Dynamics 3

(1) Momentum of the system is conserved in the n-direction. This gives (2) and (3) The momentum for each particle is conserved in the t-direction since there is no impulse on either particle in the t-direction. Thus, (4) The coefficient of restitution, as in the case of direct central impact, is the positive ratio of the recovery impulse to the deformation impulse. Equation 3/36 applies, then, to the velocity components in the n-direction. For the notation adopted with Fig. 3/20, we have Once the four final velocity components are found, the angles and of Fig. 3/20 may be easily determined.

(1) Work Associated with a Constant External Force. Consider the constant force P applied to the body as it moves from position 1 to position 2, Fig. 3/4. With the force P and the differential displacement dr written as vectors, the work done on the body by the force is (3/9) As previously discussed, this work expression may be interpreted as the force component P cos times the distance L traveled. Should be between 90 and 270, the work would be negative. The force component P sin normal to the displacement does no work.

(2) Work Associated with a Spring Force. We consider here the common linear spring of stiffness k where the force required to stretch or compress the spring is proportional to the deformation x, as shown in Fig. 3/5a. We wish to determine the work done on the body by the spring force as the body undergoes an arbitrary displacement from an initial position x1 to a final position x2. The force exerted by the spring on the body is F kxi, as shown in Fig. 3/5b. From the definition of work, we have (3/10) If the initial position is the position of zero spring deformation so that x1 0, then the work is negative for any final position x2 0. This is verified by recognizing that if the body begins at the undeformed spring position and then moves to the right, the spring force is to the left; if the body begins at x1 0 and moves to the left, the spring force is to the right. On the other hand, if we move from an arbitrary initial position x1 0 to the undeformed final position x2 0, we see that the work is positive. In any movement toward the undeformed spring position, the spring force and the displacement are in the same direction.

(3) Work Associated with Weight. Case (a) g constant. If the altitude variation is sufficiently small so that the acceleration of gravity g may be considered constant, the work done by the weight mg of the body shown in Fig. 3/6a as the body is displaced from an arbitrary altitude y1 to a final altitude y2 is

According to Newton's second law, a particle will accelerate when it is subjected to unbalanced forces. Kinetics is the study of the relations between unbalanced forces and the resulting changes in motion. In Chapter 3 we will study the kinetics of particles. This topic requires that we combine our knowledge of the properties of forces, which we developed in statics, and the kinematics of particle motion just covered in Chapter 2. With the aid of Newton's second law, we can combine these two topics and solve engineering problems involving force, mass, and motion.

According to this classical theory of impact, the value e 1 means that the capacity of the two particles to recover equals their tendency to deform. This condition is one of elastic impact with no energy loss. The value e 0, on the other hand, describes inelastic or plastic impact where the particles cling together after collision and the loss of energy is a maximum. All impact conditions lie somewhere between these two extremes

Accordingly, the power P developed by a force F which does an amount of work U is P dU/dt Because dr/dt is the velocity v of the point of application of the force, we have (3/16) Power is clearly a scalar quantity, and in SI it has the units of J/s. The special unit for power is the watt (W), which equals one joule per second (J/s). In U.S. customary units, the unit for mechanical power is the horsepower (hp). These units and their numerical equivalences are

All that we have done so far in this article is to rewrite Newton's second law in an alternative form in terms of momentum. But we are now able to describe the effect of the resultant force ΣF on the linear ΣFx G˙x ΣFy G˙y ΣFz G˙z G˙ kg m/s N s ΣF m˙ v d dt (mv) or ΣF G˙ ˙ v ˙ r Article 3/9 Linear Impulse and Linear Momentum 191 Path ΣF r2 r1 v1 t1 v2 r y m O x v = r · G = mv G · v · z t2 Figure 3/11 c03.qxd 2/9/12 7:39 PM Page 191 momentum of the particle over a finite period of time simply by integrating Eq. 3/25 with respect to the time t. Multiplying the equation by dt gives ΣF dt dG, which we integrate from time t1 to time t2 to obtain (3/27) Here the linear momentum at time t2 is G2 mv2 and the linear momentum at time t1 is G1 mv1. The product of force and time is defined as the linear impulse of the force, and Eq. 3/27 states that the total linear impulse on m equals the corresponding change in linear momentum of m. Alternatively, we may write Eq. 3/27 as

Also, it should be noted that a coefficient of restitution must be associated with a pair of contacting bodies. The coefficient of restitution is frequently considered a constant for given geometries and a given combination of contacting materials. Actually, it depends on the impact velocity and approaches unity as the impact velocity approaches zero as shown schematically in Fig. 3/19. A handbook value for e is generally unreliable.

Although the results of the ideal experiment are obtained for measurements made relative to the "fixed" primary inertial system, they are equally valid for measurements made with respect to any nonrotating reference system which translates with a constant velocity with respect to the primary system. From our study of relative motion in Art. 2/8, we know that the acceleration measured in a system translating with no acceleration is the same as that measured in the primary system. Thus, Newton's second law holds equally well in a nonaccelerating system, so that we may define an inertial system as any system in which Eq. 3/2 is valid

An increasing number of problems occur, particularly in the fields of rocket and spacecraft design, where the acceleration components of the earth are of primary concern. For this work it is essential that the fundamental basis of Newton's second law be thoroughly understood and that the appropriate absolute acceleration components be employed.

Angular momentum is the moment of linear momentum and must not be confused with linear momentum. In SI units, angular momentum has the units . In the U.S. customary system, angular momentum has the units [lb/(ft/sec2 )][ft/sec][ft] lb-ft-sec

Application of Newton's second law to each mass results in Dividing the first equation by m0, the second equation by m, and subtracting the first equation from the second give or (3/49) Equation 3/49 is a second-order differential equation which, when solved, yields the relative position vector r as a function of time. Numerical techniques are usually required for the integration of the scalar differential equations which are equivalent to the vector equation 3/49, especially if P is nonzero.

Application of the work-energy method requires isolation of the particle or system under consideration. For a single particle you should draw a free-body diagram showing all externally applied forces. For a system of particles rigidly connected without springs, draw an activeforce diagram showing only those external forces which do work (active forces) on the entire system.*

As an introduction to impact, we consider the collinear motion of two spheres of masses m1 and m2, Fig. 3/17a, traveling with velocities v1 and v2. If v1 is greater than v2, collision occurs with the contact forces directed along the line of centers. This condition is called direct central impact.

As was the case for linear-momentum problems, we encounter impulsive (large magnitude, short duration) and nonimpulsive forces in angular-momentum problems. The treatment of these forces was discussed in Art. 3/9

Because Eq. 3/25 is a vector equation, we recognize that, in addition to the equality of the magnitudes of ΣF and , the direction of the resultant force coincides with the direction of the rate of change in linear momentum, which is the direction of the rate of change in velocity. Equation 3/25 is one of the most useful and important relationships in dynamics, and it is valid as long as the mass m of the particle is not changing with time. The case where m changes with time is discussed in Art. 4/7 of Chapter 4. We now write the three scalar components of Eq. 3/25 as (3/26) These equations may be applied independently of one another.

Because the force exerted on the spring by the moving body is equal and opposite to the force F exerted by the spring on the body, it follows that the work done on the spring is the negative of the work done on the body. Therefore, we may replace the work U done by the spring on the body by Ve, the negative of the potential energy change for the spring, provided the spring is now included within the system

Before 1905 the laws of Newtonian mechanics had been verified by innumerable physical experiments and were considered the final description of the motion of bodies. The concept of time, considered an absolute quantity in the Newtonian theory, received a basically different interpretation in the theory of relativity announced by Einstein in 1905. The new concept called for a complete reformulation of the accepted laws of mechanics. The theory of relativity was subjected to early ridicule, but has been verified by experiment and is now universally accepted by scientists. Although the difference between the mechanics of Newton and that of Einstein is basic, there is a practical difference in the results given by the two theories only when velocities of the order of the speed of light (300 106 m/s) are encountered.* Important problems dealing with atomic and nuclear particles, for example, require calculations based on the theory of relativity.

Careful and consistent use of the free-body method is the most important single lesson to be learned in the study of engineering mechanics. When drawing a free-body diagram, clearly indicate the coordinate axes and their positive directions. When you write the equations of motion, make sure all force summations are consistent with the choice of these positive directions. As an aid to the identification of external forces which act on the body in question, these forces are shown as heavy red vectors in the illustrations in this book. Sample Problems 3/1 through 3/5 in the next article contain five examples of free-body diagrams. You should study these to see how the diagrams are constructed.

