8.09(F14) Chapter 1: A Review of Analytical Mechanics

Chapter 1 A Review of Analytical Mechanics 1.1 Introduction These lecture notes cover the third course in Classical Mechanics, taught at MIT since the Fall of 2012 by Professor Stewart to advanced undergraduates (course 8.09) as well as to graduate students (course 8.309). In the prerequisite classical mechanics II course the students are taught both Lagrangian and Hamiltonian dynamics, including Kepler bound motion and central force scattering, and the basic ideas of canonical transformations. This course briefly reviews the needed concepts, but assumes some familiarity with these ideas. References used for this course include • Goldstein, Poole & Safko, Classical Mechanics, 3rd edition. • Landau and Lifshitz vol.6, Fluid Mechanics. Symon, Mechanics for reading material on non-viscous fluids. • Strogatz, Nonlinear Dynamics and Chaos. • Review: Landau & Lifshitz vol.1, Mechanics. (Typically used for the prerequisite Classical Mechanics II course and hence useful here for review) 1.2 Lagrangian & Hamiltonian Mechanics Newtonian Mechanics In Newtonian mechanics, the dynamics of a system of N particles are determined by solving for their coordinate trajectories as a function of time. This can be done through the usual vector spatial coordinates ri (t) for i ∈ {1, . . . , N }, or with generalized coordinates qi (t) for i ∈ {1, . . . , 3N } in 3-dimensional space; generalized coordinates could be angles, et cetera. 1 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Velocities are represented through vi ≡ ṙi for spatial coordinates, or through q̇i for generalized coordinates. Note that dots above a symbol will always denote the total time derivative dtd . Momenta are likewise either Newtonian pi = mi vi or generalized pi . For a fixed set of masses mi Newton’s 2nd law can be expressed in 2 equivalent ways: 1. It can be expressed as N second-order equations Fi = dtd (mi ṙi ) with 2N boundary conditions given in ri (0) and ṙi (0). The problem then becomes one of determining the N vector variables ri (t). 2. It can also be expressed as an equivalent set of 2N 1st order equations Fi = ṗi & pi /mi = ṙi with 2N boundary conditions given in ri (0) and pi (0). The problem then becomes one of determining the 2N vector variables ri (t) and pi (t). Note that F = ma holds in inertial frames. These are frames where the motion of a particle not subject to forces is in a straight line with constant velocity. The converse does not hold. Inertial frames describe time and space homogeneously (invariant to displacements), isotropically (invariant to rotations), and in a time independent manner. Noninertial frames also generically have fictitious “forces”, such as the centrifugal and Coriolis effects. (Inertial frames also play a key role in special relativity. In general relativity the concept of inertial frames is replaced by that of geodesic motion.) The existence of an inertial frame is a useful approximation for working out the dynamics of particles, and non-inertial terms can often be included as perturbative corrections. Examples of approximate inertial frames are that of a fixed Earth, or better yet, of fixed stars. We can still test for how noninertial we are by looking for fictitious forces that (a) may point back to an