ESM 3124 Intermediate Dynamics 2012, HW6 Solutions. (1 + f (x) 2 ) We can first write the constraint y = f(x) in the form of a constraint

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1 ESM 314 Intermediate Dynamics 01, HW6 Solutions Roller coaster. A bead of mass m can slide without friction, under the action of gravity, on a smooth rigid wire which has the form y = f(x). (a) Find the equation for ẍ as a function of x and ẋ. y g y=f (x) (b) Find the expression for the Lagrange multiplier λ 1, related to the force maintaining the sliding constraint, as a function of x and ẋ. (c) Find the expression for the magnitude f c of the force of constraint, as a function of x and ẋ. Hint: the answers to (a) and (b) are ẍ = f (x)(f (x)ẋ + g) (1 + f (x), λ 1 = m(f (x)ẋ + g) ) m x Solution Lagrange Multiplier Method. equation equal to zero, We can first write the constraint y = f(x) in the form of a constraint ψ 1 (x, y) = f(x) + y = 0. (1) We can use the Lagrange multiplier method to write the equations of motion. This involves writing the Lagrangian function L = T V = 1 m ( ẋ + ẏ ) mgy () and then writing the Lagrange s equation of motion in the presence of one constraint, ( ) d L L = λ 1 a 1i, (i = 1, ), (3) dt q i q i where q 1 = x, q = y, and λ 1 is the Lagrange multiplier corresponding to the one constraint. Using the coefficients a 11 = ψ 1 x = f (x) and a 1 = ψ 1 y = 1, we get two equations of motion, Using these two equations to eliminate λ 1, we end up with mẍ = f (x)λ 1, mÿ + mg = λ 1. (4) ẍ + f (x)ÿ + f (x)g = 0. (5) where we have divided through by m and rearranged the order of the terms. Getting ẍ equation. To get an equation for ẍ in terms of x and ẋ, we need to write ÿ in terms of only x and its time derivatives. To do this we start with the constraint y = f(x) and take two time derivatives so we have an equation for ÿ, y = f(x) ẏ = f (x)ẋ ÿ = f (x)ẋ + f (x)ẍ. (6) Plugging this expression for ÿ into eq. (5), we get ẍ = f (x)(f (x)ẋ + g) (7) 1

2 Notice the right hand side is just a function of x and ẋ. Alternate method for getting ẍ. Going back to the Lagrangian function (), we can directly write this in terms of only x and ẋ, so x is the only generalized coordinate and only degree of freedom. L = 1 m ( ẋ + ẏ ) mgy ( (dx = 1 ) ( ) ) dy m + mgf(x) dt dt = 1 ( ) ( dx ( ) ) dy m 1 + mgf(x) dt dx (8) L(x, ẋ) = 1 mẋ ( 1 + (f (x)) ) mgf(x) and then write the Lagrangian equation for x, as usual, to recover (7). This method is more direct for getting the equation for ẍ since it skips the need for the Lagrange multiplier. Solving for the Lagrange multiplier and constraint force. of (4), we can solve for λ 1 and we get λ 1 = m(f (x)ẋ + g) Using (7) in the left-side equation (9) If we want to know the ( magnitude of) the constraint force, we proceed as follows. The constraint force is F c = λ 1 ψ 1 = λ ψ1 1 x n 1 + ψ 1 y n. Taking the norm of this vector, we get [ ψ1 ] F c = F c = λ 1 ( + x [ ] ) ψ1 y = λ 1 (f (x) + 1) ( m(f (x)ẋ ) + g) (f = (x) + 1) (10) so F c is F c = m(f (x)ẋ + g) 1 + f (x) (11)

3 θ = 0, φ = 0 Problem Find the equation of motion for a situation addressed in an earlier homework, but now include Coulomb friction; i.e., the walls of the tube are rough so there s a friction coefficient µ, where 0 < µ < 1, between the bead and the tube walls. Make no assumptions about the value of θ, although you can still assume θ is fixed. µ g m e r e θ e φ Hint: You should find that the equation of motion for r is m r mrω sin θ + mg cos θ = µnsgn(ṙ) where sgn(ṙ) = ṙ /ṙ and the normal force N of the tube acting on the bead has magnitude N = Nθ + N φ = m[4ω ṙ + (rω cos θ + g) ] 1/ sin θ θ r Figure.9. Solution Write this as a constrained Lagrangian problem, where our coordinates are the usual n = 3 spherical coordinates (r, θ, φ), but we have two constraints, i.e., number of constraints, n c, is, θ = 0 φ Ω = 0 (1) which are in the nonholonomic form, n i=1 a ji(q, t) q i + b j (q, t) = 0, so we can read off the a and b coefficients, θ = 0 = a 1r = 0, a 1θ = 1, a 1φ = 0, b 1 = 0 φ Ω = 0 = a r = 0, a θ = 0, a φ = 1, b = Ω We need to write the unconstrained Lagrangian. The unconstrained velocity is what one expects for spherical coordinates, Ṙ = ṙê r + r θê θ + r φ sin θê φ (13) so the unconstrained kinetic energy of the particle is, T = 1 m Ṙ = 1 m(ṙ + r θ + r φ sin θ). (14) The potential energy is due to gravity alone. Using the origin of the spherical coordinate system as the reference height, we have V = mgr cos θ The Lagrangian is: L = 1 m(ṙ + +r θ + r φ sin θ) mgr cos θ The Lagrangian equations of motion for a constrained system are, ( ) d dl dl = Q nc,i + C i, i = 1,, 3 (15) dt d q i dq i where equivalently, we can view the indices i as being in the range {r, θ, φ}, and where the generalized forces of constraint, C i, can be written in terms of Lagrange multipliers n c C i = λ j a ji (16) j=1 3