Consider a force field where the force F is a function of the coordinates, Fig. 3/10. The work done by F during a displacement dr of its point of application is dU . The total work done along its path from 1 to 2 is The integral is a line integral which depends, in general, on the particular path followed between any two points 1 and 2 in space. If, however, is an exact differential† dV of some scalar function V of the coordinates, then

Consider a particle of mass m, Fig. 3/21, moving under the action of the central gravitational attraction where m0 is the mass of the attracting body, which is assumed to be fixed, G is the universal gravitational constant, and r is the distance between the centers of the masses. The particle of mass m could represent the earth moving about the sun, the moon moving about the earth, or a satellite in its orbital motion about the earth above the atmosphere.

Consider again the general curvilinear motion in space of a particle of mass m, Fig. 3/11, where the particle is located by its position vector r measured from a fixed origin O. The velocity of the particle is v and is tangent to its path (shown as a dashed line). The resultant ΣF of all forces on m is in the direction of its acceleration . We may now write the basic equation of motion for the particle, Eq. 3/3, as (3/25) where the product of the mass and velocity is defined as the linear momentum G mv of the particle. Equation 3/25 states that the resultant of all forces acting on a particle equals its time rate of change of linear momentum. In SI the units of linear momentum mv are seen to be , which also equals . In U.S. customary units, the units of linear momentum mv are [lb/(ft/sec2 )][ft/sec] lb-sec.

Consider now the motion of two particles a and b which interact during an interval of time. If the interactive forces F and F between them are the only unbalanced forces acting on the particles during the interval, it follows that the moments of the equal and opposite forces about any fixed point O not on their line of action are equal and opposite. If we apply Eq. 3/33 to particle a and then to particle b and add the two equations, we obtain Ha Hb 0 (where all angular momenta are referred to point O). Thus, the total angular momentum for the system of the two particles remains constant during the interval, and we write

During a finite movement of the point of application of a force, the force does an amount of work equal to or In order to carry out this integration, it is necessary to know the relations between the force components and their respective coordinates or the relation between Ft and s. If the functional relationship is not known as a mathematical expression which can be integrated but is specified in the form of approximate or experimental data, then we can compute the work by carrying out a numerical or graphical integration as represented by the area under the curve of Ft versus s, as shown in Fig. 3/3.

Each of these expressions for angular momentum may be checked easily from Fig. 3/15, which shows the three linear-momentum components, by taking the moments of these components about the respective axes

Equation 3/13 may be restated as (3/15) which is the work-energy equation for a particle. The equation states that the total work done by all forces acting on a particle as it moves from point 1 to point 2 equals the corresponding change in kinetic energy of the particle. Although T is always positive, the change T may U1-2 T2 T1 T N m T 1 2 mv2 U1-2 2 1 F dr v2 v1 mv dv 1 2 m(v2 2 v1 2) adr U1-2 2 1 F dr 2 1 madr U1-2 2 1 F dr s2 s1 Ft ds Ft Fn F = ΣF Path 1 2 x s1 s2 z t n m dr r α y O Figure 3/7 c03.qxd 2/9/12 7:39 PM Page 159 be positive, negative, or zero. When written in this concise form, Eq. 3/15 tells us that the work always results in a change of kinetic energy. Alternatively, the work-energy relation may be expressed as the initial kinetic energy T1 plus the work done U1-2 equals the final kinetic energy T2, or (3/15a) When written in this form, the terms correspond to the natural sequence of events. Clearly, the two forms 3/15 and 3/15a are equivalent.

Equation 3/31 states that the moment about the fixed point O of all forces acting on m equals the time rate of change of angular momentum of m about O. This relation, particularly when extended to a system of particles, rigid or nonrigid, provides one of the most powerful tools of analysis in dynamics. Equation 3/31 is a vector equation with scalar components

Equations 3/25 and 3/31 add no new basic information since they are merely alternative forms of Newton's second law. We will discover in subsequent chapters, however, that the motion equations expressed in terms of the time rate of change of momentum are applicable to the motion of rigid and nonrigid bodies and provide a very general and powerful approach to many problems. The full generality of Eq. 3/31 is usually not required to describe the motion of a single particle or the plane motion of rigid bodies, but it does have important use in the analysis of the space motion of rigid bodies introduced in Chapter 7.

For given masses and initial conditions, the momentum equation contains two unknowns, and Clearly, we need an additional relationship to find the final velocities. This relationship must reflect the capacity of the contacting bodies to recover from the impact and can be expressed by the ratio e of the magnitude of the restoration impulse to the magnitude of the deformation impulse. This ratio is called the coefficient of restitution

For measurements made relative to the rotating earth, the relative value of g should be used. The internationally accepted value of g relative to the earth at sea level and at a latitude of 45 is 9.806 65 m/s2 . Except where greater precision is required, the value of 9.81 m/s2 will be used for g. For measurements relative to a nonrotating earth, the absolute value of g should be used. At a latitude of 45 and at sea level, the absolute value is 9.8236 m/s2 . The sea-level variation in both the absolute and relative values of g with latitude is shown in Fig. 1/1 of Art. 1/5.

If a particle is constrained to move along a surface, as is the hockey puck or a marble sliding on the curved surface of a bowl, only two coordinates are needed to specify its position, and in this case it is said to have two degrees of freedom. If a particle is constrained to move along a fixed linear path, as is the collar sliding along a fixed shaft, its position may be specified by the coordinate measured along the shaft. In this case, the particle would have only one degree of freedom.

If m0 m and P 0, we have the restricted two-body problem, the equation of motion of which is (3/49a) With r and expressed in polar coordinates, Eq. 3/49a becomes When we equate coefficients of like unit vectors, we recover Eqs. 3/37. Comparison of Eq. 3/49 (with P 0) and Eq. 3/49a enables us to relax the assumption that mass m0 is fixed in space. If we replace m0 by (m0 m) in the expressions derived with the assumption of m0 fixed, then we obtain expressions which account for the motion of m0. For example, the corrected expression for the period of elliptical motion of m about m0 is, from Eq. 3/44,

If the ideal experiment described were performed on the surface of the earth and all measurements were made relative to a reference system attached to the earth, the measured results would show a slight discrepancy from those predicted by Eq. 3/2, because the measured acceleration would not be the correct absolute acceleration. The discrepancy would disappear when we introduced the correction due to the acceleration components of the earth. These corrections are negligible for most engineering problems which involve the motions of structures and machines on the surface of the earth. In such cases, the accelerations measured with respect to reference axes attached to the surface of the earth may be treated as "absolute," and Eq. 3/2 may be applied with negligible error to experiments made on the surface of the earth.*

If the resultant force on a particle is zero during an interval of time, we see that Eq. 3/25 requires that its linear momentum G remain constant. In this case, the linear momentum of the particle is said to be conserved. Linear momentum may be conserved in one coordinate direction, such as x, but not necessarily in the y- or z-direction. A careful examination of the impulse-momentum diagram of the particle will disclose whether the total linear impulse on the particle in a particular direction is zero. If it is, the corresponding linear momentum is unchanged (conserved) in that direction.

If the resultant moment about a fixed point O of all forces acting on a particle is zero during an interval of time, Eq. 3/31 requires that its angular momentum HO about that point remain constant. In this case, the angular momentum of the particle is said to be conserved. Angular momentum may be conserved about one axis but not about another axis. A careful examination of the free-body diagram of the particle will disclose whether the moment of the resultant force on the particle about a fixed point is zero, in which case, the angular momentum about that point is unchanged (conserved).

If we choose the x-direction, for example, as the direction of the rectilinear motion of a particle of mass m, the acceleration in the y- and z-directions will be zero and the scalar components of Eq. 3/3 become (3/4) For cases where we are not free to choose a coordinate direction along the motion, we would have in the general case all three component equations (3/5) where the acceleration and resultant force are given by

Impact phenomena are almost always accompanied by energy loss, which may be calculated by subtracting the kinetic energy of the system just after impact from that just before impact. Energy is lost through the generation of heat during the localized inelastic deformation of the material, through the generation and dissipation of elastic stress waves within the bodies, and through the generation of sound energy.

In Fig. 3/1b we illustrate the proper units with the simplest example where we accelerate an object of mass m along the horizontal with a force F. In SI units (an absolute system), a force F 1 N causes a mass m 1 kg to accelerate at the rate a 1 m/s2 . Thus, 1 N 1 In the U.S. customary system (a gravitational system), a force F 1 lbf causes a mass m 1 lbm (1/32.2 slug) to accelerate at the rate a 32.2 ft/sec2 , whereas a force F 1 lbf causes a mass m 1 slug (32.2 lbm) to accelerate at the rate a 1 ft/sec2 .