origin with no source for the force or (b) behave in a non-standard fashion in different frames (i.e. they transform in a strange manner when going between different frames). We will use primes will denote coordinate transformations. If r is measured in an inertial frame S, and r0 is measured in frame S 0 with relation to S by a transformation r0 = f (r, t), then S 0 is inertial iff r̈ = 0 ↔ r̈0 = 0. This is solved by the Galilean transformations, r0 = r + v0 t t0 = t, which preserves the inertiality of frames, with F = mr̈ and F0 = mr̈0 implying each other. Galilean transformations are the non-relativistic limit, v  c, of Lorentz transformations which preserve inertial frames in special relativity. A few examples related to the concepts of inertial frames are: 1. In a rotating frame, the transformation is given by  0    x cos(θ) sin(θ) x = y0 − sin(θ) cos(θ) y If θ = ωt for some constant ω, then r̈ = 0 still gives r̈0 6= 0, so the primed frame is noninertial. 2 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Figure 1.1: Frame rotated by an angle θ 2. In polar coordinates, r = rr̂, gives dθˆ ˙ = −θr̂ dt   ˙ ˆ + r θθ ¨ˆ − θ˙2 r̂ . ¨r = r̈r̂ + 2ṙθθ dr̂ ˙ˆ = θθ, dt and thus (1.1) (1.2) Even if ¨r = 0 we can still have r̈ = 6 0 and θ¨ 6= 0, and we can not in general form a simple Newtonian force law equation mq̈ = Fq for each of these coordinates. This is different than the first example, since here we are picking coordinates rather than changing the reference frame, so to remind ourselves about their behavior we will call these ”non-inertial coordinates” (which we may for example decide to use in an inertial frame). In general, curvilinear coordinates are non-inertial. Lagrangian Mechanics In Lagrangian mechanics, the key function is the Lagrangian L = L(q, q̇, t). (1.3) Here, q = (q1 , . . . , qN ) and likewise q̇ = (q̇1 , . . . , q̇N ). We are now letting N denote the number of scalar (rather than vector) variables, and will often use the short form to denote dependence on these variables, as in Eq. (1.3). Typically we can write L = T − V where T is the kinetic energy and V is the potential energy. In the simplest cases, T = T (q̇) and V = V (q), but we also allow the more general possibility that T = T (q, q̇, t) and V = V (q, q̇, t). It turns out, as we will discuss later, that even this generalization does not describe all possible classical mechanics problems. The solution to a given mechanical problem is obtained by solving a set of N second-order differential equations known as Euler-Lagrange equations of motion,   d ∂L ∂L − = 0. (1.4) ∂qi dt ∂q̇i 3 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS These equations involve q̈i , and reproduce the Newtonian equations F = ma. The principle of stationary action (Hamilton’s principle), Z t2 δS = δ L(q, q̇, t) dt = 0, (1.5) t1 is the starting point for deriving the Euler-Lagrange equations. Although you have covered the Calculus of Variations in an earlier course on Classical Mechanics, we will review the main ideas in Section 1.5. There are several advantages to working with the Lagrangian formulation, including 1. It is easier to work with the scalars T and V rather than vectors like F. 2. The same formula in equation (1.4) holds true regardless of the choice of coordinates. To demonstrate this, let us