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ME6443, Test 2 Name______SOLUTION____________ Fall 2004, Dr. Ferri Instructions: This exam is closed-book and closed-notes. You can not use a calculator. The last two pages of this exam are equation sheets for your use. Honor Pledge: On my honor, I pledge that I have neither given nor received any inappropriate aid in the preparation of this test. _______________________________ Signature Problem 1 (20) ______20_________ Problem 2 (20) ______20_________ Problem 3 (20) ______20_________ Total (60) _________60________Problem 1. (20 points) A heavy cord of length L is connected from point (x1,y1,z1) to (x2,y2,z2) and rests in contact with a frictionless conical surface as shown. Under the action of gravity in the positive z-direction, the cord sags to a static-equilibrium position. Our goal is to find the location of the cord, parameterized in the form of two quantities: x = x(z) and y = y(z). Assume that the rope has a mass-per-unit length of σ. Note that a differential element of arclength of the rope would be given by: dzyxds 1)()(22+′+′= (a) If the angle of the cone is 2α, give the constraint between x(z) and y(z). (b) Set up the variational problem (functional to be minimized, any constraints that apply) that can be used to determine the shape of the cord. (c) Find the Euler-Lagrange equations for this problem. (d) Discuss how one might solve the equations including the determination of all unknown constants. Do not attempt to solve the equations. xyzcord Solution The prime objective is to minimize the potential energy: dszgσ∫− . However, we will get the same answer if we maximize the potential energy. Therefore, we define: dzyxzgdszgIzz1)()(2221+′+′==∫∫σσ 1)()(22+′+′=⇒ yxzgfσ Due to the fixed cord length, we have an isoperimetric constraint: LdzyxJzz=+′+′=∫1)()(2221 1)()(22+′+′=⇒ yxgThe isoperimetric constraint can be adjoined to the I by defining: gffµ+=*, where µ is a constant-Lagrange multiplier. In addition, there is a finite constraint: )(222zryx =+ , where r(z) is the radius of the cone as a function of z: )tan()(αzzr = . So, the constraint can be written: 0)(tan2222=−+=αzyxh This pointwise constraint can be adjoined to f* with a Lagrange multiplier, which must be a function of z: hzgfhzfF )()(*λµλ++=+= The two Euler-Lagrange Equations for this problem are given by: 0=∂∂−′∂∂xFxFzdd and 0=∂∂−′∂∂yFyFzdd These two, second-order differential equations are subject to boundary conditions, namely that the cord must pass through the points (x1,y1,z1) and (x2,y2,z2). In addition, there are two additional equations, which are needed to balance the two additional unknowns, µ and λ(z). These two equations are the isoperimetric constraint J = L and the finite constraint h = 0.Problem 2. (20 points) The simplified dynamics of a small cart can be represented by: xcx u=− + where x is the (scalar) velocity, c is a drag coefficient, and u is the driving force. Starting from rest, x(0)=0, we wish for the velocity to track a desired half-sine profile: )/sin()()( TtVtvtxdπ== The desired cost function that minimizes the difference between the desired and the actual velocity without using too much control effort is: dtvxuTxpJdTf2202)(21))((21−++=∫ Find all differential equations and boundary conditions that must be satisfied in order to find the minimizing force u. Note that vd is NOT a quantity that can be varied; it is a given and specified function. Solution Introduce comparison functions )()()( ttxtXξε+= , )()()( ttutUηε+= , where 0)0( =ξ Adjoin the constraint to the cost function: dtXcXUtvXUTXpJdTf)()()(2121))((21)(2202*&−−+−++=∫λε dtctvXUTTXpdJddTf)()()()()()(0*ξξηλξηξεε&−−+−++=∫ 0)()()()()()(00*=−−+−++=∫=dtctvxuTTxpdJddTfξξηλξηξεεε& 0)()()()()(000*=+−−+++−=∫=dtcvxuTTxpdJddTTfξλληλξλξεεε& 0 0)()()0()0()())()(()(00*=+−−++++−=∫=dtcvxuTTTxpdJddTfξλληλξλξλεεε&Set 0)( =Tξ temporarily, choose 0=+λu . Then, by the fundamental lemma of the calculus of variations, we have: 0=+−−λλ&cvxd. Now, letting )(Tξ be arbitrary, we recover the natural boundary condition: 0)()( =− TTxpfλ. Summary: 0=+−−λλ&cvxd xcx u=− + 0=+λu 0)()( =− TTxpfλ 0)0( =xProblem 3. (20 points) A mass m translates horizontally with generalized coordinate q1. The mass is resisted by a linear spring of stiffness k which is unstretched when q1 = 0. Additionally, there is a horizontal, time-varying force F(t) applied to the mass as shown. Connected to the mass is a hanging pendulum also having mass m. (a) Find the equations of motion treating the pendulum as being pinned to the mass as shown in figure (a) below. (b) Now, re-consider the problem, treating the pendulum as being separate from the mass as shown in figure (b). Introduce constraints in terms of q1, q2, q3, and q4 that require points A and A' to coincide. q1q2kF(t)mLm (a) q1q2kF(t)mAAq3q4Lm (b) Solution (a) velocity of hanging mass: ))cos()(sin(222jqiqLkqvvAmrrr&rr−×+= ; iqvAr&r1= jqLqiqLqqvmr&r&&r)sin())cos((22221++= []22222212121))sin(())cos((21212121qLqqLqqmqmvvmqmTmm&&&&rr&+++=⋅+=Kinetic Energy: 2222212121)cos( qmLqqqmLqmT&&&&++= Potential Energy: )cos(21221qmgLqkV −= (using pin A as reference height) Virtual Work: 11)()( qtFiqitFWNCδδδ=⋅=rr. Therefore, )(1tFQ = and 02=Q Take partial/total derivatives for q1: )cos(22211qqmLqmqT&&&+=∂∂; )sin()cos(22222211qqmLqqmLqmqTtdd&&&&&&−+=∂∂ 01=∂∂qT; 11qkqV=∂∂ Lagrange’s Equation for q1: )()sin()cos(21222221tFqkqqmLqqmLqm =+−+&&&&& Now, for the q2 coordinate: 22212)cos( qmLqqmLqT&&&+=∂∂; 22221212)sin()cos( qmLqqqmLqqmLqTtdd&&&&&&&+−=∂∂ )sin(2212qqqmLqT&&−=∂∂; )sin(22qmgLqV=∂∂ Lagrange’s Equation for q2: []0)sin()cos(22221=++ qmgLLqqLqm&&&& (b) Constraints: 0311=−= qqg ; 0)cos(242=−= qLqg Position of hanging mass: jqiqLqrmrrr423))sin(( −+= . Therefore, velocity is jqiqqLqtdrdvmmr&r&&rr4223))cos(( −+== .Kinetic Energy: []2422232121))cos((21212121qqqLqmqmvvmqmTmm&&&&rr&+++=⋅+= Potential Energy: 42121mgLqqkV −= Virtual Work: 11)()( qtFiqitFWNCδδδ=⋅=rr.


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