4 and also in terms of the Newtonian (i.e., physical) forces of constraint on the particle, f c, C i = f c Ṙ q i, (17) where, for this system, the force of constraint is equal to the reaction force of the tube walls on the particle, so f c = N, which has components N θ, and N φ, i.e., N = N θ ê θ + N φ ê φ. The non-conservative Newtonian force applied to the mass is Coulomb friction, f nc = µn sgn(ṙ)ê r, where N = N, so the generalized non-conservative forces are, Q nc,r = f nc Ṙ ṙ = f nc ê r = µn sgn(ṙ) Q nc,θ = f nc Ṙ θ = f nc rê θ = 0 Q nc,φ = f nc Ṙ φ = f nc r sin θê φ = 0 At this point, most of the hard work is done. Now we just work out the derivatives and do some algebra. Remember, we do not yet apply the constraints. The Lagrange s equations (15) give us, m r mr θ mr sin θ φ + mg cos θ = µn sgn(ṙ) mr θ mr sin θ cos θ φ mgr sin θ = λ 1 mr sin θ φ + mrṙ sin θ φ + mr sin θ cos θ θ φ = λ Applying the constraints (1), this reduces to m r mr sin θω + mg cos θ = µn sgn(ṙ) mr sin θ cos θω mgr sin θ = λ 1 mrṙ sin θω = λ (18) From (16)-(17), we can get the relationship between the λ s and the components N θ, and N φ of the constraint force N. Using (16), we have C θ = λ 1 C φ = λ (19) but using (17), we have C θ = N Ṙ θ = (N θê θ + N φ ê φ ) rê θ = rn θ C φ = N Ṙ φ = (N (0) θê θ + N φ ê φ ) r sin θê φ = r sin θn φ So, the magnitude of N can be written in terms of the Lagrange multipliers N = and using the last two equations of (18), this gives N θ + N φ = ( λ1 r ) + ( λ r sin θ ) (1) N = [(mr sin θ cos θω + mg sin θ) + (mṙ sin θω) ] 1/ = m sin θ[4ω ṙ + (rω cos θ + g) ] 1/ () We therefore obtain the r equation of motion from the first equation of (18) m r mrω sin θ + mg cos θ = µnsgn(ṙ) where N is written completely in terms of r and ṙ via (). 4

5 Problem 3 A dumbbell consists of two particles, each of mass m, connected by a massless rigid rod of length l, and moving in the horizontal plane. Assume there is a knife-edge constraint at particle 1, which constrains the velocity of particle 1 to be along the instantaneous axis of the dumbbell. Think of it this way; there s a wheel or ice skate underneath particle 1 which is along the dumbbell axis. Find the differential equations of motion. 1 There are 4 unknowns (ẍ, ÿ, θ, λ 1 ), where λ 1 is the Lagrange multiplier related to the knife-edge constraint. So you should get 4 equations. [Hint: begin by writing the velocity constraint equation in the standard form. ] Solution. The positions of particles 1 and are, respectively, Their velocities are: The unconstrained Lagrangian is: r 1 = xˆn 1 + yˆn r = (x + l cos θ)ˆn 1 + (y + l sin θ)ˆn r 1 = ẋˆn 1 + ẏˆn r = (ẋ l θ sin θ)ˆn 1 + (ẏ + l θ cos θ)ˆn L = T V = T 0 The constraint imposed by the knife-edge is: = m(ẋ + ẏ + ẏ θl cos θ ẋ θl sin θ + 1 l θ ) r 1 ê θ = 0 = ẋ cos θ + ẏ sin θ = 0 (3) This is a nonholonomic constraint. Writing it in the standard form, we see that n a ji (q, t) q i + b j (q, t) = 0, j = 1,..., n c = a 1x = sin θ, a 1y = cos θ, a 1θ = 0, b 1 = 0, i=1 since the number of constraints, n c = 1. The Lagrange s equations of motion are ( ) d L L dt ẋ x = λ 1a 1x Equations (4) and (3) give: d dt d dt ( ) L ẏ ( L θ L y = λ 1a 1y ) L θ = λ 1a 1θ ẍ θ sin θ l θ cos θ = λ 1 m sin θ ÿ + l θ cos θ l θ sin θ = λ 1 m cos θ l θ ẍ sin θ + ÿ cos θ = 0 ẋ cos θ + ẏ sin θ = 0 (4) 5

6 OPTIONAL, EXTRA CREDIT Two masses, m 1 and m, are connected by a string: m 1 moves on the horizontal table, while m hangs suspended by the string below the table. The string passes through a hole in the center of the table. (See sketch). You may assume that m moves only in the vertical direction, and that neither m 1 or m pass through the hole. We ve seen this problem in class before, but now I want you to tackle it using a Lagrangian formulation of the problem. (a) Reduce the equations of motion to a single second-order differential equation in the variable r (distance from hole to m 1 ). m 1 g (b) This second-order equation can be interpreted as an equivalent 1-D system with an effective potential V eff (r). Find V eff (r), and make a qualitative sketch of V eff (r). m (c) What is the condition for m 1 to move in a circle? Hint: think in terms of V eff (r). Solution The Lagrangian for this system is: Since L θ L = 1 m 1(ṙ + r θ ) + 1 m ṙ + m g(l r) = 0, we see that L θ = const. = m 1r θ = H. Hence the required EoM is: (m 1 + m ) r = m 1 r θ m g The effective potential is defined as V eff = rdr. Hence, V eff = gr 1 r θ The condition for the mass m 1 to move in a circle is that both ṙ = 0 and r = 0. Thus, r 3 = H m 1 m g 6