consider new coordinates Qi = Qi (q1 , . . . , qN , t). (1.6) This particular sort of transformation is called a point transformation. Defining the new Lagrangian by ˙ t) = L(q, q̇, t), L0 = L0 (Q, Q, (1.7) we claim that the equations of motion are simply   d ∂ L0 ∂ L0 = 0. − dt ∂Q˙ i ∂Qi (1.8) Proof: (for N = 1, since the generalization is straightforward) ˙ t) = L(q, q̇, t) with Q = Q(q, t) then Given L0 (Q, Q, Q̇ = d ∂Q ∂Q Q(q, t) = q̇ + . dt ∂q ∂t (1.9) Therefore ∂Q˙ ∂Q = , ∂q̇ ∂q (1.10) a result that we will use again in the future. Then ∂L ∂L0 ∂L0 ∂Q ∂L0 ∂Q˙ = = + , ∂q ∂q ∂Q ∂q ∂Q˙ ∂q ∂L ∂L0 ∂L0 ∂Q˙ ∂L0 ∂Q = = = . ∂q̇ ∂q̇ ∂Q˙ ∂q̇ ∂Q˙ ∂q 4 (1.11) CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Since ∂Q ∂q̇ ∂L0 ∂Q in ∂q̇  ∂Q  = dtd ∂∂q̇L = 0 there is no term Plugging these results into 0 the last line. − ∂L ∂q gives      0  d ∂L0 ∂ Q ∂L0 d ∂ Q ∂ L ∂Q ∂L0 ∂Q˙ 0= + − + dt ∂Q˙ ∂q ∂Q ∂q ∂Q˙ ∂q ∂Q˙ dt ∂q    0 ∂ L0 ∂ Q d ∂L − , = dt ∂Q˙ ∂Q ∂q  (1.12) ˙ ∂ ∂ ∂Q ∂ ∂ ∂ )Q = ∂Q since dtd ∂Q = (q̇ ∂q + ∂t ) ∂q = ∂q (q̇ ∂q + ∂t so that the second and fourth terms ∂q ∂q cancel. Finally for non-trivial transformation where ∂Q 6= 0 we have, as expected, ∂q d 0= dt  ∂ L0 ∂Q˙  − ∂ L0 . ∂Q (1.13) Note two things: • This implies we can freely use the Euler-Lagrange equations for noninertial coordinates. • We can formulate L in whatever coordinates are easiest, and then change to convenient variables that better describe the symmetry of a system (for example, Cartesian to spherical). 3. Continuing our list of advantages for using L, we note that it is also easy to incorporate constraints. Examples include a mass constrained to a surface or a disk rolling without slipping. Often when using L we can avoid discussing forces of constraint (for example, the force normal to the surface). Lets discuss the last point in more detail (we will also continue to discuss it in the next section). The method for many problems with constraints is to simply make a good choice for the generalized coordinates to use for the Lagrangian, picking N − k independent variables qi for a system with k constraints. Example: For a bead on a helix as in Fig. 1.2 we only need one variable, q1 = z. Example: A mass m2 attached by a massless pendulum to a horizontally sliding mass m1 as in Fig. 1.3, can be described with two variables q1 = x and q2 = θ. Example: As an example using non-inertial coordinates consider a potential V = V (r, θ) ˙ ˆ we have in polar coordinates for
a fixed mass m at position r = rr̂. Since ṙ = ṙr̂ + rθθ T = m2 ṙ2 = m2 ṙ2 + r2 θ˙2 , giving L=  m 2 ṙ + r2 θ˙2 − V (r, θ). 2 5 (1.14) CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Figure 1.2: Bead on a helix Figure 1.3: Pendulum of mass m2 hanging on a rigid bar of length ` whose support m1 is a frictionless horizontally sliding bead For r the Euler-Lagrange equation is   d ∂L ∂L d ∂V (mṙ) − mrθ˙2 + . 