Chapter 2bMore on the Momentum Principle2b.1 Derivative form of the Momentum Principle 1142b.1.1 An approximate result: F = ma 1152b.2 Momentum not changing 1162b.2.1 Static equilibrium 1162b.2.2 Uniform motion 1162b.2.3 Momentarily at rest vs.static equilibrium 1172b.2.4 Finding the rate of change of momentum 1182b.2.5 Example: Hanging block (static equilibrium) 1192b.2.6 Preview of the Angular Momentum Principle 1202b.3 Curving motion 1202b.3.1 Example: The Moon around the Earth (curving motion) 1212b.3.2 Example: Tarzan swings from a vine (curving motion) 1222b.3.3 The Momentum Principle relates different things 1232b.3.4 Example: Sitting in an airplane (curving motion) 1232b.3.5 Example: A turning car (curving motion) 1252b.4 A special case: Circular motion at constant speed 1262b.4.1 The initial conditions required for a circular orbit 1272b.4.2 Nongravitational situations 1282b.4.3 Period 1292b.4.4 Example: Circular pendulum (curving motion) 1292b.4.5 Comment: Forward reasoning 1312b.4.6 Example: An amusement park ride (curving motion) 1312b.5 Systems consisting of several objects 1322b.5.1 Collisions: Negligible external forces 1332b.5.2 Momentum flow within a system 1342b.6 Conservation of momentum 1342b.7 Predicting the future of complex gravitating systems 1362b.7.1 The three-body problem 1362b.7.2 Sensitivity to initial conditions 1372b.8 Determinism 1372b.8.1 Practical limitations 1382b.8.2 Chaos 1382b.8.3 Breakdown of Newton’s laws 1392b.8.4 Probability and uncertainty 1392b.9 *Derivation of the multiparticle Momentum Principle 1402b.10 Summary 1422b.11 Review questions 1432b.12 Problems 1442b.13 Answers to exercises 151114 Chapter 2: The Momentum PrincipleChapter 2bMore on the Momentum PrincipleIn this chapter we continue to apply the Momentum Principle to a varietyof systems.The major topics in this chapter are:• The derivative form of the Momentum Principle, with applications• Applying the Momentum Principle to multiparticle systems• Conservation of momentum2b.1 Derivative form of the Momentum PrincipleThe forms of the Momentum Principle given so far, and, are particularly useful when we know the momentum at aparticular time and want to predict what the momentum will be at a latertime. We use these forms in repetitive computer calculations to predict fu-ture motion. Another important form is obtained by dividing by the time interval ∆t:Just as we did in finding the instantaneous velocity of an object (see Section1.7.6), we can find the instantaneous rate of change of momentum by let-ting the time interval ∆t approach zero (an infinitesimal time interval). Theratio of the infinitesimal momentum change (written as ) to the infini-tesimal time interval (written as dt) is the instantaneous rate of change ofthe momentum. As the time interval gets very small, , the ratio of themomentum change to the time interval approaches the ratio of the infini-tesimal quantitiesand we obtain the following form:This form of the Momentum Principle is useful when we know somethingabout the rate of change of the momentum at a particular instant. Knowingthe rate of change of momentum, we can use this form of the MomentumPrinciple to deduce the net force acting on the object, which is numericallyequal to the rate of change of momentum. Knowing the net force, we maybe able to figure out particular contributions to the net force.∆pFnet∆t=pfpiFnet∆t+=∆p∆t------- Fnet=dp∆t 0→∆p∆t-------∆t 0→limdpdt------=THE MOMENTUM PRINCIPLE (DERIVATIVE FORM)In words, “the instantaneous time rate of change of the momen-tum of an object is equal to the net force acting on the object.”Or in calculus terms, since the limit of a ratio of infinitesimalquantities is called a derivative, we can say that “the derivativeof the momentum with respect to time is equal to the net forceacting on the object.”dpdt------ Fnet=2b.1: Derivative form of the Momentum Principle 115 Ex. 2b.1 At a certain instant the z component of the momentum ofan object is changing at a rate of per second. At thatinstant, what is the z component of the net force on the object? Ex. 2b.2 If an object is sitting motionless, what is the rate of changeof its momentum? What is the net force acting on the object? Ex. 2b.3 If an object is moving with constant momentum, what is the rate of change of momentum? What is the net force acting on the object?2b.1.1 An approximate result: F = maIn simple cases the derivative form of the Momentum Principle reduces toa form that may be familiar to you from a previous course. If the mass is con-stant (the usual situation), and the speed is small enough compared to thespeed of light that the momentum is well approximated by , we havethis: (nonrelativistic form; constant mass)The quantity is called acceleration, and is often given the symbol .This form of the Momentum Principle (Newton’s second law) says that masstimes acceleration (time rate of change of velocity, ) is equal to thenet force, or in simplified, scalar form “ma=F” or “F=ma”.The approximate form is not valid in situations where an object’s massisn’t constant. One example is a rocket with exhaust gases ejecting out theback; as a result, the rocket has decreasing mass. In such cases the momen-tum-based formula gives the correct results, whereas the con-stant-mass formula cannot be used. The approximate form is also not valid for objects moving at speeds closeto the speed of light, because in these circumstances the approximation is not valid. When the principle is written in terms of momentum, it is valid even forobjects whose mass changes or which are moving at speeds close to thespeed of light, as long as we use the relativistically correct definition of mo-mentum. So is the more general and more powerful form.The extra symbols on compared to the simplified versionF=ma are important:• The Momentum Principle is a vector principle, so the arrows over thesymbols are extremely important; they remind us that there are reallythree separate component equations, for x, y, and z. • It is also important to remember that we have to add up all vector forc-es to give the “net” force, so we write , not just F. Ex. 2b.4 The velocity of a 80 gram ball changes from to in 0.01 s, due to thegravitational attraction of the Earth and to air resistance. What isthe acceleration of the ball? What is the rate of change ofmomentum of the ball? What is the net force acting on the ball?4