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Our cylinder has radius c and height Now change to polar coordinates, so that At this point, two of the tangent vectors to the surface of the Therefore, the total force acting on the particle is r rF ? eThe acceleration of the particle is If the time it takes to cycle Describe one of If she were late This time is given by ( ) 030 m 0.68 s v. The vector expression is: This means that the bale This solution is what is The time error she is allowed to make is the same as in the previous No air resistanceTherefore, 1 x 2 xThe equation of motion for this case is: Using the given values, the plots are shown in the figure. Of course, the reader will not be able The reader will have to So we find the highest point by A C u uUsing Eq. (E.17a), Appendix E, we obtain This minimum is No air resistanceThis trajectory is shown in Figure (b). The distance from the point of projection is 0Therefore, the maximum value of.Assume a coordinate system in which the projectile moves in the 2 x x 3? plane. Then, Now, the force acting on the projectile is This is a centripetal In this case we have also so that ? v BRewriting this as component equations: It is found upon substitution into (6), however, We may now evaluate all the C ’s and. This gives us Showing this explicitly, Using this equation with the initial and final points being the top and bottom of the incline So The track counters the gravitational force and provides centripetal acceleration N mgSubstituting into (2): After 2 m of floor, the block has energy. Use of the quadratic formula gives The total amount of work that needs to be W dz g z zWe may therefore use the elementary Conservation of momentum gives Integrate again: Using the quadratic formula, we get: Proceeding from there we have The equation of motion is (calling downward positive). The forces on the other mass are FfBoth the ping-pong ball and Since the baseball has a higher Take the y -axis to be positive downwards.The equations for the x and y motion are then The y equation then gives The optimum ? for a given x is plotted in the figure, along with its F t mna n atIt follows that That N is normal to T follows from A tdvUsing na taWe therefore have an equilibrium at U x F dx kx kWe find the x values by setting 1 E 0Let 1 1 1 x x ? ? ? and 2 2 2 x x 0? ? ?. In the calculations, Equation (3) then The equilibrium is therefore stable, when it U x xF x xF d Ft v tF cF xU U vU U a xx USo if F is conservative, its components satisfy the following relations U bxFrom the second equality of (1), we have We can eliminate ft, and obtain an Therefore, On the other hand, when T and T tU tThen, by taking time average of (9), we U kx dx xT A xT m AIf we notice that the reduced mass of the system is defined as 1 xThe force responsible for the motion of the pendulum bob is the component of the gravitational It is convenient to define Ce De tSince the amplitude decreases to 1 after n periods, we have eAn exploded (or From this equation, we see that the motion is not bounded, irrespective of the relative values of This gives 0 2 Q ? ?. To obtain the total energy, we use the We then have We also can approximate R.The dashed line is the path that all paths go to asymptotically as When damping is high, The amplitude of These figures are shown in figure (a).Equation (3.43) is used instead of (3.40) as the complementary solution. The distortion due to The latter fact is because there is no resonance in this case. F t a c n t ) n ? ?In fact, from (4), Analytically, such a change of scale can be F t a bF t tThus, we add these H tThus, if we write 22 0Since the oscillator is undamped, and since the impulse A B eA B ?? ?A B A B e ?? ?A aD e e ?? ? ?? ?This can be insured by using as a solution for H tH tG tF tF t a a n t b n t ? ?F t t n tA xUsing initial conditions of x ( t ) and v ( t ), we find A xThe most general solution is Then the effective potential is (see for example. Chapter 2 and Equation (8.14)): From Equation (3.40) we have In order to attach the mass m, each When the mass is moved a distance x, F x x x dF x kxThe substitution, If the result is not within this amount of the starting This procedure The potential energy as a function of.Solving for ? and taking time derivatives, we obtain When written in terms of, the above equation becomes (with Therefore, from (2) we have U xThis occurs faster for some values of ? than The box encloses the In addition, we also show This value is approximate, however, and depends on the The behavior returns to a three-cycle near the They are created with the logistic The first plot has the seed value as asked for in the text. Only one additional seed has Here we use the Its accuracy is limited by the fact that the map does not converge very rapidly near the Equation (4.47) to compute the Feigenbaum constant, we can obtain the following: This is a perfectly Using their algorithm, one obtains the following: The above values are only good to about Another alternative in computing the. Feigenbaum constant, which is not requested in the problem, is to use the so-called Frontiers of Science by Peitgen, Jurgens and Saupe. As a result, the estimates for.In this problem the initial For the case. The phase space plot (line) and. Poincare section (boxes) for this case are overlaid and shown in figure (a). All integrations are One can experiment with B and determine that the If one further experiments with The second plot Circumventing the actual task of computing where these transition points are, we do know that What look to be phase paths in the figures are actually just different points that come from For example, in figure (a), an ellipse about the origin (just Above the ellipses is chaotic orbit, However, if then the This situation is similar to that for two This fact implies G dr mU F dx mkUsing Eq. (E.9), Appendix E, we find Therefore, the time for the particle to travel from x ? x ? to x T dtT dtI dThe contribution to the force from a The contribution from a small G a z z zR R RR a aRR sin ?Since the average value of the potential The potential d ? due to a small amount of mass dm Writing this out with the help of the figure, we have RR r Rur ? ?The force is in the radially inward The equation of motion is F mr rThe period is F dFF GMhThe expression for and xg yg are given by The origin of the coordinate system is at the center Equation (5.55). Turning to the exact solution of (3), we obtain Inclusion of the centrifugal term in does not change this answer significantly. xgThus S dxBecause ? is a S dx dyThen the total distance is Substituting these constants into (7), we find 1 C 2 3The curve described by this Rearranging gives Alternatively, Fermat’s principle can be proven by the method This may be extended to several variables and to The result is Then, the equations for the solution are Combining (7) with (1) and The radius of curvature of a parabola, The Euler equations with undetermined multipliers (6.69) tell us that Eliminating the factor ?, we obtain This looks to be in the simplest form we can make it, but is it a plane. Take the equation of a The substitution yields (7) exactly.