0= − = dt ∂r dt ∂ṙ ∂r (1.15) This gives ∂V mr̈ − mrθ˙2 = − = Fr , ∂r from which we see that Fr = 6 mr̈. For θ the Euler-Lagrange equation is   d ∂L ∂L d  2 ˙ ∂ V 0= − = mr θ + . dt ∂θ˙ ∂θ dt ∂θ (1.16) (1.17) This gives d  2 ˙ ∂V mr θ = − = Fθ , (1.18) dt ∂θ which is equivalent to the relation between angular momentum and torque perpendicular to the plane, L˙ z = Fθ = τz . (Recall L = r × p and τ = r × F.) 6 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Figure 1.4: Particle on the inside of a cone Example: Let us consider a particle rolling due to gravity in a frictionless cone, shown in p Fig. 1.4, whose opening angle α defines an equation for points on the cone tan(α) = x2 + y 2 /z. There are 4 steps which we can take to solve this problem (which are more general than this example): 1. Formulate T and V by N = 3 generalized coordinates. Here it is  most convenient  to m 2 2 ˙2 2 choose cylindrical coordinates denoted (r, θ, z), so that T = 2 ṙ + r θ + ż and V = mgz. 2. Reduce the problem to N − k = 2 independent coordinates and determine the new Lagrangian L = T − V . In this case we eliminate z = r cot(α) and ż = ṙ cot(α), so i
mh L= 1 + cot2 α ṙ2 + r2 θ˙2 − mgr cot α. (1.19) 2
3. Find the Euler-Lagrange equations. For r, 0 = dtd ∂∂ṙL − ∂∂rL , which here is 0=
 d  m 1 + cot2 α ṙ − mrθ˙2 + mg cot α dt (1.20) giving
1 + cot2 α r̈ − rθ˙2 + g cot α = 0. For θ we have 0 = d dt  ∂L ∂θ˙  − ∂L , ∂θ (1.21) so d  2  0= mr θ̇ − 0, dt (1.22) ¨ = 0. (2ṙθ˙ + rθ)r (1.23) giving 4. Solve the system analytically or numerically, for example using Mathematica. Or we might be only interested in determining certain properties or characteristics of the motion without a full solution. 7 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Hamiltonian Mechanics ∂L In Hamiltonian mechanics, the canonical momenta pi ≡ ∂q̇ are promoted to coordinates i on equal footing with the generalized coordinates qi . The coordinates (q, p) are canonical variables, and the space of canonical variables is known as phase space. ∂L . These need not equal the kinematic momenta The Euler-Lagrange equations say ṗi = ∂q i mi q̇i if V = V (q, q̇). Performing the Legendre transformation H(q, p, t) = q̇i pi − L(q, q̇, t) (1.24) (where for this equation, and henceforth, repeated indices will imply a sum unless otherwise specified) yields the Hamilton equations of motion ∂H ∂pi ∂H ṗi = − ∂qi (1.25) q̇i = which are 2N 1st order equations. We also have the result that ∂H ∂L . =− ∂t ∂t (1.26) Proof: (for N = 1) Consider ∂H ∂H ∂H dq + dp + dt ∂q ∂p ∂t ∂L ∂L ∂L dq − dq̇ − = pdq̇ + q̇dp − dt . ∂q ∂q̇ ∂t dH = Since we are free to independently vary dq, dp, and dt this implies ∂H = − ∂L . ∂t ∂t We can interpret the two Hamilton equations as follows: • q̇i = ∂H ∂pi is an inversion of pi = ∂L ∂q̇i (1.27) (1.28) ∂L ∂q̇ = p, ∂L ∂q = ṗ, and = pi (q, q̇, t). • ṗi = − ∂H provides the Newtonian dynamics. ∂qi However, these two equation have an have equal footing in Hamiltonian mechanics, since ∂L the coordinates and momenta are treated on a common ground. We can use pi = ∂q̇ to i construct H from L and then forget about L. 8 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS As an example of the manner in which we will usually consider transformations between Lagrangians and Hamiltonians, consider again the variables relevant for the particle on a cone from Fig. 1.4: ˙ ż) L(r, θ, z, ṙ, θ, l H(r, θ, z, pr , pθ , pz ) z=r cot α −→ not here =⇒ ˙ new L(r, θ, ṙ, θ) l −→ Euler-Lagrange Eqtns. l H(r, θ, pr , pθ ) −→ Hamilton Eqtns. (1.29) Here we consider transforming between L and H either before or after removing the redundant coordinate z, but in this course we will only consider constraints imposed on Lagrangians and not in the Hamiltonian