In U.S. customary units, on the other hand, the units of mass (slugs) are derived from the units of force (pounds force, lb) divided by acceleration (feet per second squared, ft/sec2 ). Thus, the mass units are slugs lb-sec2 /ft. This system is known as a gravitational system since mass is derived from force as determined from gravitational attraction

In U.S. customary units, we frequently speak of the weight of a body when we really mean mass. It is entirely proper to specify the mass of a body in pounds (lbm) which must be converted to mass in slugs before substituting into Newton's second law. Unless otherwise stated, the pound (lb) is normally used as the unit of force (lbf).

In addition to the equations of linear impulse and linear momentum, there exists a parallel set of equations for angular impulse and angular momentum. First, we define the term angular momentum. Figure 3/14a shows a particle P of mass m moving along a curve in space. The particle is located by its position vector r with respect to a convenient origin O of fixed coordinates x-y-z. The velocity of the particle is v , and its linear momentum is G mv. The moment of the linear momentum vector mv about the origin O is defined as the angular momentum HO of P about O and is given by the cross-product relation for the moment of a vector

In applying these motion equations to a body treated as a particle, you should follow the general procedure established in the previous article on rectilinear motion. After you identify the motion and choose the coordinate system, draw the free-body diagram of the body. Then obtain the appropriate force summations from this diagram in the usual way. The free-body diagram should be complete to avoid incorrect force summations.

In discussing particle motion relative to moving reference systems, we should note the special case where the reference system has a constant velocity and no rotation. If the x-y-z axes of Fig. 3/25 have a constant velocity, then aB 0 and the acceleration of the particle is aA arel. Therefore, we may write Eq. 3/50 as (3/51) which tells us that Newton's second law holds for measurements made in a system moving with a constant velocity. Such a system is known as an inertial system or as a Newtonian frame of reference. Observers in the moving system and in the fixed system will also agree on the designation of the resultant force acting on the particle from their identical free-body diagrams, provided they avoid the use of any so-called "inertia forces."

In solving problems, you may wonder how to get started and what sequence of steps to follow in arriving at the solution. This difficulty may be minimized by forming the habit of first recognizing some relationship between the desired unknown quantity in the problem and other quantities, known and unknown. Then determine additional relationships between these unknowns and other quantities, known and unknown. Finally, establish the dependence on the original data and develop the procedure for the analysis and computation. A few minutes spent organizing the plan of attack through recognition of the dependence of one quantity on another will be time well spent and will usually prevent groping for the answer with irrelevant calculations.

In the U.S. customary system, the standard value of g relative to the rotating earth at sea level and at a latitude of 45 is 32.1740 ft/sec2 . The corresponding value relative to a nonrotating earth is 32.2230 ft/sec2 .

In the general case, of course, neither x1 nor x2 is zero. The magnitude of the work is equal to the shaded trapezoidal area of Fig. 3/5a. In calculating the work done on a body by a spring force, care must be U1-2 2 1 F dr 2 1 (kxi)dxi x2 x1 kx dx 1 2 k(x1 2 x2 2) x2 x1 P cos dx P cos (x2 x1) PL cos U1-2 2 1 F dr 2 1 [(P cos )i (P sin )j]dxi 156 Chapter 3 Kinetics of Particles P dr x y α L 1 2 Figure 3/4 c03.qxd 2/9/12 7:39 PM Page 156 Article 3/6 Work and Kinetic Energy 157 taken to ensure that the units of k and x are consistent. If x is in meters (or feet), k must be in N/m (or lb/ft). In addition, be sure to recognize that the variable x represents a deformation from the unstretched spring length and not the total length of the spring.

In the previous article on work and kinetic energy, we isolated a particle or a combination of joined particles and determined the work done by gravity forces, spring forces, and other externally applied forces acting on the particle or system. We did this to evaluate U in the workenergy equation. In the present article we will introduce the concept of potential energy to treat the work done by gravity forces and by spring forces. This concept will simplify the analysis of many problems.

In the previous two articles, we applied Newton's second law F ma to various problems of particle motion to establish the instantaneous relationship between the net force acting on a particle and the resulting acceleration of the particle. When we needed to determine the change in velocity or the corresponding displacement of the particle, we integrated the computed acceleration by using the appropriate kinematic equations.

In the second type of problem, the forces acting on the particle are specified and we must determine the resulting motion. If the forces are constant, the acceleration is also constant and is easily found from Eq. 3/3. When the forces are functions of time, position, or velocity, Eq. 3/3 becomes a differential equation which must be integrated to determine the velocity and displacement.

It is customary to take k equal to unity in Eq. 3/2, thus putting the relation in the usual form of Newton's second law [1/1] A system of units for which k is unity is known as a kinetic system. Thus, for a kinetic system the units of force, mass, and acceleration are not independent. In SI units, as explained in Art. 1/4, the units of force (newtons, N) are derived by Newton's second law from the base units of mass (kilograms, kg) times acceleration (meters per second squared, m/s2 ). Thus, N This system is known as an absolute system since the unit for force is dependent on the absolute value of mass.

Let Fr and Fd represent the magnitudes of the contact forces during the restoration and deformation periods, respectively, as shown in Fig. 3/18. For particle 1 the definition of e together with the impulsemomentum equation give us Similarly, for particle 2 we have We are careful in these equations to express the change of momentum (and therefore v) in the same direction as the impulse (and thus the force). The time for the deformation is taken as t0 and the total time of contact is t. Eliminating v0 between the two expressions for e gives us

Now consider the energies of particle m. The system is conservative, and the constant energy E of m is the sum of its kinetic energy T and potential energy V. The kinetic energy is T mv2 and the potential energy from Eq. 3/19 is V mgR2 /r. Recall that g is the absolute acceleration due to gravity measured at the surface of the attracting body, R is the radius of the attracting body, and Gm0 gR2 . Thus, This constant value of E can be determined from its value at 0, where 0, 1/r C gR2 /h2 from Eq. 3/40, and h/r from Eq. 3/38. Substituting this into the expression for E and simplifying yield Now C is eliminated by substitution of Eq. 3/42, which may be written as h2 C egR2 , to obtain (3/45) The plus value of the radical is mandatory since by definition e is positive. We now see that for the These conclusions, of course, depend on the arbitrary selection of the datum condition for zero potential energy (V 0 when r ). The expression for the velocity v of m may be found from the energy equation, which is 1 2 mv2 mgR2 r E hyperbolic orbit e 1, E is positive parabolic orbit e 1, E is zero elliptical orbit e 1, E is negative 2E m h2C2 g2R4 h2 ˙r r ˙ E 1 2 m(˙r2 r2 ˙2) mgR2 r 1 2 m(˙r2 r2 ˙2) 1 2 1 r 1 d cos ( ) 1 ed or 1 r 1 ed cos d Article 3/13 Central-Force Motion 233 I θ 1 -θ 1 II Figure 3/23 e 1 2Eh2 mg2R4 c03.qxd 2/9/12 7:39 PM Page 233 The total energy E is obtained from Eq. 3/45 by combining Eq. 3/42 and 1/C d a(1 e2 )/e to give for the elliptical orbit (3/46) Substitution into the energy equation yields (3/47) from which the magnitude of the velocity may be computed for a particular orbit in terms of the radial distance r. Next, combining the expressions for rmin and rmax corresponding to perigee and apogee, Eq. 3/43, with Eq. 3/47 results in a pair of expressions for the respective velocities at these two positions for the elliptical orbit:

Once you assign reference axes, you must use the expressions for both the forces and the acceleration which are consistent with that assignment. In the first of Eqs. 3/7, for example, the positive sense of the n-axis is toward the center of curvature, and so the positive sense of our force summation ΣFn must also be toward the center of curvature to agree with the positive sense of the acceleration an v2 /.

