PHYSICS 110A : CLASSICAL MECHANICS MIDTERM EXAM #2

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1 PHYSICS 110A : CLASSICAL MECHANICS MIDTERM EXAM #2 [1] Two blocks connected by a spring of spring constant k are free to slide frictionlessly along a horizontal surface, as shown in Fig. 1. The unstretched length of the spring is a. Figure 1: Two asses connected by a spring sliding horizontally along a frictionless surface. a Identify a set of generalized coordinates and write the Lagrangian. [15 points] Solution : As generalized coordinates I choose X and u, where X is the position of the right edge of the block of ass M, and X + u + a is the position of the left edge of the block of ass, where a is the unstretched length of the spring. Thus, the extension of the spring is u. The Lagrangian is then L = 1 2 MẊ Ẋ + u2 1 2 ku2 = 1 2 M + Ẋ u2 + Ẋ u 1 2 ku2. 1 b Find the equations of otion. [15 points] Solution : The canonical oenta are p X L Ẋ = M + Ẋ + u, p u L = Ẋ + u. 2 u The corresponding equations of otion are then ṗ X = F X = L X ṗ u = F u = L u c Find all conserved quantities. M + Ẍ + ü = 0 3 Ẍ + ü = ku. 4 Solution : There are two conserved quantities. One is p X itself, as is evident fro the fact that L is cyclic in X. This is the conserved charge Λ associated with the continuous syetry X X + ζ. i.e. Λ = p X. The other conserved quantity is the Hailtonian H, since L is cyclic in t. Furtherore, because the kinetic energy is hoogeneous of degree two in the generalized velocities, we have that H = E, with E = T + U = 1 2 M + Ẋ u2 + Ẋ u ku2. 5 1

2 It is possible to eliinate Ẋ, using the conservation of Λ: This allows us to write E = Ẋ = Λ u M +. 6 Λ 2 M u2 + 2M + 2M ku2. 7 d Find a coplete solution to the equations of otion. As there are two degrees of freedo, your solution should involve 4 constants of integration. You need not atch initial conditions, and you need not choose the quantities in part c to be aong the constants. Solution : Using conservation of Λ, we ay write Ẍ in ters of ẍ, in which case M ü = ku ut = A cosωt + B sinωt, 8 M + where M + k Ω = M. 9 For the X otion, we integrate eqn. 6 above, obtaining Xt = X 0 + Λt M + M + A cosωt A + B sinωt. 10 There are thus four constants: X 0, Λ, A, and B. Note that conservation of energy says E = Λ 2 2M ka2 + B Alternate solution : We could choose X as the position of the left block and x as the position of the right block. In this case, L = 1 2 MẊ ẋ2 1 2 kx X b2. 12 Here, b includes the unstretched length a of the spring, but ay also include the size of the blocks if, say, X and x are easured relative to the blocks idpoints. The canonical oenta are p X = L Ẋ = MẊ, p x = L = ẋ. ẋ 13 The equations of otion are then ṗ X = F X = L X ṗ x = F x = L x MẌ = kx X b 14 ẍ = kx X b. 15 2

3 The one-paraeter faily which leaves L invariant is X X + ζ and x x + ζ, i.e. siultaneous and identical displaceent of both of the generalized coordinates. Then Λ = MẊ + ẋ, 16 which is siply the x-coponent of the total oentu. Again, the energy is conserved: E = 1 2 MẊ ẋ k x X b2. 17 We can cobine the equations of otion to yield which yields Fro the conservation of Λ, we have M d2 dt 2 x X b = k M + x X b, 18 xt Xt = b + A cosωt + B sinωt, 19 MXt + xt = Λt + C, 20 were C is another constant. constants: A, B, Λ, and C: Thus, we have the otion of the syste in ters of four Xt = M+ b + A cosωt + B sinωt + Λt + C M + 21 xt = M Λt + C M+ b + A cosωt + B sinωt + M

4 [2] A uniforly dense ladder of ass and length 2l leans against a block of ass M, as shown in Fig. 2. Choose as generalized coordinates the horizontal position X of the right end of the block, the angle θ the ladder akes with respect to the floor, and the coordinates x, y of the ladder s center-of-ass. These four generalized coordinates are not all independent, but instead are related by a certain set of constraints. Recall that the kinetic energy of the ladder can be written as a su T CM + T rot, where T CM = 1 2 ẋ2 + ẏ 2 is the kinetic energy of the center-of-ass otion, and T rot = 1 2 I θ 2, where I is the oent of inertial. For a uniforly dense ladder of length 2l, I = 1 3 l2. Figure 2: A ladder of length 2l leaning against a assive block. All surfaces are frictionless.. a Write down the Lagrangian for this syste in ters of the coordinates X, θ, x, y, and their tie derivatives. Solution : We have L = T U, hence L = 1 2 MẊ ẋ2 + ẏ I θ 2 gy. 23 b Write down all the equations of constraint. Solution : There are two constraints, corresponding to contact between the ladder and the block, and contact between the ladder and the horizontal surface: G 1 X, θ, x, y = x l cos θ X = 0 24 G 2 X, θ, x, y = y l sin θ = c Write down all the equations of otion. Solution : Two Lagrange ultipliers, λ 1 and λ 2, are introduced to effect the constraints. We have for each generalized coordinate q σ, d L L = dt q σ q σ k j=1 λ j G j q σ Q σ, 26 4

5 where there are k = 2 constraints. We therefore have MẌ = λ 1 27 ẍ = +λ 1 28 ÿ = g + λ 2 29 I θ = l sin θ λ 1 l cos θ λ These four equations of otion are suppleented by the two constraint equations, yielding six equations in the six unknowns {X, θ, x, y, λ 1, λ 2 }. d Find all conserved quantities. Solution : The Lagrangian and all the constraints are invariant under the transforation X X + ζ, x x + ζ, y y, θ θ. 31 The associated conserved charge is Λ = L q σ q σ ζ = MẊ + ẋ. 32 ζ=0 Using the first constraint to eliinate x in ters of X and θ, we ay write this as Λ = M + Ẋ l sin θ θ. 33 The second conserved quantity is the total energy E. This follows because the Lagrangian and all the constraints are independent of t, and because the kinetic energy is hoogeneous of degree two in the generalized velocities. Thus, E = 1 2 MẊ ẋ2 + ẏ I θ 2 + gy 34 Λ 2 = 2M I + l 2 M+ l2 sin θ 2 θ2 + gl sin θ, 35 where the second line is obtained by using the constraint equations to eliinate x and y in ters of X and θ. e What is the condition that the ladder detaches fro the block? You do not have to solve for the angle of detachent! Express the detachent condition in ters of any quantities you find convenient. Solution : The condition for detachent fro the block is siply λ 1 = 0, i.e. the noral force vanishes. 5

6 Further analysis : It is instructive to work this out in detail though this level of analysis was not required for the exa. If we eliinate x and y in ters of X and θ, we find We can now write x = X + l cos θ y = l sin θ 36 ẋ = Ẋ l sin θ θ ẏ = l cos θ θ 37 ẍ = Ẍ l sin θ θ l cos θ θ 2 ÿ = l cos θ θ l sin θ θ λ 1 = ẍ = Ẍ l sin θ θ l cos θ θ 2 = MẌ, 39 which gives and hence We also have M + Ẍ = l sin θ θ + cos θ θ 2, 40 Q x = λ 1 = M + l sin θ θ + cos θ θ Q y = λ 2 = g + ÿ = g + l cos θ θ sin θ θ We now need an equation relating θ and θ. This coes fro the last of the equations of otion: I θ = l sin θ λ 1 l cos θλ 2 = M M+ l2 sin 2 θ θ + sin θ cos θ θ 2 gl cos θ l 2 cos 2 θ θ sin θ cos θ θ 2 = gl cos θ l 2 1 M+ sin2 θ θ + M+ l2 sin θ cos θ θ Collecting ters proportional to θ, we obtain I + l 2 M+ sin2 θ θ = M+ l2 sin θ cos θ θ 2 gl cos θ. 44 We are now ready to deand Q x = λ 1 = 0, which entails θ = cos θ sin θ θ Substituting this into eqn. 44, we obtain I + l 2 θ2 = gl sin θ. 46 Finally, we substitute this into eqn. 35 to obtain an equation for the detachent angle, θ Λ 2 E 2M + = 3 M + l 2 I + l 2 sin2 θ 1 2 gl sin θ. 47 6

7 If our initial conditions are that the syste starts fro rest 1 with an angle of inclination θ 0, then the detachent condition becoes sin θ 0 = 3 2 sin θ 2 1 l 2 M+ sin 3 θ I+l 2 = 3 2 sin θ 1 2 α 1 sin 3 θ, 48 where α 1 + M 1 + I l Note that α 1, and that when M/ = 2, we recover θ = sin sin θ 0. For finite α, the ladder detaches at a larger value of θ. A sketch of θ versus θ 0 is provided in Fig. 3. Note that, provided α 1, detachent always occurs for soe unique value θ for each θ 0. Figure 3: Plot of θ versus θ 0 for the ladder-block proble eqn. 48. Allowed solutions, shown in blue, have α 1, and thus θ θ 0. Unphysical solutions, with α < 1, are shown in agenta. The line θ = θ 0 is shown in red. 1 Rest eans that the initial velocities are Ẋ = 0 and θ = 0, and hence Λ = 0 as well. 2 I ust satisfy I l 2. 7