formalism (the step indicated by =⇒). For the curious, the topic of imposing constraints on Hamiltonians, including even more general constraints than those we will consider, is covered well in Dirac’s little book “Lectures on Quantum Mechanics”. Although Hamiltonian and Lagrangian mechanics provide equivalent formalisms, there is often an advantage to using one or the other. In the case of Hamiltonian mechanics potential advantages include the language of phase space with Liouville’s Theorem, Poisson Brackets and the connection to quantum mechanics, as well as the Hamilton-Jacobi transformation theory (all to be covered later on). Special case: Let us consider a special case that is sufficient to imply that the Hamiltonian is equal to the energy, H = E ≡ T + V . If we only have quadratic dependence on velocities in the kinetic energy, T = 21 Tjk (q)q̇j q̇k , and V = V (q) with L = T − V , then q̇i pi = q̇i ∂L 1 1 = q̇i T ikq̇k + q̇j Tji q̇i = 2T. 2 ∂q̇i 2 (1.30) Hence, H = q̇i pi − L = T + V = E (1.31) which is just the energy. Another Special case: Consider a class of Lagrangians given as 1 L(q, q̇, t) = L0 + aj q̇j + q̇j Tjk q̇k 2 (1.32) where L0 = L0 (q, t), aj = aj (q, t), and Tjk = Tkj = Tjk (q, t). We can write this in shorthand as 1 L = L0 + ~a · ~q˙ + ~q˙ · T̂ · ~q˙ . 2 (1.33) Here the generalized coordinates, momenta, and coefficients have been collapsed into vectors, like ~q (rather than the boldface that we reserve for Cartesian vectors), and dot products of 9 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS vectors from the left imply transposition of that vector. Note that ~q is an unusual vector, since its components can have different dimensions, eg. ~q = (x, θ), but nevertheless this notation is useful. To find H, ∂L pj = = aj + Tjk q̇k , (1.34) ∂q̇j meaning p~ = ~a + Tˆ · ~q˙. Inverting this gives ~q˙ = Tˆ−1· (p~ − ~a), where Tˆ−1 will exist because of the positive-definite nature of kinetic energy, which implies that Tˆ is a postive definite matrix. Thus, H = ~q˙ · p~ − L yields 1 H = (p~ − ~a) · Tˆ−1 · (p~ − ~a) − L0 (q, t) (1.35) 2 as the Hamiltonian. So for any Lagrangian in the form of Eq. (1.32), we can find Tˆ−1 and write down the Hamiltonian as in Eq. (1.35) immediately. Example: let us consider L = 21 mv2 − eφ + eA · v, where e is the electric charge and SI units are used. In Eq. (1.32), because the coordinates are Cartesian, a = eA, Tˆ = m1, and L0 = −eφ, so 1 H= (p − eA)2 + eφ . (1.36) 2m As you have presumably seen in an earlier course, this Hamiltonian does indeed reproduce the Lorentz force equation e(E + v × B) = mv̇. A more detailed Example. Find L and H for the frictionless pendulum shown in Fig. 1.3. This system has two constraints, that m1 is restricted to lie on the x-axis sliding without friction, and that the rod between m1 and m2 is rigid, giving y1 = 0 , (y1 − y2 )2 + (x1 − x2 )2 = `2 . (1.37) Prior to imposing any constraints the Lagrangian is m1 2 m2 2 ẋ + (ẋ + ẏ22 ) − m2 gy2 − m1 gy1 . (1.38) L=T −V = 2 1 2 2 Lets choose to use x ≡ x1 and the angle θ as the independent coordinates after imposing the constraints in Eq. (1.37). This allows us to eliminate y1 = 0, x2 = x + ` sin θ and ˙ ẏ2 = ` sin θ θ, ˙ ẋ1 = ẋ. The Lagrangian with y2 = −` cos θ, together with ẋ2 = ẋ + ` cos θ θ, constraints imposed is
