Variation Principle in Mechanics

Transcription

1 Section 2 Variation Principle in Mechanics Hamilton s Principle: Every mechanical system is characterized by a Lagrangian, L(q i, q i, t) or L(q, q, t) in brief, and the motion of he system is such that the action t 2 S = L(q, q, t)dt t computed between the starting point (q, t ) and (q 2, t 2 ) is minimum. Here q i represent the generalized coordinates, chosen in such a way so as to eliminate the constraints. The number of generalized coordinates is 3N C, where N is the number of particles, and C is the number of constraints. For a rigid body we need 6 generalized coordinates: 3 for the CM, and 3 for the Euler angles. Using the derivation of the Section 3., we deduce the equation of motion of the system: d dt ( q i ) = q i Law of Inertia Homogeneity of space: Every mechanical system and its properties are invariant under a space translation. Homogeneity of time: Every mechanical system and its properties are invariant under a time translation. Isotropy of space: Every mechanical system and its properties are invariant under a space rotation. Inertial frame: A reference frame can always be chosen in which space is homogeneous and isotropic, and time is homogeneous. Law of inertia holds in such a frame. Due to homogeneity of space and time, L can only depend on the velocity of the particle v. Due to the isotropy of space, it can depend only on a function of v 2 : L = L(v 2 ) () 28

2 Since L is independent of r, / v = constant, hence v = constant. This is the law of inertia. Galilean invariance: Reference frames moving with constant velocities relative to one another are equivalent. We cannot differentiate one from another. Lagrangian of a free particle In a reference frame moving with a velocity -ε with respect to the reference to that of Eq. (), the velocity of the particle would be v = v + ε. Therefore, the Lagrangian of the particle in the new frame is L = L[(v + ϵ) 2 ] = L[v 2 + 2v ϵ + ϵ 2 ] Assuming ε to be small, L (v 2 ) = L(v 2 ) + v 2 2v ϵ The additional term can be represented as df (r, t)/dt only when v 2 = 2 m = const. Therefore the additional term is mv ϵ = d [mr ϵ]. dt Thus we deduce that the Lagrangian of the free particle is L = 2 mv2, where m is the mass of the particle. Action S = 2 mv2 (t 2 t ) = m 2 (Δr) 2 (t 2 t ). For action to be minimum m > 0. Lagrangian of a system of particles Additive property for noninteracting systems implies that for N free particles, L = L a = a 2 m a v2 a Interaction among these particles is introduced using a potential function U: L = a 2 m a v2 a U(r, r 2,..., r n ) 29

3 Homogeneity of space requires that U is dependent on the distance between the particles r i r j, but not on their absolute positions. The equation of motion of the particle is m v a = U r a This is the Newton s law. Example: Two masses m and m2 coupled by a spring of a spring constant k. The Lagrangian of the system would be L = 2 m x m 2 x k(x 2 x )2 that yields the equations of motions as m x = k(x x 2 ) m x 2 = k(x x 2 ). We could also write L in terms of the CM and relative coordinate x = x 2 x as L = 2 M X 2 CM + 2 μ x 2 2 k x2 where μ is the reduced mass. The above equation yield X CM = constant and μ x = k x whose solution is with ω = x(t) = c cos ωt + c 2 sin ωt convenient. k /μ. Clearly, the later coordinates are more Sometimes, we consider a system under the influence of an external potential. L = 2 mv2 U(r, t) Example: The Lagrangian of a forced oscillator is L = 2 m x 2 2 k x2 + F 0 sin ω f t An important note: Interaction speed either infinite or a finite constant for all the reference frame. Cyclic coordinates and conservation If a variable q does not appear explicitly in the Lagrangian (but q appears), then the equation of motion yields = constant in time q Hence it is conserved. Here q is called a cyclic variable. 30

4 Example:. In the spring mass system, L is independent of X CM, hence X = m X CM = constant, which is the linear momentum of the CM CM. 2. Double pendulum: figure T = 2 m l ϕ 2 Ml 2 ϕ + m 2 l l 2 cos(ϕ ) ϕ 2 m 2 l l 2 sin(ϕ ) ϕ = Mgl sin ϕ m 2 l l 2 sin(ϕ ) ϕ m 2 l2 2 + m 2 l l 2 cos(ϕ ) ϕ + m 2 l l 2 sin(ϕ ) ϕ = m 2 gl 2 sin + m 2 l l 2 sin(ϕ ) ϕ 3. Cone problem: constraint: z = ρ cot α L = 2 m[ ρ 2 + z 2 + ρ 2 ] mgz T 2 = 2 m 2 ( x y 2 2 ) where x 2 = l cos ϕ + l 2 cos and y 2 = l sin ϕ + l 2 sin. Therefore, T 2 = 2 m 2 [l + l l l 2 cos(ϕ ) ϕ ] U = m gl cos ϕ m 2 g(l cos ϕ + l 2 cos ) Hence L = 2 Ml + 2 m 2l m 2l l 2 cos(ϕ ) ϕ ] Mgl cos ϕ m 2 gl 2 cos The equations of motion for the bobs are = 2 m[ ρ 2 cosec 2 α + ρ 2 ] mgρ cot α Since L is not an explicit function of ϕ, ϕ is a cyclic variable. Hence ϕ = mρ2 ϕ = L z = const The equation of motion for ρ is m ρcosec 2 α = mρ mg cot α or m ρcosec 2 α = L2 z mg cot α mρ3 We will solve this equation later in Chapter 5. 3

5 Constrained Mechanical Systems Examples: () Simple pendulum: We use the constraint that r = l. The Lagrangian of the pendulum is L = 2 mr2 θ2 + 2 m r 2 + mgr cos θ + λ(r l) The equations of motion of the system are m θ = g sin θ m r = λ mg sin θ + mr θ 2 r = l Note that λ is the constraint force, which is tension at present. Exercises: c. A double pendulum consisting of two bars of lengths l and l 2, and masses m and m 2. d. A vertical spring-mass system (mass m and spring constant k) whose base is being vibrated vertically with a frequency of ω f. e. A pendulum whose length is l, and whose bob s mass is bob m. Make the base of the bob oscillate (i) horizontally as a cos ω f t and (ii) vertically as a cos ω f t. 2. Suppose a system depends on the generalized coordinate q and its first two derivatives, that is L = L(q, q, q, t). Show that the equation of motion for such system is dt 2 ( q ) d dt ( q ) + q = 0 d 2 Apply this result to the Lagrangian L = 2 mq q 2 kq2.. Compute the Lagrangian for the following systems. After that derive the equations of motion and then solve them. a. Two pendulums of equal lengths (l) and masses m and m2 that are coupled together by a spring of a spring constant k. b. A bar pendulum. 32