Opinion differs concerning the original interpretation of D'Alembert's principle, but the principle in the form in which it is generally known is regarded in this book as being mainly of historical interest. It evolved when understanding and experience with dynamics were extremely limited and was a means of explaining dynamics in terms of the principles of statics, which were more fully understood. This excuse for using an artificial situation to describe a real one is no longer justified, as today a wealth of knowledge and experience with dynamics strongly supports the direct approach of thinking in terms of dynamics rather than statics. It is somewhat difficult to understand the long persistence in the acceptance of statics as a way of understanding dynamics, particularly in view of the continued search for the understanding and description of physical phenomena in their most direct form

Problems of this second type are often more formidable, as the integration may be difficult to carry out, particularly when the force is a mixed function of two or more motion variables. In practice, it is frequently necessary to resort to approximate integration techniques, either numerical or graphical, particularly when experimental data are involved. The procedures for a mathematical integration of the acceleration when it is a function of the motion variables were developed in Art. 2/2, and these same procedures apply when the force is a specified function of these same parameters, since force and acceleration differ only by the constant factor of the mass.

The SI units of work are those of force (N) times displacement (m) or This unit is given the special name joule (J), which is defined as the work done by a force of 1 N acting through a distance of 1 m in the direction of the force. Consistent use of the joule for work (and energy) rather than the units will avoid possible ambiguity with the units of moment of a force or torque, which are also written In the U.S. customary system, work has the units of ft-lb. Dimensionally, work and moment are the same. In order to distinguish between the two quantities, it is recommended that work be expressed as foot pounds (ft-lb) and moment as pound feet (lb-ft). It should be noted that work is a scalar as given by the dot product and involves the product of a force and a distance, both measured along the same line. Moment, on the other hand, is a vector as given by the cross product and involves the product of force and distance measured at right angles to the force.

The acceleration of points attached to the earth as measured in the primary system are quite small, however, and we normally neglect them for most earth-surface measurements. For example, the acceleration of the center of the earth in its near-circular orbit around the sun considered fixed is 0.00593 m/s2 (or 0.01946 ft/sec2 ), and the acceleration of a point on the equator at sea level with respect to the center of the earth considered fixed is 0.0339 m/s2 (or 0.1113 ft/sec2 ). Clearly, these accelerations are small compared with g and with most other significant accelerations in engineering work. Thus, we make only a small error when we assume that our earth-attached reference axes are equivalent to a fixed reference system.

The angular momentum then is a vector perpendicular to the plane A defined by r and v. The sense of HO is clearly defined by the right-hand rule for cross products. The scalar components of angular momentum may be obtained from the expansion

The basic principles and methods of particle kinetics were developed and illustrated in the first three sections of this chapter. This treatment included the direct use of Newton's second law, the equations of work and energy, and the equations of impulse and momentum. We paid special attention to the kind of problem for which each of the approaches was most appropriate. Several topics of specialized interest in particle kinetics will be briefly treated in Section D: 1. Impact 2. Central-force motion 3. Relative motion These topics involve further extension and application of the fundamental principles of dynamics, and their study will help to broaden your background in mechanics.

The basic relation between force and acceleration is found in Newton's second law, Eq. 1/1, the verification of which is entirely experimental. We now describe the fundamental meaning of this law by considering an ideal experiment in which force and acceleration are assumed to be measured without error. We subject a mass particle to the action of a single force F1, and we measure the acceleration a1 of the particle in the primary inertial system.* The ratio F1/a1 of the magnitudes of the force and the acceleration will be some number C1 whose value depends on the units used for measurement of force and acceleration. We then repeat the experiment by subjecting the same particle to a different force F2 and measuring the corresponding acceleration a2. The ratio F2/a2 of the magnitudes will again produce a number C2. The experiment is repeated as many times as desired.

The capacity of a machine is measured by the time rate at which it can do work or deliver energy. The total work or energy output is not a measure of this capacity since a motor, no matter how small, can deliver a large amount of energy if given sufficient time. On the other hand, a large and powerful machine is required to deliver a large amount of energy in a short period of time. Thus, the capacity of a machine is rated by its power, which is defined as the time rate of doing work.

The choice of an appropriate coordinate system depends on the conditions of the problem and is one of the basic decisions to be made in solving curvilinear-motion problems. We now rewrite Eq. 3/3 in three ways, the choice of which depends on which coordinate system is most appropriate.

The choice of an appropriate coordinate system is frequently indicated by the number and geometry of the constraints. Thus, if a particle is free to move in space, as is the center of mass of the airplane or rocket in free flight, the particle is said to have three degrees of freedom since three independent coordinates are required to specify the position of the particle at any instant. All three of the scalar components of the equation of motion would have to be integrated to obtain the space coordinates as a function of time

The corresponding work done by the gravitational force on the particle is mgh. Thus, the work done by the gravitational force is the negative of the change in potential energy. When large changes in altitude in the field of the earth are encountered, Fig. 3/8b, the gravitational force Gmme/r2 mgR2 /r2 is no longer constant. The work done against this force to change the radial position of the particle from r1 to r2 is the change (Vg)2 (Vg)1 in gravitational potential energy, which is It is customary to take (Vg)2 0 when r2 , so that with this datum we have (3/19) In going from r1 to r2, the corresponding change in potential energy

The expression F kx is actually a static relationship which is true only when elements of the spring have no acceleration. The dynamic behavior of a spring when its mass is accounted for is a fairly complex problem which will not be treated here. We shall assume that the mass of the spring is small compared with the masses of other accelerating parts of the system, in which case the linear static relationship will not involve appreciable error.

The force may also be written as the vector (3/24) where the symbol stands for the vector operator "del", which is The quantity V is known as the potential function, and the expression V is known as the gradient of the potential function. When force components are derivable from a potential as described, the force is said to be conservative, and the work done by F between any two points is independent of the path followed.

The foregoing analysis is based on three assumptions: 1. The two bodies possess spherical mass symmetry so that they may be treated as if their masses were concentrated at their centers, that is, as if they were particles. 2. There are no forces present except the gravitational force which each mass exerts on the other. 3. Mass m0 is fixed in space. Assumption (1) is excellent for bodies which are distant from the central attracting body, which is the case for most heavenly bodies. A significant class of problems for which assumption (1) is poor is that of artificial satellites in the very near vicinity of oblate planets. As a comment on assumption (2), we note that aerodynamic drag on a lowaltitude earth satellite is a force which usually cannot be ignored in the orbital analysis. For an artificial satellite in earth orbit, the error of assumption (3) is negligible because the ratio of the mass of the satellite to that of the earth is very small. On the other hand, for the earth-moon system, a small but significant error is introduced if assumption (3) is invoked—note that the lunar mass is about 1/81 times that of the earth.

The foregoing angular-impulse and angular-momentum relations have been developed in their general three-dimensional forms. Most of the applications of interest to us, however, can be analyzed as plane-motion problems where moments are taken about a single axis normal to the plane of motion. In this case, the angular momentum may change magnitude and sense, but the direction of the vector remains unaltered

The free-body diagram serves the same key purpose in dynamics as it does in statics. This purpose is simply to establish a thoroughly reliable method for the correct evaluation of the resultant of all actual forces acting on the particle or body in question. In statics this resultant equals zero, whereas in dynamics it is equated to the product of mass and acceleration. When you use the vector form of the equation of motion, remember that it represents several scalar equations and that every equation must be satisfied.

The interpretation of Eq. 3/40 requires a knowledge of the equations for conic sections. We recall that a conic section is formed by the locus of a point which moves so that the ratio e of its distance from a point (focus) to a line (directrix) is constant. Thus, from Fig. 3/21, e r/(d r cos ), which may be rewritten as (3/41) which is the same form as Eq. 3/40. Thus, we see that the motion of m is along a conic section with d 1/C and ed h2 /(Gm0), or (3/42) The three cases to be investigated correspond to e 1 (ellipse), e 1 (parabola), and e 1 (hyperbola). The trajectory for each of these cases is shown in Fig. 3/22.

The kinetic energy T of the particle is defined as (3/14) and is the total work which must be done on the particle to bring it from a state of rest to a velocity v. Kinetic energy T is a scalar quantity with the units of or joules (J) in SI units and ft-lb in U.S. customary units. Kinetic energy is always positive, regardless of the direction of the velocity.

The magnitude of this dot product is dU F ds cos , where is the angle between F and dr and where ds is the magnitude of dr. This expression may be interpreted as the displacement multiplied by the force component Ft F cos in the direction of the displacement, as represented by the dashed lines in Fig. 3/2b. Alternatively, the work dU may be interpreted as the force multiplied by the displacement component ds cos in the direction of the force, as represented by the full lines in Fig. 3/2b.