m1 2 m2 2 L= ẋ + ẋ + 2` cos θ ẋθ˙ + `2 cos2 θ θ˙2 + `2 sin2 θ θ˙2 + m2 g` cos θ . (1.39) 2 2 Next we determine the Hamiltonian. First we find ∂L ˙ = (m1 + m2 )ẋ + m2 ` cos θ θ˙ , px = = m1 ẋ + m2 (ẋ + ` cos θθ) (1.40) ∂ẋ ∂L pθ = = m2 ` cos θ ẋ + m2 `2 θ˙ . ∂θ˙ 10 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Note that px is notsimply pr to ẋ here (actually px is the center-of-mass momen oportional ẋ px gives tum). Writing = Tˆ · pθ θ̇   m1 + m2 m2 ` cos θ T̂ = , (1.41) m2 ` cos θ m2 `2   1 ˙ ˆ ẋ with L = (ẋ θ) · T · ˙ + L0 where L0 = m2 g` cos θ. Computing θ 2   1 m2 `2 −m2 ` cos θ −1 , (1.42) T̂ = m1 m2 `2 + m2 `2 sin2 θ −m2 ` cos θ m1 + m2 we can simply apply Eq. (1.35) to find the corresponding Hamiltonian   1 px −1 ˆ − m2 g` cos θ (1.43) H = (px pθ ) ·T · pθ 2 h i 1 2 2 2 = m2 ` px + (m1 + m2 )pθ − 2m2 ` cos θpx pθ − m2 g` cos θ . 2m2 `2 (m1 + m2 sin2 θ) Lets compute the Hamilton equations of motion for this system. First for (x, px ) we find ∂H px cos θ pθ = , − 2 ∂px m1 + m2 sin θ `(m1 + m2 sin2 θ) ∂H ṗx = − = 0. ∂x ẋ = As we might expect, the CM momentum is time independent. Next for (θ, pθ ): h i ∂H 1 θ̇ = = (m + m )p − m ` cos θp , 1 2 θ 2 x ∂pθ m2 `2 (m1 + m2 sin2 θ) h i ∂H sin θ cos θ 2 2 2 ṗθ = − = 2 m ` p + (m + m )p − 2m ` cos θp p 2 1 2 2 x θ x θ ∂θ ` (m1 + m2 sin2 θ)2 sin θpx pθ − m2 g` sin θ − . `(m1 + m2 sin θ) (1.44) (1.45) These non-linear coupled equations are quite complicated, but could be solved in mathematica or another numerical package. To test our results for these equations of motion analytically, we can take the small angle limit, approximating sin θ ' θ, cos θ ' 1 to obtain h i px pθ 1 ẋ = − , ṗx = 0 , θ˙ = ( m + m )p − m `p 1 2 θ 2 x , m1 m2 `2 m1 `m1 i θp p θ h x θ 2 2 2 ṗθ = 2 2 m2 ` px + (m1 + m2 )pθ − 2m2 ` cos θpx pθ − − m2 g`θ . (1.46) ` m1 `m1 11 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS To simplify it further we can work in the CM frame, thus setting px = 0, and linearize the equations by noting that pθ ∼ θ̇ should be small for θ to remain small, and hence θp2θ is a higher order term. For the non-trivial equations this leaves ẋ = − pθ , `m1 pθ θ˙ = 2 , µ` ṗθ = −m2 g`θ , (1.47) where µ = m1 m2 /(m1 + m2 ) is the reduced mass for the two-body system. Thus θ¨ = ṗθ /(µ`2 ) = − mµ2`g θ as expected for simple harmonic motion. 1.3 Symmetry and Conservation Laws A cyclic coordinate is one which does not appear in the Lagrangian, or equivalently in the Hamiltonian. Because H(q, p, t) = q̇i pi − L(q, q̇, t), if qj is absent in L for some particular j, it will be absent in H as well. The absence of that qj corresponds with a symmetry in the dynamics. In this context, Noether’s theorem means that a symmetry implies a cyclic coordinate, which in turn produces a conservation law. If qj is a cyclic coordinate for some j, then we can change that coordinate without changing the dynamics given by the Lagrangian or Hamiltonian, and hence there is a symmetry. Furthermore the corresponding canonical momentum pj is conserved, meaning it is a constant through time. ∂L ∂L ∂H ∂L = 0 then ṗj = dtd ∂q̇ = ∂q = 0, or even more simply, ∂q =0 The proof is simple. If ∂q j j j j is equivalent to ṗj = 0, so pj is a constant in time. Special cases and examples of this abound. Lets consider a few important ones: 1. Consider a system of N particles where