Review Louisiana State University Qualifier Exam Jeff Kissel October 22, 2006 A particle of mass m. at rest initially, slides without friction on a wedge of angle θ and and mass M that can move without friction on a smooth horizontal surface. Figure 1: A mass m slides in the +ˆx direction down a frictionless wedge M, which can slide across a frictionless plane in the ˆx direction. Note, this picture does not represent the actual motion; realistically Ẋ is in the ˆx direction as will be shown later. a What is the Hamiltonian? First we must find the Lagrangian L = T V, to determine whether it is explicitly independent of time. If this is so, then the Hamiltonian is simply the total energy H = T + V. We can define a coordinate system where the origin is the top of the wedge at some initial time as in Figure 1, such that the position vector for the top corner of the wedge at time, t will be and particle s position vector will be R = X ˆx + 0 ŷ 1 r = R + x ˆx + y ŷ r = X + l cosθ ˆx + l sinθ ŷ 2 1
where l is the distance the mass has traveled down the ramp. While the particle is on the ramp, it is constrained to move along l, or = y x sin θ = y Y cosθ x X x Xsinθ = y 0cosθ f : x Xsinθ y cosθ = 0 3 which is an equation of constraint related to the normal force. The kinetic energy of the system is that of two free particles, since there is no friction: T = 1 2 mẋ2 + ẏ 2 + 1 2 MẊ2 4 The only potential in the system is gravitational, with g = g ŷ. We can ignore the potential on the wedge because it is constant, so the system s potential is just the gravitational potential of the particle, V g m g r V = mgy 5 The Lagrangian is then potential subtracted from the kinetic energy, L T V = 1 2 mẋ2 + ẏ 2 + 1 2 MẊ2 + mgy L = 1 2 mẋ2 + 1 2 mẏ2 + 1 2 MẊ2 + mgy 6 which does not explicitly depend on time, so the Hamiltonian is the sum of the kinetic and potential energies, H = 1 2 mẋ2 + 1 2 mẏ2 + 1 2 MẊ2 mgy 7 b Derive the equation of motion from the Lagrangian. With Lagrange multipliers, The Euler-Lagrange equation in general is d L L = λ f + µ f 8 dt q i q i q i q i yet the second term on the right-hand side is zero because Eq. 3 is not a function of any q i. From Eq. 8, the equations of motion for each coordinate are, d mẋ = λ sin θ dt 2
mẍ = λ sin θ 9 d mẏ mg dt = λ cosθ mÿ mg = λcosθ 10 d mẋ dt mẍ = λ sin θ = λsin θ 11 Immediately, if we add Eq. 11 to Eq. 9, we see why Figure 1 has the direction of motion for the wedge incorrect, mẍ + MẌ = 0 ẍ = M m Ẍ 12 Also, this tells us that the ˆx position of the system s center of mass stays stationary because the only external force on the system is gravity, which is in the ŷ direction. We ve assumed that the particle s initial position is at the origin, but if we also assume that the initial velocity of the center of mass is also zero then from Eq. 12, x = M X t t m t t x t t = M X t m t x 0 x = M m X 0 X x = M m X or X = m M x 13 giving us an equation relating X in terms of x. We can use the equation of constraint Eq. 3 to get y in terms of x, x Xsinθ = y cosθ x + m xsin θ = y cosθ M from which only y and x are a function of time, so y = x1 + m 14 M ẏ = ẋ1 + m M ÿ = ẍ1 + m 15 M 3
OK, OK, let s finally get the equations of motion. Starting with Eq. 9, mẍ = λsin θ mÿ g Eq. 10: mÿ mg = λcosθ λ = cosθ mẍ = mg ÿ 16 Eq. 15: ÿ = ẍ1 + m M mẍ + mẍtan 2 θ + mẍ m M tan2 θ = mg tan 2 θ m1 + tan 2 θ + m M tan2 θẍ = mg cos2 θ cos 2 θ + sin2 θ cos 2 θ + m M mẍ = mg mẍ1 + m M tan2 θ sin 2 θ sin θ cos 2 ẍ = g θ cosθ 1 cos 2 θ + sin 2 θ + m/msin 2 θ ẍ = g sinθ cosθ ẍ = g sin θ cosθ 1 + m/msin 2 θ ẍ = M sinθ cosθ g 17 Since the acceleration in the ˆx direction is a constant in time, a x Eq. 17 has the usual 1-D kinematic solution, xt = x 0 + v 0,x t + 1 2 a xt 2 set x 0 0, and v 0,x 0 xt = 1 2 g M sin θ cosθ t 2 18 For the ŷ equation of motion, we plug Eq. 17 back into Eq. 15, ÿ = ẍ1 + m M M sin θ cosθ = g ÿ = g M + msin2 θ 1 M M + m sinθ cosθ 19 which is also constant, so with the same initial conditions y 0 0, y 0,y 0 we get a similar free-fall equation, y = 1 M + msin 2 2 g θ M + m sin 2 t 2 20 θ 4
Finally we get the solution for X plugging Eq. 17 into Eq. 12, Ẍ = m M ẍ = m M g M sinθ cosθ m sin θ cosθ Ẍ = g X 0 0, and v 0,X 0 X = 1 m sin θ cosθ 2 g M + m sin 2 t 2 21 θ One could also solve for the normal force λ via Eq. 9, but the problem does not ask for it, so it is left as an exercise to the reader. Ew. I feel so dirty saying that. c What are the constants of motion? To find the constants of motion, i.e. conserved quantities, we need to express the Lagrangian Eq. 6, in terms of independent coordinates. We can use the constraint equation Eq. 3 to solve for X, x Xsinθ y cosθ = 0 x X y tan θ = 0 X = x y Ẋ = ẋ ẏ L = 1 2 mẋ2 + 1 2 mẏ2 + 1 2 MẊ2 + mgy L = 1 2 mẋ2 + 1 2 mẏ2 + 1 2 M ẋ ẏ 2 + mgy 22 tan θ Because this Lagrangian is explicitly independent of x, then the L x term in the Euler-Lagrange equation will be zero. Thus, the canonical momentum in the ˆx direction is a conserved quantity. p x = L ẋ p x = mẋ + M ẋ ẏ 1 = m + Mẋ Mẏ 23 General Rule: If the Lagrangian is independent of any generalized coordinate, that coordinate is cyclic, and therefore it s respective canonical momentum p i L/ q i is conserved. 5
Also, the Lagrangian does not explicitly depend on time, so as previously stated, the total energy is conserved which is why we could write the Hamiltonian as the sum of kinetic and potential energies. d Describe the motion of the system. Figure 2: An alternate view of the particle-wedge system at times t = 1, 2, 3 and 4. Since friction is not involved in the system, the ˆx coordinate of the center of mass between the two objects will remain stationary as has been shown in part b. The ŷ coordinate will accelerate as though in slow free-fall a la Eq. 20, until the particle reaches the surface on which the wedge slides. From then on, it will simply travel according to Eq. 18. Thus, as in Figure 2 from a reference point following the center of mass, the wedge and particle will fly apart from each other. If M m, it will not move much Ẍ 0, and the system approximates that of a mass sliding down a fixed incline. 6