The most convenient coordinate system to use is polar coordinates in the plane of motion since F will always be in the negative r-direction and there is no force in the -direction. Equations 3/8 may be applied directly for the r- and -directions to give (3/37) The second of the two equations when multiplied by r/m is seen to be the same as d(r2 )/dt 0, which is integrated to give (3/38) The physical significance of Eq. 3/38 is made clear when we note that the angular momentum r mv of m about m0 has the magnitude mr2 . Thus, Eq. 3/38 merely states that the angular momentum of m about m0 remains constant (is conserved). This statement is easily deduced from Eq. 3/31, which shows that the angular momentum HO remains constant (is conserved) if there is no moment acting on the particle about a fixed point O.

The only reliable way to account accurately and consistently for every force is to isolate the particle under consideration from all contacting and influencing bodies and replace the bodies removed by the forces they exert on the particle isolated. The resulting free-body diagram is the means by which every force, known and unknown, which acts on the particle is represented and thus accounted for. Only after this vital step has been completed should you write the appropriate equation or equations of motion

The particle acceleration we measure from a fixed set of axes X-Y-Z, Fig. 3/26a, is its absolute acceleration a. In this case the familiar relation ΣF ma applies. When we observe the particle from a moving ΣF m(aB arel) aA aB arel ¨ r 244 Chapter 3 Kinetics of Particles ΣF Z Y X z y x m A rA rA/B = rrel O rB B Figure 3/25 y x Y X ΣF a m (a) (b) m - ma Y X ΣF a Figure 3/26 c03.qxd 2/9/12 7:40 PM Page 244 system x-y-z attached to the particle, Fig. 3/26b, the particle necessarily appears to be at rest or in equilibrium in x-y-z. Thus, the observer who is accelerating with x-y-z concludes that a force ma acts on the particle to balance ΣF. This point of view, which allows the treatment of a dynamics problem by the methods of statics, was an outgrowth of the work of D'Alembert contained in his Traité de Dynamique published in 1743.

The principles of impulse and momentum have important use in describing the behavior of colliding bodies. Impact refers to the collision between two bodies and is characterized by the generation of relatively large contact forces which act over a very short interval of time. It is important to realize that an impact is a very complex event involving material deformation and recovery and the generation of heat and sound. Small changes in the impact conditions may cause large changes in the impact process and thus in the conditions immediately following the impact. Therefore, we must be careful not to rely heavily on the results of impact calculations.

The ratio of the work done by a machine to the work done on the machine during the same time interval is called the mechanical efficiency em of the machine. This definition assumes that the machine operates uniformly so that there is no accumulation or depletion of energy within it. Efficiency is always less than unity since every device operates with some loss of energy and since energy cannot be created within the machine. In mechanical devices which involve moving parts, there will always be some loss of energy due to the negative work of kinetic friction forces. This work is converted to heat energy which, in turn, is dissipated to the surroundings. The mechanical efficiency at any instant of time may be expressed in terms of mechanical power P by (3/17) In addition to energy loss by mechanical friction, there may also be electrical and thermal energy loss, in which case, the electrical efficiency ee and thermal efficiency et are also involved. The overall efficiency e in such instances is

The second conclusion we draw from this ideal experiment is that the acceleration is always in the direction of the applied force. Thus, Eq. 3/1 becomes a vector relation and may be written (3/2) Although an actual experiment cannot be performed in the ideal manner described, the same conclusions have been drawn from countless accurately performed experiments. One of the most accurate checks is given by the precise prediction of the motions of planets based on Eq. 3/2.

The second example of potential energy occurs in the deformation of an elastic body, such as a spring. The work which is done on the spring to deform it is stored in the spring and is called its elastic potential energy Ve. This energy is recoverable in the form of work done by the spring on the body attached to its movable end during the release of the deformation of the spring. For the one-dimensional linear spring of stiffness k, which we discussed in Art. 3/6 and illustrated in Fig. 3/5, the force supported by the spring at any deformation x, tensile or compressive, from its undeformed position is F kx. Thus, we define the elastic potential energy of the spring as the work done on it to deform it an amount x, and we have (3/20) If the deformation, either tensile or compressive, of a spring increases from x1 to x2 during the motion, then the change in potential energy of the spring is its final value minus its initial value or

The second type is constrained motion where the path of the particle is partially or totally determined by restraining guides. An icehockey puck is partially constrained to move in the horizontal plane by the surface of the ice. A train moving along its track and a collar sliding along a fixed shaft are examples of more fully constrained motion. Some of the forces acting on a particle during constrained motion may be applied from outside sources, and others may be the reactions on the particle from the constraining guides. All forces, both applied and reactive, which act on the particle must be accounted for in applying Eq. 3/3.

The shape of the path followed by m may be obtained by solving the first of Eqs. 3/37, with the time t eliminated through combination with Eq. 3/38. To this end the mathematical substitution r 1/u is useful. Thus, (1/u2 ) , which from Eq. 3/38 becomes h( ) or h(du/d). The second time derivative is h(d2 u/d2 ) , which by combining with Eq. 3/38, becomes h2 u2 (d2 u/d2 ). Substitution into the first of Eqs. 3/37 now gives

The three general approaches to the solution of kinetics problems are: (A) direct application of Newton's second law (called the forcemass-acceleration method), (B) use of work and energy principles, and 3/1 Introduction Section A Force, Mass, and Acceleration 3/2 Newton's Second Law 3/3 Equation of Motion and Solution of Problems 3/4 Rectilinear Motion 3/5 Curvilinear Motion Section B Work and Energy 3/6 Work and Kinetic Energy 3/7 Potential Energy Section C Impulse and Momentum 3/8 Introduction 3/9 Linear Impulse and Linear Momentum 3/10 Angular Impulse and Angular Momentum Section D Special Applications 3/11 Introduction 3/12 Impact 3/13 Central-Force Motion 3/14 Relative Motion 3/15 Chapter Review CHAPTER OUTLINE 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 117 (C) solution by impulse and momentum methods. Each approach has its special characteristics and advantages, and Chapter 3 is subdivided into Sections A, B, and C, according to these three methods of solution. In addition, a fourth section, Section D, treats special applications and combinations of the three basic approaches. Before proceeding, you should review carefully the definitions and concepts of Chapter 1, because they are fundamental to the developments which follow.

There are cases where a force acting on a particle varies with the time in a manner determined by experimental measurements or by other approximate means. In this case a graphical or numerical integration must be performed. If, for example, a force F acting on a particle in a given direction varies with the time t as indicated in Fig. 3/13, then the impulse, F dt, of this force from t1 to t2 is the shaded area under the curve.

There are two general classes of problems in which the cumulative effects of unbalanced forces acting on a particle are of interest to us. These cases involve (1) integration of the forces with respect to the displacement of the particle and (2) integration of the forces with respect to the time they are applied. We may incorporate the results of these integrations directly into the governing equations of motion so that it becomes unnecessary to solve directly for the acceleration. Integration with respect to displacement leads to the equations of work and energy, which are the subject of this article. Integration with respect to time leads to the equations of impulse and momentum, discussed in Section C.

There are two physically distinct types of motion, both described by Eq. 3/3. The first type is unconstrained motion where the particle is free of mechanical guides and follows a path determined by its initial motion and by the forces which are applied to it from external sources. An airplane or rocket in flight and an electron moving in a charged field are examples of unconstrained motion.

These three scalar impulse-momentum equations are completely independent. Whereas Eq. 3/27 clearly stresses that the external linear impulse causes a change in the linear momentum, the order of the terms in Eqs. 3/27a and 3/27b corresponds to the natural sequence of events. While the form of Eq. 3/27 may be best for the experienced dynamicist, the form of Eqs. 3/27a and 3/27b is very effective for the beginner.

This approach merely amounts to rewriting the equation of motion as ΣF ma 0, which assumes the form of a zero force summation if ma is treated as a force. This fictitious force is known as the inertia force, and the artificial state of equilibrium created is known as dynamic equilibrium. The apparent transformation of a problem in dynamics to one in statics has become known as D'Alembert's principle.

