Exam_2 - Name: 2 Mar 11 EM232 Exam 2 PROBLEM 1: 20 Points...

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Unformatted text preview: Name: 2 Mar 11 EM232 Exam 2 PROBLEM 1: 20 Points Given: if the rod of negligible mass is subjected to a couple moment of M = (30t2) Nm and the engine of the car supplies a traction force of F = (15t) N to the wheels, where t is in seconds, determine the speed of the car lat the instant t = S s. The car starts from rest. The total mass of the car and rider is 150 kg. Neglect the size of the car. FIND: @ ttgfi SOLUTION: frat: .5 fl ‘3 ‘ ($31» :3}, M90“; 10,331 0 PAM“ i0 New {fl-“Amie = oh. M'VL $5; 2 t o 3% alt + gamma 0”“ (way/50:45) v; 3 9 z 5 lot [Q + 30»: )0 = éOO’vL zooo - 400 «Q [VLz 3.33'H/sl Name: 2 Mar 11 EM232 Exam 2 PROBLEM 2: 40 Points Given: The stone A used in the sport ofcurling slides over the ice and strikes another stone B as shown. If each stone is smooth and has a weight of 47 lb, and the coefficient of restitution between the stones is e = 0.8, determine each stones velocity vectorjnst after collision, Initially A has a velocity of8 fi/s and B is at rest. Neglect friction. ._~, —-, FIND: VAL tvrsl SOLUTION: B: 7 M47 Velocier 'Vecyor—S - MA!" _..\, l VA. = (~ 3cos307§ + sum 30 33 W; a t 30. —'> S) VAltééoqst Co‘s’giCiC/V‘+ 0'5: “ship-Hoyt: Omlx valid o‘low LIWL 08' lMPAL-‘F / :\ & : VBx "‘ VAfiL :3 : IVER —’\/A7< W" ‘V'Sx. (-5.413 ’V/D'O ~5M5 = m — m G) COVWCVVGU'I'OVI DS— MoMnM‘i" 33m- Hu. 575+cv~4 .‘5 comgeuvtai axlongbr'ru- GJLl/Mpac‘i' 0 MA IVAK' 4 -; MAVAXL + MrsN/Biz- I C‘ é-‘lZS’H/sx ’- VA; + ’Vrs>< Q) Z 57 '35 Z QVIIL, izm' Ma’- Mrs fl/fir = - 04993 t 5‘7: VB, " ' 6.142 575 Coytyivaom OJ, MomwM a} emu“ 5+0“ ’3 (ongvvcd A\°K$?law¢4 Cot/‘4‘“)? 4,“ I45 MAW/A7. = Mn ’VAYL juice Mass clots “0+ ckautt; / Name: 2 Mar 11 EM232 Exam 2 PROBLEM 3: [I d ’f 40 Points » ‘i Given: The 0.75 kg bob of a pendulum is fired from rest at position A by a spring which has a stiffness of k = 6 kN/m and is compresssed 125 mm. 1 4: Determine the speed of the bob and the tension in the cord when the bob is at V . h position C. ’ FIND: m as bob as me P+ a SOLUTION: MED Q, I4 X Cowutvodflow a? gagrqéj T o o F; /,’+ 'V. +(Z/l/Léfimwsr '7; +V1 WV J— L n L 1 L5. = “awn/g + M657, L N Z I 7, (£000 /w)(.125w§ : Z (O.?—5k®( 14;) + (0.?5 p907.“ “752)(16’M3 at: ‘/é.5?’.5' '- O..3$"—51/z + [3 24,, .oii‘fi/ * ‘ ’ 1 z S’Q,45«LI 1/", \fi— Vat ‘ 4.4; “4/5 I’Bb @ c W F, t ((— KCI'IL DIACSTAM rt Q, (L - ggwfit T May, ZFH:MQV1 W4 7— : gar aw : L2 P (O‘BILQMsiN/sz) +7— : (0}5KQ (won-“7,31 (’3va gL'XSS) + 7— : 56.05, “\‘N ...
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This note was uploaded on 02/09/2012 for the course MECHANICAL EM232 taught by Professor Zollars during the Spring '11 term at Naval Academy.

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Exam_2 - Name: 2 Mar 11 EM232 Exam 2 PROBLEM 1: 20 Points...

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