exam1s - ME 861 Advanced Dynamics Mid-Term Exam Wednesday...

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Unformatted text preview: ME 861 Advanced Dynamics Mid-Term Exam Wednesday, March 21, 2007 Name __ ________ g. q_i_‘/@_n::--___--_____._- Instructions: You may use one 8.5” x 11” page of notes, but no other reference materials. No use of calculators is allowed. There are three problems worth a total of 90 points; they are equally weighted. Read each problem carefully and do only what is asked — this will save you some work! Indicate your methods used and show all of your work — this will help with partial credit. Clearly indicate your answers; for example, vectors must be expressed in terms of unit vectors, etc. Good Luck!! 1. (30 points total, 15 each part) Consider the system shown below, consisting of a disk that rotates at a constant rate 9 about a fixed point 0, and a mass m that slides along a diametrical slot machined in the disk. The mass is connected via a spring of stiffness k and free length 2 to the center of rotation, such that its radial displacement is given by (Z + 2:). You may ignore friction and gravity in this problem. NOTE: there a few ways to independently check your answers to this problem. (3) Use Newton’s second law to derive the equation of motion for the mass in terms of x. (b) Compute Ho, the angular momentum of m about point 0, for a general state of motion, that is, for a general a: and :t. Then use Ho = Mo to compute the normal force W acting on the mass (that is, the constraint force that keeps the mass in the slot). A J :1: i€,+&+Q “ G . - A 17) Eat N" “(1%) ,M W H” '2": garage tr m; A , p A “Q : (firx)?pxwlga+[fl+k)é€aj l“ 2*8 59 e 2., A *K "“r 5 Mafia 14 64w ° L .1. ‘(J 52:0 H =2M [“96" 5 WA 2. (30 points, 15 each part) The system shown below consists of two rigid massless bars each of length 23 connected at their midpoints by a frictionless pin, with equal masses m fixed to both ends of both bars (four masses in total). The lower two masses are restricted to move along a smooth horizontal surface. (a) Consider the case in which the system is released from rest in the configuration shown, with 0 = 90, and falls due to gravity. Determine the angular speed 6 at the instant when the upper two masses reach the surface. Assume the system remains planar throughout its motion. (b) Determine an expression for the (equal) normal force acting on each of the lower two masses. Express your result in terms of e, m, g, 6, and time derivatives of 9. , 1'0) qHern‘a-k \II' v 74/19:: Nicol? :9 N A_zfl,|/(Q 1. " - a ‘2. fl 8 Mg Bar )7 in ’l‘N fl: “:5 ”M me any Ha“ I” ’ D L‘j‘al’LL‘ ,‘f/héi “0% 60m} 6 ”Na/6'44"], 3. (30 points total, 15 each part) Consider the two degree of freedom system shown below. ass m1 moves without friction along a horizontal surface and is attached to ground via a spring of stiffness Is. An inextensible cable is attached to ground at one end and to a mass m2 at the other end, forming a pendulum as shown. The cable passes through m1 so that the length of the pendulum depends on the displacement of ml; the coordinate :1: is chosen such that for a; = 0 the pendulum length is given by Z. Use generalized coordinates :1; and 0 as shown. (a) Show that the total system kinetic energy is given by T = émlzbz + $777.22 (2i:2(1 — sin 6) + 2(e — z)éa'; c039 + (e — @262) (b) Use the T given above to derive the equation of motion for z using Lagrange’s approach. Mam-k LJVV— f‘fifi .. - . ,. 21 X ~ .' y.‘ . ° .fllvx)gt019 0L. ’1)“ «a; + m ZV‘“~“‘“93”"“” WW9 4 ( 31 V 9~Ea Mb" , MW)"; Maj @ / a, . d. . 27 : .‘m i’liéwwrllfi'gl‘be] <9 )1 v I a by : (2.“ngng @ E350 2%: \(A vecrmJ ,— 45‘1”“ 1x5 LEOWV: (D ’6) +© landn/ " \ “.6163 132 (14%?) alga £019 4. .(L'X)€960JD , (ll 99 r Ms; +V"‘1[ I ””59 :00 Wow U \ . Jr ”(my +1“: 4’“; ...
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