This example should help clarify the relation between the scalar and vector forms of the angular impulse-momentum relations. Whereas Eq. 3/33 clearly stresses that the external angular impulse causes a change in the angular momentum, the order of the terms in Eqs. 3/33a and 3/33b corresponds to the natural sequence of events. Equation 3/33a is analogous to Eq. 3/27a, just as Eq. 3/31 is analogous to Eq. 3/25

Thus, for a particle of mass m moving along a curved path in the x-y plane, Fig. 3/16, the angular momenta about O at points 1 and 2 have the magnitudes mv1d1 and r2 mv2 mv2d2, respectively. In the illustration both and are represented in the counterclockwise sense in accord with the direction of the moment of the linear momentum. The scalar form of Eq. 3/33a applied to the motion between points 1 and 2 during the time interval t1 to t2 becomes

To help clarify the difference between the use of Eqs. 3/15 and 3/21, Fig. 3/9 shows schematically a particle of mass m constrained to move along a fixed path under the action of forces F1 and F2, the gravitational force W mg, the spring force F, and the normal reaction N. In Fig. 3/9b, the particle is isolated with its free-body diagram. The work done by each of the forces F1, F2, W, and the spring force F kx is evaluated, say, from A to B, and equated to the change T in kinetic energy using Eq. 3/15. The constraint reaction N, if normal to the path, will do no work. The alternative approach is shown in Fig. 3/9c, where the spring is included as a part of the isolated system. The work done during the interval by F1 and F2 is the -term of Eq. 3/21 with the changes in elastic and gravitational potential energies included on the energy side of the equation.

To help visualize angular momentum, we show in Fig. 3/14b a twodimensional representation in plane A of the vectors shown in part a of the figure. The motion is viewed in plane A defined by r and v. The magnitude of the moment of mv about O is simply the linear momentum mv times the moment arm r sin or mvr sin , which is the magnitude of the cross product HO r mv.

Up to this point in our development of the kinetics of particle motion, we have applied Newton's second law and the equations of workenergy and impulse-momentum to problems where all measurements of motion were made with respect to a reference system which was considered fixed. The nearest we can come to a "fixed" reference system is the primary inertial system or astronomical frame of reference, which is an imaginary set of axes attached to the fixed stars. All other reference systems then are considered to have motion in space, including any reference system attached to the moving earth.

We are now ready to relate the moment of the forces acting on the particle P to its angular momentum. If ΣF represents the resultant of all forces acting on the particle P of Fig. 3/14, the moment MO about the origin O is the vector cross product

We assume that any forces acting on the spheres during impact, other than the large internal forces of contact, are relatively small and produce negligible impulses compared with the impulse associated with each internal impact force. In addition, we assume that no appreciable change in the positions of the mass centers occurs during the short duration of the impact.

We cite only one simple example of the method known as D'Alembert's principle. The conical pendulum of mass m, Fig. 3/27a, is swinging in a horizontal circle, with its radial line r having an angular velocity . In the straightforward application of the equation of motion ΣF man in the direction n of the acceleration, the free-body diagram in part b of the figure shows that T sin mr2 . When we apply the equilibrium requirement in the y-direction, T cos mg 0, we can find the unknowns T and . But if the reference axes are attached to the particle, the particle will appear to be in equilibrium relative to these axes. Accordingly, the inertia force ma must be added, which amounts to visualizing the application of mr2 in the direction opposite to the acceleration, as shown in part c of the figure. With this pseudo free-body diagram, a zero force summation in the n-direction gives T sin mr2 0 which, of course, gives us the same result as before

We consider first the motion of a particle of mass m in close proximity to the surface of the earth, where the gravitational attraction (weight) mg is essentially constant, Fig. 3/8a. The gravitational potential energy Vg of the particle is defined as the work mgh done against the gravitational field to elevate the particle a distance h above some arbitrary reference plane (called a datum), where Vg is taken to be zero. Thus, we write the potential energy as (3/18) This work is called potential energy because it may be converted into energy if the particle is allowed to do work on a supporting body while it returns to its lower original datum plane. In going from one level at h h1 to a higher level at h h2, the change in potential energy becomes

We consider now a system of two particles joined together by a connection which is frictionless and incapable of any deformation. The forces in the connection are equal and opposite, and their points of application necessarily have identical displacement components in the direction of the forces. Therefore, the net work done by these internal forces is zero during any movement of the system. Thus, Eq. 3/15 is applicable to the entire system, where U1-2 is the total or net work done on the system by forces external to it and T is the change, T2 T1, in the total kinetic energy of the system. The total kinetic energy is the sum of the kinetic energies of both elements of the system. We thus see that another advantage of the work-energy method is that it enables us to analyze a system of particles joined in the manner described without dismembering the system

We draw two important conclusions from the results of these experiments. First, the ratios of applied force to corresponding acceleration all equal the same number, provided the units used for measurement are not changed in the experiments. Thus, We conclude that the constant C is a measure of some invariable property of the particle. This property is the inertia of the particle, which is its resistance to rate of change of velocity. For a particle of high inertia (large C), the acceleration will be small for a given force F. On the other hand, if the inertia is small, the acceleration will be large. The mass m is used as a quantitative measure of inertia, and therefore, we may write the expression C km, where k is a constant introduced to account for the units used. Thus, we may express the relation obtained from the experiments as F kma (3/1) F1 a1 F2 a2 F a C, a constant *The primary inertial system or astronomical frame of reference is an imaginary set of reference axes which are assumed to have no translation or rotation in space. See Art. 1/2, Chapter 1. c03.qxd 2/9/12 7:38 PM Page 118 where F is the magnitude of the resultant force acting on the particle of mass m, and a is the magnitude of the resulting acceleration of the particle.

We encounter two types of problems when applying Eq. 3/3. In the first type, the acceleration of the particle is either specified or can be determined directly from known kinematic conditions. We then determine the corresponding forces which act on the particle by direct substitution into Eq. 3/3. This problem is generally quite straightforward.

We have observed that the work done against a gravitational or an elastic force depends only on the net change of position and not on the particular path followed in reaching the new position. Forces with this characteristic are associated with conservative force fields, which possess an important mathematical property

We may conclude that no advantage results from this alternative formulation. The authors recommend against using it since it introduces no simplification and adds a nonexistent force to the diagram. In the case of a particle moving in a circular path, this hypothetical inertia force is known as the centrifugal force since it is directed away from the center and is opposite to the direction of the acceleration. You are urged to recognize that there is no actual centrifugal force acting on the particle. The only actual force which may properly be called centrifugal is the horizontal component of the tension T exerted by the particle on the cord.

We need to use both SI units and U.S. customary units, so we must have a clear understanding of the correct force and mass units in each system. These units were explained in Art. 1/4, but it will be helpful to illustrate them here using simple numbers before applying Newton's second law. Consider, first, the free-fall experiment as depicted in Fig. 3/1a where we release an object from rest near the surface of the earth. We allow it to fall freely under the influence of the force of gravitational attraction W on the body. We call this force the weight of the body. In SI units for a mass m 1 kg, the weight is W 9.81 N, and the corresponding downward acceleration a is g 9.81 m/s2 . In U.S. customary units for a mass m 1 lbm (1/32.2 slug), the weight is W 1 lbf and the resulting gravitational acceleration is g 32.2 ft/sec2 . For a mass m 1 slug (32.2 lbm), the weight is W 32.2 lbf and the acceleration, of course, is also g 32.2 ft/sec2

We note that in SI units where the mass is expressed in kilograms (kg), the weight W of the body in newtons (N) is given by W mg, where g 9.81 m/s2 . In U.S. customary units, the weight W of a body is expressed in pounds force (lbf), and the mass in slugs (lbf-sec2 /ft) is given by m W/g, where g 32.2 ft/sec2

We note that the center diagram is very much like a free-body diagram, except that the impulses of the forces appear rather than the forces themselves. As with the free-body diagram, it is necessary to include the effects of all forces acting on the body, except those forces whose magnitudes are negligible.

We note with the first approach that the work done by F kx could require a somewhat awkward integration to account for the changes in magnitude and direction of F as the particle moves from A U 1-2 T1 V1 U 1-2 T2 V2 Article 3/7 Potential Energy 177 System (a) (c) h N F = kx W = mg U1-2 = ΔT U′ 1-2 = ΔT + ΔV A B F1 F2 F1 F2 (b) F1 F2 Vg = 0 Vg = mgh Figure 3/9 c03.qxd 2/9/12 7:39 PM Page 177 to B. With the second approach, however, only the initial and final lengths of the spring are required to evaluate Ve. This greatly simplifies the calculation. For problems where the only forces are gravitational, elastic, and nonworking constraint forces, the U-term of Eq. 3/21a is zero, and the energy equation becomes (3/22) where E T V is the total mechanical energy of the particle and its attached spring. When E is constant, we see that transfers of energy between kinetic and potential may take place as long as the total mechanical energy T V does not change. Equation 3/22 expresses the law of conservation of dynamical energy.

