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e1_02m_s - ME 364 Kinematics Summer Term 2002 Exam 1 100...

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Unformatted text preview: ME 364 - Kinematics Summer Term, 2002 Exam 1, 100 Points, Open Book NAME 501,0 Z [04; WW ’1 l. (15 Points) The quick return mechanism is show one half of full siz/. In the position shown, the slider 6 is moving to the right at a speed of 10 inches per s We the required angular velocity of the crank (link 2) and the resulting angular velocities of links 4 and 5 V004: Ale/mmfl‘ 4 cu ; EM : ‘W/ M? 4 a m WM (cw/67 VMNWW’” 5 Va Q57 ; 46,- n t, w; 0 , Um .M7 015% MAE 364 - Kinematics, Summer,2002 2. (20 Points) Determine the locations of all of the instant centers in the mechanism below. Clearly identify the location and identity of each one. Location accuracy is part of the grade. 6% ¥@ 3. (10 Points) The disk shown is made from aluminum ( p= 0.101b/in3). Determine the mass moment of inertia of the member about the axis GG (going through the mass center). ‘ J 1,—— gmz‘: {(4,444}: -— may; L:%H Naif Ind) 7,1” 95/475372”) + ”62705) MMM +,/¢7// : ,/5')>7 H M: “1-7/0“: 4,Jq/' 9/,é’6é2/ 33,732/j 4. (7 Points) If the disk in the previous problem is suspended on a knife edge through one of its holes, what will the period of oscillation be? 0’, 7; V: / 975’ /’7 A 10 : If M V 4: 6,77 I— 3.774 (Ar/75') = 301,993 I: 3.}, awry/é: m :ol/7(/0::) « 667 T: (679% MAE 364 - Kinematics, Summer,2002 Exam 1 Page 3 5. (25 Points) The angular velocity of link 2 is 2.50 radians per second clockwise. Determine the angular velocity of links 4 and 6. The mechanism is shown Lalf of full size. Work Here: (l/Mfijé ('5 A [two/27 {/5261 “” Pam’s on ? I'Iavf #6 542416 ye/aqy‘ X/ J/ /)( V/ '@ Vfi 7’ (/6 +yA/Q : a“) ZM’VE V74 7 *0/47? 89 {A M) l/ £4 A lw:_,L/£_: :2; '9 Ac #177 C , 0 1" = 1 m/sec V. or 1 mm = 1 mm/sec MAE 364 - Kinematics, Summer,2002 Exam 1 Page 4 6. (15 PointS) Determine the corrections that must be placed in balancing planes A and B to dynamically balance the system shown. A / Balancing Planes B (11103 = 4929’" l V g ;_ - o #2 r9 I 9a = /7 4. ‘f | # | a 15 10 : 60 | | (mr) 2 = 0.10 kg in bearing 0° 150° 9b = fl | | #1 | (mr)l = 0.25 kg m #1 | Alldimensionsinmm. QM, Mflmmé «I‘M/A) amM 4 # Mr 9 a M- Mf‘a’5 fl/{f/flé’ i w M’fi (05 5. ”in; :/”€ 1 ,4!“ —/;0° 90 /f 43.7%; -7,m +30 2; —c,€%’ «3,75? ”7( .1 I 13W v/o ~/ .707 mm W I0 ~707/ 7. "0.0193 —g,.w - @565 3.33 I / can m: u; 3 ,’. cal/”77': fl; fl ’l m, m, 5’. «W m 773 427” /3, m} — 9, 3.2 / /3, 7;; 41/243 ”If ,7]? mra a”; _ ‘ W 245/”; = M W: 4 : 7 4’” 7. (8 Points) Determine the degrees of freedom in the following kinematic chains (show your work for credit). ...
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