ME3133: DYNAMICS EXAM #1- SPRING POINTS - CLOSED BOOKFORMULA SHEET ALLOWED

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1 ME3133: DYNAMICS EXAM #1- SPRING 1 1 POINTS - CLOSED BOOKFORMULA SHEET ALLOWED CLASS ID: Instructions: 1. SET CELL PHONES OFF/SILENT -- THEY ARE NOT TO BE USED DURING EXAM. FREE BODY DIAGRAMS or ACTIVE FORCE DIAGRAMS are required in order to receive ANY credit for KINETICS problems (i.e. those involving forces in the analysis. 3. DO NOT OPEN this exam until you are told to do so. The exam pages will be separated for grading; therefore, your CLASS ID # must appear at the top of every page you want graded. You will have exactly 15 minutes (1 hour 45 minutes to complete this examination. Use only your CLASS ID on exam pages as this preserves anonymity in grading. 4. Work only on the front side of any page. Extra paper will be available from the proctor if necessary. Each extra sheet of paper should relate to one and only one exam problem. Be certain you indicate on the original question sheet that your solution is continued on another page and clearly label each sheet at the top with your Class ID # and the problem number. 5. Define all acronyms abbreviations the first time you use them in every problem. Use illustrations, figures and other pictorial representations where appropriate to assist with your explanation. This alone, however, does not constitute a complete solution unless explicitly requested as such in the exam question. Be certain you concisely explain your work if you hope to receive full or partial credit. You are expected and encouraged to utilize your technical insight and intuition in the formulation and resolution of these questions. Remember, however, that others may not share your insights; therefore, briefly explain any unusual shortcuts or assumptions you use in solving the problems. 6. If you get stuck on any problem, move on to the next one and return to it when you have completed the remainder of the exam. 7. You may use pencils, erasers, a straight edge, protractor, and/or compass to complete the examination. 8. Calculators: You may use a calculator ONLY for basic algebraic operations and trigonometric functions. NO OTHER PRE-PROGRAMMED routines are to be used in the generating the final numbers for a solution. Your instructor is more interested in your ability to devise a solution strategy than in punching keys on a calculator. Therefore, formulate the problem as far as you can using variables or symbols. Then only your last step is plug and chug or number crunching. It also localizes numerical errors to the last step. 9. Portions of the questions contained in this examination may come directly from your textbooks. Total 1. (13. (16 3. (35 4. (35 (1=99 Points + 1 Test Eval Below Test Evaluation (Circle One In Each Row (NOTE: WORTH ONE POINT ON EXAMINATION: CONTENT: Too Easy 1 (1 (1 3 (55 4 (63 5 (15 Too Difficult LENGTH: Too Short 1 (1 (5 3 (11 4 (18 5 (9 Too Long

eθ Class ID #: (1 CONCEPT QUESTIONS (13 PTS (A For each of the following statements, indicate TRUE or FALSE (Circle one.

coordinates are always horizontal/vertical: TRUE FALSE v Path coordinates are set at your discretion TRUE FALSE (B Particle P travels the curvilinear path illustrated in the figure below to the right.

For instant shown, CIRCLE one response which best describes the parameters listed in (i (iv.

2 eθ Class ID #: (1 CONCEPT QUESTIONS (13 PTS (A For each of the following statements, indicate TRUE or FALSE (Circle one. i Constant speed means zero acceleration: TRUE FALSE ii Zero acceleration means constant speed: TRUE FALSE iii Polar coordinates are not valid with projectile motion: TRUE FALSE iv Cartesian coordinates are always horizontal/vertical: TRUE FALSE v Path coordinates are set at your discretion TRUE FALSE (B Particle P travels the curvilinear path illustrated in the figure below to the right. At the instant shown, the acceleration A and position vectors are orthogonal, and the velocity vector V is as depicted. For instant shown, CIRCLE one response which best describes the parameters listed in (i (iv. i r < r = r > ii " < " = " > iii v < v = v > iv r < r = r > e r (C Particle P travels the fixed, circular path illustrated in the figure below to the right. At the instant shown, the velocity V and position vectors are aligned, and the acceleration vector A is as depicted. For instant shown, CIRCLE the one answer which best describes the parameters listed in (i (iv. i r < r = r > e r ii " < " = " > iii v < v = v > eθ iv " < " = " > ME 3133: Dynamics Spring 1 - Exam #1

3 ( SHORT ANSWER QUESTIONS (16 POINTS GIVEN: (A A jet aircraft is catapulted from the deck of an aircraft carrier by a hydraulic ram and accelerates uniformly from rest to 16 miles per hour in a distance of 3 feet. i Determine the ACCELERATION of the jet; ii SOLUTION: Determine the TIME it takes to leave the deck. (i BC s given/requested => distance, uniform acceleration, speed (s, a, v (ii v f => suggests using the alternate differential form v f s 3 " vdv = " a c ds = a c " ds s 3 1 (v f # = a c (s 3 # or formula sheet a c = v f = s 3 ( 16 mph 58 ft /mile + 36 seconds/hour* * 3 ft = 91.8 ft /s (answer Knowing the uniform acceleration and the velocity boundary conditions => a c =dv/dt " dv = a c dt t f " # v f = a c t f # t f = v f a c = 58 ft /mile 16 mph 36 seconds/hour( =.56 seconds (answer 91.8 ft /second (B Engineers analyzing the motion of a linkage locate an attachment point P which travels along straight line for a short segment and whose speed in that part of the trajectory is given by: v = A + 4 s (feet/second where A is a constant. For the instant where s= feet, the acceleration of the point is measured and determined to be a=3 (ft/sec. At this instant i Determine the SPEED of the particle P ; ii Determine the constant A ; SOLUTION: (i a=dv/dt since a(s => need to apply chain rule dv/dt = dv/ds * ds/dt = v v a = v = d dt ( A + 4s = 8ss " 3 ft /s = ft /s ( ft v " v = = ft /s (answer ft * s 16 1/s (ii Now knowing s= v = A + 4s " A = v # 4s = # 4( ft /s = 4 ft /s (answer ME 3133: Dynamics Spring 1 - Exam #1