no external or internal force acts P on the center 1 of mass (CM) coordinate R = M mi ri , where the total mass M = i mi . Then the CM momentum P is conserved. This is because FR = −∇R V = 0 (1.48) P so V is independent of R. Meanwhile, T = 21 i mi ṙi2 , which when using coordinates relative to the center of mass, r0i ≡ ri − R, becomes ! ! X 1 X d 1X 1 ˙2 1X 0 2 ˙ +R ˙ · T = mi R mi ri + mi ṙ02 mi ṙ02 i = MR + i . (1.49) 2 dt 2 i 2 2 i i i P Note that i mi r0i = 0 from the definitions of M , R, and r0i , so T splits into two terms, one for the CM motion and one for relative motion. We also observe that T is ˙ independent of R. This means that R is cyclic for the full Lagrangian L, so P = M R is a conserved quantity. In our study of rigid bodies we will also need the forms of M and R for a conRtinuous body with massR distribution ρ(r), which for a three dimensional body are M = d3 r ρ(r) and R = M1 d3 r ρ(r) r. 12 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS P = Note that Ṗ = 0 is satisfied by having no total external force, so Fext = i Fext i rd 0, and by the internal forces obeying Newton’s 3 law Fi→j = −Fj→i . Hence, X X ¨ = MR Fext Fi→j = 0. (1.50) i + i i,j 2. Let us consider a system that is invariant with respect to rotations of angle φ about a symmetry axis. This has a conserved angular momentum. If we pick φ as a generalized coordinate, then L = T −V is independent of φ, so ṗφ = ∂L = 0 meaning pφ is constant. ∂φ In particular, for a system where V is independent of the angular velocity φ˙ we have pφ = X ∂T ∂ṙi X ∂ ri = mi ṙi · = mi vi · . ∂ϕ̇ ∂ϕ̇ ∂ϕ i i Simplifying further using the results in Fig. 2.2 yields X X pϕ = mi vi · (n̂ × ri ) = n̂ · ri × mi vi = n̂ · Ltotal . i (1.51) (1.52) i Figure 1.5: Rotation about a symmetry axis Note about the CM is conserved for systems with no external torque, P that L ext τ = i ri ×Fi = 0 and internal forces that are all central. Defining rij ≡ ri −rj and its magnitude appropriately, this means Vij = Vij (rij ). This implies that Fji = −∇i Vij (no sum on the repeated index) is parallel to rij . Hence, ext X X dL X = ri × ṗi = ri × Fext + ri × Fji . i dt i i i,j 13 (1.53) CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS However, P i = 0, so ri × Fext i dL X = rij × Fji = 0. (1.54) dt i arccos 32 ? Now we no longer have the constraint so d  2 ˙ 2 ˙ mr θ − mgr sin(θ) = 0. mr̈ − mrθ + mg cos(θ) = 0 and dt
2g The initial conditions are r1 = a, θ1 = arccos 23 , ṙ1 = 0, and θ˙12 = 3a from before. Simpler coordinates are x = r sin(θ) and z = r cos(θ), giving L=
m 2 ẋ + ż 2 − mgz, 2 (1.95) √ , x1 = 35a , and the initial velocities so ẍ = 0 and z̈ = −g with initial conditions z1 = 2a 3 simply left as ż1 and ẋ1 for simplicity in writing (though the actual values follow from ż1 = −a sin θ1 θ˙1 and ẋ1 = a cos θ1 θ˙1 ). This means x(t) = ẋ1 (t − t1 ) + x1 , g z(t) = − (t − t1 )2 + ż1 (t − t1 ) + z1 , 2 where t1 is the time when the mass leaves the sphere. That can be found from   2g 4g θ 2 2 θ̇ = , (1 − cos(θ)) = sin 2 a a so t1 = q a 4g R arccos( 23 ) θ0 dθ sin( θ2 ) (1.96) (1.97) (1.98) where θ0 is the small initial angular displacement from the top of the sphere. Example: Consider a hoop of radius a and mass m rolling down an inclined plane of angle φ without slipping as shown in Fig. 1.11, where we define the x̂ direction as being parallel to the ramp as shown. What is the friction force of constraint, and how does the acceleration compare to the case where the hoop is sliding rather than rolling? The no-slip constraint means aθ˙ = ẋ, so h = aθ˙ − ẋ = a, which can be made holonomic but which we will treat as semi-holonomic. Then T = TCM + Trotation = 12 mẋ2 + 12 ma2 θ˙2 as Ihoop = ma2 . Meanwhile, V = mg(l − x) sin(φ) so that V (x = l) = 0. This means L=T −V = m 2 ma2 2 ẋ + θ̇ + mg(x − l) sin(φ). 