg and the total elastic potential energy Ve and has the form dV d(Σmi ghi Σ 1 2kjxj 2) Σmi g dhi Σkjxj dxj MGi si ΣMGi dT ΣRi di si ΣI dT Σmi ai i i di ¨i , ai (ai )t si (a , i )t s ai i dsi Ii i di I , i i di s mi ai i m i vi dvi vi dT d(Σ1 2mi vi 2 Σ1 2 Ii i 2) Σmi vi dvi ΣIi i di dU dT dV Article 6/7 Acceleration from Work-Energy; Virtual Work 477 c06.qxd 2/10/12 2:13 PM Page 477 where hi represents the vertical distance of the center of mass of the representative body of mass mi above any convenient datum plane and where xj stands for the deformation, tensile or compressive, of a representative elastic member of the system (spring) whose stiffness is kj. The complete expression for dU may now be written as d (6/11) When Eq. 6/11 is applied to a system of one degree of freedom, the terms d and will be positive if the accelerations are in the same direction as the respective displacements and negative if they are in the opposite direction. Equation 6/11 has the advantage of relating the accelerations to the active forces directly, which eliminates the need for dismembering the system and then eliminating the internal forces and reactive forces by simultaneous solution of the force-mass-acceleration equations for each member.

The work done by a force F has been treated in detail in Art. 3/6 and is given by where dr is the infinitesimal vector displacement of the point of application of F, as shown in Fig. 3/2a. In the equivalent scalar form of the integral, is the angle between F and the direction of the displacement, and ds is the magnitude of the vector displacement dr.

Upon occasion, in problems dealing with two or more connected rigid bodies whose motions are related kinematically, it is convenient to analyze the bodies as an entire system. Figure 6/7 illustrates two rigid bodies hinged at A and subjected to the external forces shown. The forces in the connection at A are internal to the system and are not disclosed. The resultant of all external forces must equal the vector sum of the two resultants and and the sum of the moments about some arbitrary point such as P of all external forces must equal the moment of the resultants, Thus, we may state (6/5) ΣMP ΣI Σmad ΣF Σma m2a2d2. m1a1d1 I22 I11 m2a2 m , 1a1 ax ay 418 Chapter 6 Plane Kinetics of Rigid Bodies P Kinetic Diagram of System Free-Body Diagram of System A A G1 G1 G2 m1a - 1 a - a 2 - 1 d1 d2 m2a - 2 I - 2 α2 α 2 α 1 I - 1 α1 ≡ G2 ≡ Figure 6/7 c06.qxd 2/10/12 2:13 PM Page 418 where the summations on the right-hand side of the equations represent as many terms as there are separate bodies.

We frequently need to evaluate the work done by a couple M which acts on a rigid body during its motion. Figure 6/11 shows a couple M Fb acting on a rigid body which moves in the plane of the couple. During time dt the body rotates through an angle d, and line AB moves to AB. We may consider this motion in two parts, first a translation to AB and then a rotation d about A. We see immediately that during the translation the work done by one of the forces cancels that done by the other force, so that the net work done is dU F(b d) M d due to the rotational part of the motion. If the couple acts in the sense opposite to the rotation, the work done is negative. During a finite rotation, the work done by a couple M whose plane is parallel to the plane of motion is, therefore,

We may now write where is a constant property of the body. This property is a measure of the rotational inertia, which is the resistance to change in rotational velocity due to the radial distribution of mass around the z-axis through G. With this substitution, our moment equation, Eq. 4/9, becomes where is the angular acceleration of the body. We may now express the moment equation and the vector form of the generalized Newton's second law of motion, Eq. 4/1, as (6/1) Equations 6/1 are the general equations of motion for a rigid body in plane motion. In applying Eqs. 6/1, we express the vector force equation ΣMG I ΣF ma ˙ ΣMG H˙G I˙ I I HG I I Σi 2mi . Σi i 2mi 2. ˙i ˙i mi ˙i ˙ a, 414 Chapter 6 Plane Kinetics of Rigid Bodies y x mi G a - F4 F3 F2 F1 β α ω ρi Figure 6/2 c06.qxd 2/10/12 2:13 PM Page 414 in terms of its two scalar components using x-y, n-t, or r- coordinates, whichever is most convenient for the problem at hand.

We now apply the foregoing relationships to the case of plane motion. Figure 6/2 represents a rigid body moving with plane motion in the x-y plane. The mass center G has an acceleration and the body has an angular velocity k and an angular acceleration k, both taken positive in the z-direction. Because the z-direction of both and remains perpendicular to the plane of motion, we may use scalar notation and to represent the angular velocity and angular acceleration

We now use the familiar expression for the kinetic energy of a particle to develop expressions for the kinetic energy of a rigid body for each of the three classes of rigid-body plane motion illustrated in Fig. 6/12. (a) Translation. The translating rigid body of Fig. 6/12a has a mass m and all of its particles have a common velocity v. The kinetic energy of any particle of mass mi of the body is Ti so for the entire body T or (6/7) This expression holds for both rectilinear and curvilinear translation. (b) Fixed-axis rotation. The rigid body in Fig. 6/12b rotates with an angular velocity about the fixed axis through O. The kinetic energy of a representative particle of mass mi is Ti Thus, for the entire body T But the moment of inertia of the body about O is IO so (6/8) Note the similarity in the forms of the kinetic energy expressions for translation and rotation. You should verify that the dimensions of the two expressions are identical. (c) General plane motion. The rigid body in Fig. 6/12c executes plane motion where, at the instant considered, the velocity of its mass center G is and its angular velocity is . The velocity vi of a representative particle of mass mi may be expressed in terms of the mass-center velocity and the velocity i relative to the mass center as shown. With the aid of the law of cosines, we write the kinetic energy of the body as the sum ΣTi of the kinetic energies of all its particles. Thus, Because and are common to all terms in the third summation, we may factor them out. Thus, the third term in the expression for T becomes since Σmi yi 0. The kinetic energy of the body is then T or (6/9) where is the moment of inertia of the body about its mass center. This expression for kinetic energy clearly shows the separate contributions to the total kinetic energy resulting from the translational velocity of the mass center and the rotational velocity about the mass center. v I T 1 2mv 2 1 2 I2 1 22Σmi i 1 2 2 v 2Σmi my v Σmi i cos v Σmiyi 0 v T Σ 1 2 mi vi 2 Σ 1 2 mi (v 2 i 22 2vi cos ) v v T 1 2 IO2 Σmi ri 2, 1 2 2Σmi ri 2. 1 2 mi (ri )2. T 1 2 mv2 1 2 v2 Σ Σmi 1 2 mi v2 1 2 mi v2, 460 Chapter 6 Plane Kinetics of Rigid Bodies (a) Translation mi vi = v v (c) General Plane Motion mi G yi vi -v -v (b) Fixed-Axis Rotation mi ri vi = riω ρiω ω ρi θ θ G O ω Figure 6/12 c06.qxd 2/10/12 2:13 PM Page 460 The kinetic energy of plane motion may also be expressed in terms of the rotational velocity about the instantaneous center C of zero velocity. Because C momentarily has zero velocity, the proof leading to Eq. 6/8 for the fixed point O holds equally well for point C, so that, alternatively, we may write the kinetic energy of a rigid body in plane motion as (6/10) In Art. 4/3 we derived Eq. 4/4 for the kinetic energy of any system of mass. We now see that this expression is equivalent to Eq. 6/9 when the mass system is rigid. For a rigid body, the quantity in Eq. 4/4 is the velocity of the representative particle relative to the mass center and is the vector i, which has the magnitude i. The summation term in Eq. 4/4 becomes which brings Eq. 4/4 into agreement with Eq. 6/9.

ector sum of the linear momenta of all its particles and wrote G Σmivi. With ri representing the position vector to mi, we have vi and G which, for a system whose total mass is constant, may be written as G d(Σmiri)/dt. When we substitute the principle of moments Σmiri to locate the mass center, the momentum becomes G , where is the velocity of the mass center. Therefore, as before, we find that the linear momentum of any mass system, rigid or nonrigid, is [4/5] In the derivation of Eq. 4/5, we note that it was unnecessary to employ the kinematic condition for a rigid body, Fig. 6/13, which is vi i. In that case, we obtain the same result by writing G i). The first sum is and the second sum becomes Σmii 0 since i is measured from the mass center, making zero.

this choice. The impulse-momentum diagram (see Art. 3/9) is again essential. See the Sample Problems which accompany this article for examples of these diagrams. With the moments about G of the linear momenta of all particles accounted for by HG it follows that we may represent the linear momentum G as a vector through the mass center G, as shown in Fig. 6/14a. Thus, G and HG have vector properties analogous to those of the resultant force and couple. With the establishment of the linear- and angular-momentum resultants in Fig. 6/14a, which represents the momentum diagram, the angular momentum HO about any point O is easily written as (6/15) This expression holds at any particular instant of time about O, which may be a fixed or moving point on or off the body. When a body rotates about a fixed point O on the body or body extended, as shown in Fig. 6/14b, the relations and d may be substituted into the expression for HO, giving HO But IO so that (6/16) In Art. 4/2 we derived Eq. 4/7, which is the moment-angularmomentum equation about a fixed point O. This equation, written in scalar notation for plane motion along with its integrated form, is (6/17) Note that you should not add linear momentum and angular momentum for the same reason that force and moment cannot be added directly.

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