We now account for the motion of both masses and allow the presence of other forces in addition to those of mutual attraction by considering the perturbed two-body problem. Figure 3/24 depicts the major mass m0, the minor mass m, their respective position vectors r1 and r2 measured relative to an inertial frame, the gravitation forces F and F, and a non-two-body force P which is exerted on mass m. The force P may be due to aerodynamic drag, solar pressure, the presence of a third body, on-board thrusting activities, a nonspherical gravitational field, or a combination of these and other sources.

We now apply the concepts discussed in Arts. 3/2 and 3/3 to problems in particle motion, starting with rectilinear motion in this article and treating curvilinear motion in Art. 3/5. In both articles, we will analyze the motions of bodies which can be treated as particles. This simplification is possible as long as we are interested only in the motion of the mass center of the body. In this case we may treat the forces as concurrent through the mass center. We will account for the action of nonconcurrent forces on the motions of bodies when we discuss the kinetics of rigid bodies in Chapter 6.

We now consider a particle A of mass m, Fig. 3/25, whose motion is observed from a set of axes x-y-z which translate with respect to a fixed reference frame X-Y-Z. Thus, the x-y-z directions always remain parallel to the X-Y-Z directions. We postpone discussion of motion relative to a rotating reference system until Arts. 5/7 and 7/7. The acceleration of the origin B of x-y-z is aB. The acceleration of A as observed from or relative to x-y-z is arel aA/B A/B, and by the relative-motion principle of Art. 2/8, the absolute acceleration of A is Thus, Newton's second law ΣF maA becomes (3/50) We can identify the force sum ΣF, as always, by a complete free-body diagram. This diagram will appear the same to an observer in x-y-z or to one in X-Y-Z as long as only the real forces acting on the particle are represented. We can conclude immediately that Newton's second law does not hold with respect to an accelerating system since ΣF marel.

We now consider the work done on a particle of mass m, Fig. 3/7, moving along a curved path under the action of the force F, which stands for the resultant ΣF of all forces acting on the particle. The position of m is specified by the position vector r, and its displacement along its path during the time dt is represented by the change dr in its position vector. The work done by F during a finite movement of the particle from point 1 to point 2 is where the limits specify the initial and final end points of the motion. When we substitute Newton's second law F ma, the expression for the work of all forces becomes But at ds, where at is the tangential component of the acceleration of m. In terms of the velocity v of the particle, Eq. 2/3 gives at ds v dv. Thus, the expression for the work of F becomes (3/13) where the integration is carried out between points 1 and 2 along the curve, at which points the velocities have the magnitudes v1 and v2, respectively.

We now develop the quantitative meaning of the term "work."* Figure 3/2a shows a force F acting on a particle at A which moves along the path shown. The position vector r measured from some convenient origin O locates the particle as it passes point A, and dr is the differential displacement associated with an infinitesimal movement from A to A. The work done by the force F during the displacement dr is defined as

We now extend the relationships developed for direct central impact to the case where the initial and final velocities are not parallel, Fig. 3/20. Here spherical particles of mass m1 and m2 have initial velocities v1 and v2 in the same plane and approach each other on a collision course, as shown in part a of the figure. The directions of the velocity vectors are measured from the direction tangent to the contacting surfaces, Fig. 3/20b. Thus, the initial velocity components along the t- and n-axes are (v1)n v1 sin 1, (v1)t v1 cos 1, (v2)n v2 sin 2,

We now introduce the concept of the impulse-momentum diagram. Once the body to be analyzed has been clearly identified and isolated, we construct three drawings of the body as shown in Fig. 3/12. In the first drawing, we show the initial momentum mv1, or components thereof. In m(v1)z t2 t1 ΣFz dt m(v2)z m(v1)y t2 t1 ΣFy dt m(v2)y m(v1)x t2 t1 ΣFx dt m(v2)x G1 t2 t1 ΣF dt G2 t2 t1 ΣF dt G2 G1 G 192 Chapter 3 Kinetics of Particles ΣF dt t 1 t 2 G1 = mv1 G2 = mv2 + = Figure 3/12 c03.qxd 2/9/12 7:39 PM Page 192 the second or middle drawing, we show all the external linear impulses (or components thereof). In the final drawing, we show the final linear momentum mv2 (or its components). The writing of the impulse-momentum equations 3/27b then follows directly from these drawings, with a clear one-to-one correspondence between diagrams and equation terms. We note that the center diagram is very much like a free-body di

We now see from Eq. 3/15 that a major advantage of the method of work and energy is that it avoids the necessity of computing the acceleration and leads directly to the velocity changes as functions of the forces which do work. Further, the work-energy equation involves only those forces which do work and thus give rise to changes in the magnitude of the velocities.

We observe that during time dt, the radius vector sweeps out an area, shaded in Fig. 3/21, equal to dA (r d). Therefore, the rate at which area is swept by the radius vector is , which is constant according to Eq. 3/38. This conclusion is expressed in Kepler's second law of planetary motion, which states that the areas swept through in equal times are equal.

We see that horizontal movement does not contribute to this work. We also note that if the body rises (perhaps due to other forces not shown), then ( y2 y1) 0 and this work is negative. If the body falls, ( y2 y1) 0 and the work is positive. Case (b) g constant. If large changes in altitude occur, then the weight (gravitational force) is no longer constant. We must therefore use the gravitational law (Eq. 1/2) and express the weight as a variable force of magnitude F as indicated in Fig. 3/6b. Using the radial coordinate shown in the figure allows the work to be expressed as

We turn our attention now to the kinetics of particles which move along plane curvilinear paths. In applying Newton's second law, Eq. 3/3, we will make use of the three coordinate descriptions of acceleration in curvilinear motion which we developed in Arts. 2/4, 2/5, and 2/6.

We will now examine the parallel question concerning the validity of the work-energy equation and the impulse-momentum equation relative to a constant-velocity, nonrotating system. Again, we take the x-y-z axes of Fig. 3/25 to be moving with a constant velocity vB relative to the fixed axes X-Y-Z. The path of the particle A relative to x-y-z is governed by rrel and is represented schematically in Fig. 3/28. The work done by ΣF relative to x-y-z is dUrel . But ΣF maA marel since aB 0. Also for the same reason that at ds v dv in Art. 2/5 on curvilinear motion. Thus, we have We define the kinetic energy relative to x-y-z as Trel so that we now have (3/52) which shows that the work-energy equation holds for measurements made relative to a constant-velocity, nonrotating system. Relative to x-y-z, the impulse on the particle during time dt is ΣF dt maA dt marel dt. But marel dt m dvrel d(mvrel) so We define the linear momentum of the particle relative to x-y-z as Grel mvrel, which gives us ΣF dt dGrel. Dividing by dt and integrating give (3/53) Thus, the impulse-momentum equations for a fixed reference system also hold for measurements made relative to a constant-velocity, nonrotating system. Finally, we define the relative angular momentum of the particle about a point in x-y-z, such as the origin B, as the moment of the ΣF G˙rel and ΣF dt Grel ΣF dt d(mvrel) dUrel dTrel or Urel Trel 1 2 mvrel 2 dUrel marel drrel mvrel dvrel d( 1 2 mvrel 2) arel drrel vrel dvrel ΣF drrel ˙ rB ΣF marel 246 Chapter 3 Kinetics of Particles vrel rrel aA = arel drrel z y m Path relative to x-y-z x Z Y O B X ΣF Figure 3/28 c03.qxd 2/9/12 7:40 PM Page 246 relative linear momentum. Thus, . The time derivative gives . The first term is nothing more than vrel mvrel 0, and the second term becomes rrel ΣF ΣMB, the sum of the moments about B of all forces on m. Thus, we have

When a particle moves under the influence of a force directed toward a fixed center of attraction, the motion is called central-force motion. The most common example of central-force motion is the orbital movement of planets and satellites. The laws which govern this motion were deduced from observation of the motions of the planets by J. Kepler (1571-1630). An understanding of central-force motion is required to design high-altitude rockets, earth satellites, and space vehicles

When a particle of mass m is subjected to the action of concurrent forces F1, F2, F3, . . . whose vector sum is ΣF, Eq. 1/1 becomes (3/3) When applying Eq. 3/3 to solve problems, we usually express it in scalar component form with the use of one of the coordinate systems developed in Chapter 2. The choice of an appropriate coordinate system depends on the type of motion involved and is a vital step in the formulation of any problem. Equation 3/3, or any one of the component forms of the forcemass-acceleration equation, is usually called the equation of motion. The equation of motion gives the instantaneous value of the acceleration corresponding to the instantaneous values of the forces which are acting

When applying any of the force-mass-acceleration equations of motion, you must account correctly for all forces acting on the particle. The only forces which we may neglect are those whose magnitudes are negligible compared with other forces acting, such as the forces of mutual attraction between two particles compared with their attraction to a celestial body such as the earth. The vector sum ΣF of Eq. 3/3 means the vector sum of all forces acting on the particle in question. Likewise, the corresponding scalar force summation in any one of the component directions means the sum of the components of all forces acting on the particle in that particular direction.

When work must be calculated, we may always begin with the definition of work, U insert appropriate vector expressions for the force F and the differential displacement vector dr, and carry out the required integration. With some experience, simple work calculations, such as those associated with constant forces, may be performed by inspection. We now formally compute the work associated with three frequently occurring forces: constant forces, spring forces, and weights.

With the elastic member included in the system, we now modify the work-energy equation to account for the potential-energy terms. If stands for the work of all external forces other than gravitational forces and spring forces, we may write Eq. 3/15 as (Vg) (Ve) T or (3/21) where V is the change in total potential energy, gravitational plus elastic. This alternative form of the work-energy equation is often far more convenient to use than Eq. 3/15, since the work of both gravity and spring forces is accounted for by focusing attention on the end-point positions of U 1-2 T V U 1-2 U 1-2 Ve 1 2 k(x2 2 x1 2) Ve x 0 kx dx 1 2 kx2 176 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 176 the particle and on the end-point lengths of the elastic spring. The path followed between these end-point positions is of no consequence in the evaluation of Vg and Ve.

With this definition of work, it should be noted that the component Fn F sin normal to the displacement does no work. Thus, the work dU may be written as Work is positive if the working component Ft is in the direction of the displacement and negative if it is in the opposite direction. Forces which dU Ft ds dU F dr 154 Chapter 3 Kinetics of Particles *The concept of work was also developed in the study of virtual work in Chapter 7 of Vol. 1 Statics. O A A′ F r r + dr dr α α (a) (b) α Ft = F cos ds = |dr| α ds cos F Fn Figure 3/2 c03.qxd 2/9/12 7:39 PM Page 154 do work are termed active forces. Constraint forces which do no work are termed reactive forces.

and (v2)t v2 cos 2. Note that (v1)n is a negative quantity for the particular coordinate system and initial velocities shown. The final rebound conditions are shown in part c of the figure. The impact forces are F and F, as seen in part d of the figure. They vary from zero to their peak value during the deformation portion of the impact and back again to zero during the restoration period, as indicated in part e of the figure where t is the duration of the impact interval. For given initial conditions of m1, m2, (v1)n, (v1)t, (v2)n, and (v2)t, there will be four unknowns, namely, and The four needed equations are obtained as follows:

kes place until the contact area between the spheres ceases to increase. At this instant, both spheres, Fig. 3/17b, are moving with the same velocity v0. During the remainder of contact, a period of restoration occurs during which the contact area decreases to zero. In the final condition shown in part c of the figure, the spheres now have new velocities and where must be less than All velocities are arbitrarily assumed positive to the right, so that with this scalar notation a velocity to the left would carry a negative sign. If the impact is not

ment and the time rate of change of angular momentum. To obtain the effect of the moment ΣMO on the angular momentum of the particle over a finite period of time, we integrate Eq. 3/31 from time t1 to time t2. Multiplying the equation by dt gives ΣMO dt dHO, which we integrate to obtain (3/33) where (HO)2 r2 mv2 and (HO)1 r1 mv1. The product of moment and time is defined as angular impulse, and Eq. 3/33 states that the total angular impulse on m about the fixed point O equals the corresponding change in angular momentum of m about O. Alternatively, we may write Eq. 3/33 as

ose magnitudes are negligible. In some cases, certain forces are very large and of short duration. Such forces are called impulsive forces. An example is a force of sharp impact. We frequently assume that impulsive forces are constant over their time of duration, so that they can be brought outside the linear-impulse integral. In addition, we frequently assume that nonimpulsive forces can be neglected in comparison with impulsive forces. An example of a nonimpulsive force is the weight of a baseball during its collision with a bat—the weight of the ball (about 5 oz) is small compared with the force (which could be several hundred pounds in magnitude) exerted on the ball by the bat.

overly severe and if the spheres are highly elastic, they will regain their original shape following the restoration. With a more severe impact and with less elastic bodies, a permanent deformation may result. Because the contact forces are equal and opposite during impact, the linear momentum of the system remains unchanged, as discussed in Art. 3/9. Thus, we apply the law of conservation of linear momentum and write

rved) in that direction. Consider now the motion of two particles a and b which interact during an interval of time. If the interactive forces F and F between them are the only unbalanced forces acting on the particles during the interval, it follows that the linear impulse on particle a is the negative of the linear impulse on particle b. Therefore, from Eq. 3/27, the change in linear momentum Ga of particle a is the negative of the change Gb in linear momentum of particle b. So we have Ga Gb or (Ga Gb) 0. Thus, the total linear momentum G Ga Gb for the system of the two particles remains constant during the interval, and we write

where Newton's second law ΣF m has been substituted. We now differentiate Eq. 3/29 with time, using the rule for the differentiation of a cross product (see item 9, Art. C/7, Appendix C) and obtain The term v mv is zero since the cross product of parallel vectors is identically zero. Substitution into the expression for ΣMO gives

where the equivalence Gme gR2 was established in Art. 1/5, with g representing the acceleration of gravity at the earth's surface and R representing the radius of the earth. The student should verify that if a body rises to a higher altitude (r2 r1), this work is negative, as it was in case (a). If the body falls to a lower altitude (r2 r1), the work is positive. Be sure to realize that r represents a radial distance from the center of the earth and not an altitude h r R above the surface of the earth. As in case (a), had we considered a transverse displacement in addition to the radial displacement shown in Fig. 3/6b, we would have concluded that the transverse displacement, because it is perpendicular to the weight, does not contribute to the work.

which depends only on the end points of the motion and which is thus independent of the path followed. The minus sign before dV is arbitrary but is chosen to agree with the customary designation of the sign of potential energy change in the gravity field of the earth. If V exists, the differential change in V becomes

which is positive. Conversely, if the deformation of a spring decreases during the motion interval, then the change in potential energy of the spring becomes negative. The magnitude of these changes is represented by the shaded trapezoidal area in the F-x diagram of Fig. 3/5a.

which shows that the moment-angular momentum relation holds with respect to a constant-velocity, nonrotating system. Although the work-energy and impulse-momentum equations hold relative to a system translating with a constant velocity, the individual expressions for work, kinetic energy, and momentum differ between the fixed and the moving systems. Thus, Equations 3/51 through 3/54 are formal proof of the validity of the Newtonian equations of kinetics in any constant-velocity, nonrotating system. We might have surmised these conclusions from the fact that ΣF ma depends on acceleration and not velocity. We are also ready to conclude that there is no experiment which can be conducted in and relative to a constant-velocity, nonrotating system (Newtonian frame of reference) which discloses its absolute velocity. Any mechanical experiment will achieve the same results in any Newtonian system

which states that the initial angular momentum of the particle plus the angular impulse applied to it equals its final angular momentum. The units of angular impulse are clearly those of angular momentum, which are or in SI units and lb-ft-sec in U.S. customary units. As in the case of linear impulse and linear momentum, the equation of angular impulse and angular momentum is a vector equation where changes in direction as well as magnitude may occur during the interval of integration. Under these conditions, it is necessary to express ΣMO

which, again, is the negative of the work done by the gravitational force. We note that the potential energy of a given particle depends only on its position, h or r, and not on the particular path it followed in reaching that position.

Dynamics 3

संबंधित स्टडी सेट्स

Midterm, Midterm pt2 (ling 100)

Chemistry Law of Definite and Multiple Proportions

Selenium/TestNG/Azure/Jenkins

LMFT Clinical Exam - Practice Exam VI (TDC-II) ヽ(〃·◇·)ﾉ

Chapter 31 Mastering Bio

US HISTORY II-TEST 1

JAVA INTERVIEW QUESTIONS 1-) BEHAVIORAL QUESTIONS

Sociology Exams 1 & 2

Psych 1010 Chapter 4

Microbiology Chapter 13

台語三、住址及號碼

Unit 3 Populations

Retail Operations Exam 3 - Langston 2020

Psych Assessment I

World History- Ch. 25

Types of Government

Cognitive Neuropsych

Ch. 4: Measures of Variability

Criminology Unit 2 AC2.2 INDIVIDUALISTIC THEORIES

Final Exam Review