2 2 26 (1.99) CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS Figure 1.11: Hoop rolling on inclined plane The equations of motion from d dt ∂L ∂ẋ
− ∂L ∂x = λ ∂h ∂ẋ and d dt  ∂L ∂θ˙  − ∂L ∂θ = λ ∂h are ∂θ˙ mẍ − mg sin(φ) = λ and ma2 θ̈ = λa, (1.100) ˙ Taking a time derivative of the constraint gives ẍ = aθ, ¨ so mẍ = λ, and along with ẋ = aθ. g ẍ = 2 sin(φ). This is one-half of the acceleration of a sliding mass. Plugging this back in we find that 1 λ = mg sin(φ) (1.101) 2 is the friction force in the −x̂ direction for the no-sliding constraint, and also θ¨ = g sin(φ). 2a Example: Consider a wedge of mass m2 and angle α resting on ice and moving without friction. Let us also consider a mass m1 sliding without friction on the wedge and try to find the equations of motion and constraint forces. The constraints are that y2 = 0 so the Figure 1.12: Wedge sliding on ice 2 wedge is always sitting on ice, and xy11 −y = tan(α) so the point mass is always sitting on the −x2 wedge. (We will ignore the constraint force for no rotation of the wedge, and only ask about these two.) The kinetic energy is simply T = m21 (ẋ21 + ẏ12 ) + m22 (ẋ22 + ẏ22 ), while the potential 27 CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS energy is V = m1 gy1 + m2 g(y2 + y0 ), where y0 is the CM of the wedge taken from above its bottom. Then L = T − V , with the constraints f1 = (y1 − y2 ) − (x1 − x2 ) tan(α) = 0 and f2 = y2 = 0. The equations of motion from the Euler-Lagrange equations with holonomic constraints are ∂L ∂f1 ∂f2 d ∂L − = λ1 + λ2 dt ∂ẋ1 ∂x1 ∂x1 ∂x1 d ∂L ∂L ∂f1 ∂f2 − = λ1 + λ2 dt ∂ẏ1 ∂y1 ∂y1 ∂y1 d ∂L ∂L ∂f1 ∂f2 − = λ1 + λ2 dt ∂ẋ2 ∂x2 ∂x2 ∂x2 ∂f1 ∂f2 d ∂L ∂L = λ1 + λ2 − ∂y2 dt ∂ ẏ2 ∂y2 ∂ y2 =⇒ =⇒ m1 ẍ1 = −λ1 tan(α), (1.102) m1 ÿ1 + m1 g = λ1 , =⇒ m2 ẍ2 = λ1 tan(α), =⇒ m2 ÿ2 + m2 g = −λ1 + λ2 , which in conjunction with y1 − y2 = (x1 − x2 ) tan(α) and y2 = 0 is six equations. We number them (1) to (6). Equation (6) gives ÿ2 = 0 so (4) gives m2 g = λ2 − λ1 where λ2 is the force of the ice on the wedge and λ1 is the vertical force (component) of the wedge on the point mass. Adding (1) and (3) gives m1 ẍ1 + m2 ẍ2 = 0 meaning that the CM of m1 and m2 has no overall force acting on it. Additionally, as (5) implies ÿ1 = (ẍ1 − ẍ2 ) tan(α), then using (1), (2), and (3) we find the constant force g . (1.103) λ1 = 2 1 + tanm2(α) m1 cos2 (α) With this result in hand we can use it in (1), (2), and (3) to solve for the trajectories. Since tan(α) λ1 , m2 tan(α) λ1 , ẍ1 = − m1 λ1 ÿ1 = − g, m1 ẍ2 = (1.104) the accelerations are constant. As a check on our results, if m2 → ∞, then ẍ2 = 0 so indeed the wedge is fixed; and for this case, ẍ1 = −g sin(α) cos(α) and ÿ1 = −g sin2 (α) which both vanish as α → 0 as expected (since in that limit the wedge disappears, flattening onto the icy floor below it). 28