4 (3 GIVEN: (35 POINTS The pin P is driven to the right by the constant velocity slider B (v= m/s and follows the sinusoidal slot shown in the figure. The slot is closely approximated by the following expression for y. y = (/π sin ( π x (meters For the instant when x=1/6 meter, (A The VELOCITY of the pin P; (B The ACCELERATION of the pin P; (C The RATE OF CHANGE OF SPEED ( v for the pin. SOLUTION: * instantaneous, y=f(x * Cartesian coordinates with X elements defined: position x=1/6m, v x =m/s constant, thus a x = * Constrained motion relates Y to X (position use chain rule to differentiate position constraint (dy/dt=dy/dx*dx/dt to get velocity constraint answer (A. Then apply chain rule again to get an acceleration constraint answer (B while noting that x =. y = d # dt " sin ("x( = " " sin("xcos("x x = 4sin("xcos("x x => y = 4sin(" /6cos(" /6m /s = = 3 m /s = 3.5 m /s x=1/ 6 d # y = 4sin("xcos("x x ( = 4" ( cos ("x sin ("x(x dt => y = 4" ( cos (" /6 # sin (" /6( =16" x=1/ 6 3 ( # 1 = 8" = 5 m /s ( ( (A V P = v x i + v y j = x i + y j = i + 3j m /s = i + 3.5j m /s (answer (B A P = a x i + a y j = x i + y j = i + 8" j m /s = 5j m /s (answer (C v = a t = A P e t = A P V P v = A P V P i + 3j m /s = ( 8" j m /s V P V P 1+ 3 = ( 8" * 3 /4 m /s = 4" 3 m /s = 1.8 m /s (answer ME 3133: Dynamics Spring 1 - Exam #1

5 (4 GIVEN: (35 POINTS A ball of negligible dimensions is dropped from the top of a 6 meter tower. A horizontal wind causes a horizontal component of acceleration defined by the relation a x = Ke "t where K is a constant, t is given in seconds and a x in meters per second. The ball strikes the ground 13.6 meters out from its release point. The vertical effects of air resistance are neglible as are the rotation effects of the earth. At the point where the ball strikes the ground: (A Determine the ACCELERATION of the ball. (B Determine the VELOCITY of the ball. (C Determine the ANGLE (θ at which the ball impacts the ground. 6 m! SOLUTION: * Cartesian axes set hor/vert with the origin on the ground directly beneath the drop point. * Y: a y =-g is constant, X acceleration is given as function of time. Known BC s: Initial: Position (x,y=(,6 m, Velocity (v x,v y =(, Impact: Position (x,y=(13.6, m * Integrate kinematic constraints twice Accel => Velocity => Position {Note: since a y =-g is constant, you could simply use the constant acceleration formula} Starting with A=dV/dt integrate once for the velocity and again to get the position vector. ( A = Ke!t,!g v f t f t f " dv = " Adt # V f = " (Ke t,gdt = (Ke t t f,gt = K(e t 1,gt r f ( t f t f " dr = " Vdt # r f r i = " ( K(e t 1,gtdt = K(e t t, g t ( * r i => (13.6, " (,6 # [ ] = K (e "t "1 + t 13.6 m ME 3133: Dynamics Spring 1 - Exam #1 t f = K ((e t 1 + t, g t ( * (," g t ( (meters (continued on extra sheet

6 EXTRA WORKSHEET (ONLY ONE PROBLEM PER PAGE PROBLEM # 4 Solving the y component equation for t "6 = " g t # t = 1/g = 1m /( 9.81m /s = 3.5 seconds With t known, the x equation can be used to resolve K 13.6 m = K ((e "t "1 + t # K = t= 3.5s Knowing t and K, one can now resolve the Velocity V = "K(e "t "1,"gt 13.6m ((e "3.5 " s = 13.6 = 5.37 m /s.53 ( = "5.37(e "3.5 "1,"9.81(3.5 t= 3.5s ( = 5.1,"34.3 and use K to complete the acceleration from earlier ( m /s =.16,#9.81 " A = 5.37e #3.5,#9.81 ( m /s (answer B ( m /s (answer Answer (A (C The angle θ shown and its opposite interior angle are defined by the velocity vector at impact " = # tan v #1 y #34.3 = # tan #1 =1.38radians or 79! (answer ( 5.1 ( v x ME 3133: Dynamics Spring 1 - Exam #1

MLC Practice Final Exam. Recitation Instructor: Page Points Score Total: 200.

MLC Practice Final Exam. Recitation Instructor: Page Points Score Total: 200. Name: PID: Section: Recitation Instructor: DO NOT WRITE BELOW THIS LINE. GO ON TO THE NEXT PAGE. Page Points Score 3 20 4 30 5 20 6 20 7 20 8 20 9 25 10 25 11 20 Total: 200 Page 1 of 11 Name: Section: