分析力学第一章作业解答

1.1 Spherical coordinates. Many of the problems which we will solve will possess spherical symmetry, so that the most natural coordinates are spherical coordinates. We'll need to know how to express the position, velocity, acceleration, and kinetic energy of a particle in this coordinate system.
er = θ eθ + sinθφeφ eθ = −θ er + cosθφeφ eφ = − sinθφer − cosθφeθ
(d) In spherical coordinates the position vector is r =rer . Use your results above to calculate the velocity v = r , the kinetic energy T = mv2 / 2 , and the acceleration a = r of a particle in spherical coordinates.
v = rer + rθ eθ + r sinθφeφ T = 1 m(r2 + r2 θ 2 + r2 sin2 θ φ 2 )
2
a = (r − r θ 2 − r sin2 θ φ 2 )er + (r θ + 2 r θ − r sinθ cosθ φ 2 )eθ
+(r sinθ φ + 2sinθ r φ + 2r cosθ θ φ) eφ
ex er × eθ = sinθ cosφ
cosθ cosφ
ey sz cosθ = − sin φex + cosφey = eφ − sinθ
The others are straightforward.
(c) er = (cosθθ cosθ − sinθ sinφφ)ex + (cosθθ sinφ + sinθ cosφφ)ey − sinθθ ez =θ (cosθ cosθ ex + cosθ sinφey − sinθ ez ) + sinθφ(− sinφex + cosφey ) = θ eθ + sinθφeφ
er = sinθ cosφex + sinθ sin φey + cosθ ez eθ = cosθ cosφex + cosθ sinφey − sinθ ez eφ = − sin φex + cosφey (a) Draw a careful figure which illustrates the meaning of the spherical coordinates and their relation to the rectangular coordinates. (b) Show that this is indeed on orthogonal coordinate system. In other words, show that er ⋅ eθ = 0, er ⋅ eφ = 0, eθ ⋅ eφ = 0 . Also show that each of the vectors er , eθ , eφ has a length of one. Finally, show that the coordinate system is “right-handed”, so that er × eθ = eφ , eθ × eφ = er , eφ × er = eθ (c) Use these relations to show that
Multiply both sides by r; we can then write this as d (mr2φ) = 0 dt
So that mr2φ = l , with l a constant (independent of time). If we recognize mrφ as the tangential component of the particle momentum, then mrφ is the angular momentum about the center of force. Therefore, the condition mr2φ = l is simply a statement of conservation of angular momentum about the center of force. The fact that θ = π / 2 means that the motion occurs in a plane (c) Substitute θ = π / 2 , θ = 0 , and φ = l / mr2 into m(r − rθ 2 − r sin2 φ 2 ) = f (r) ,
The others are straightforward.
(d) This closely follows the derivation in class for cylindrical coordinates. The key point to remember is that the unit vectors are time dependent, so you need to use equations above and the product rule when carrying out the differentiations. The results are
θ+
g b
sinθ
− pφ2 m2b4
cosθ sin3 θ
=0
Solution: (a) The figure is below
(b) Resolving the forces into components along the unit vectors, we have F = (−T + mg cosθ )er − mg sinθ eθ
we can get the result m r− l2 = f (r) mr 3
1.4 Spherical pendulum. A mass m is attached to the ceiling by an inextensible (i.e. fixed length), mass less string of length b, as shown in the figure.
The spherical coordinates (r,θ ,φ) are related to the rectangular coordinates (x, y, z) through x = r sinθ cosφ , y = r sinθ sinφ , z = r cosθ . The unit vectors are
1.2 Central forces. A central force is a force directed along the line connecting the two interacting particles which only depends on the separation between the particles. Assuming that the center of the force is at the origin (r = 0), then the central force can be written as F(r) = f (r) er, with r the distance from the origin. Important examples of central forces are the gravitational force, the electrostatic force (Coulomb's law), and the force exerted by a spring on a mass. (a) Assume that a particle of mass m is acted upon by a central force. The natural coordinates to describe the motion of the particle are spherical coordinates (r,θ ,φ) . Separate Newton's Second Law in spherical coordinates, and write down the three equations of motion (one for each component). (b) Show that for any f (r), a solution of theθ and φ components of these equations has θ = π / 2 and mr2φ = l , with l a constant which is independent of time. What is the physical significance of these results?
Use F = ma and the expression for acceleration in spherical coordinates along with the constraint that r = b and r = r = 0 to obtain the equations of motion
m(r − rθ 2 − r sin2 φ 2 ) = f (r) m(rθ + 2 rθ − r sinθ cosθ φ 2 ) = 0 m(r sinθ φ + 2sinθ r φ + 2r cosθ θ φ) = 0