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Unformatted text preview: ME 364  Kinematics NAME fﬂb W Summer Term, 2001
Exam 1, 100 Points, Open Book l. (15 Points) The quick return mechanism is shown 0 '51
full size. The crank (link 2) is rotating clockwis .35 rate of 50 radians per second. Determine the velocity of slider
6, the angular velocity of link 5, and the linear velocity of
point B. Be sure to include both magnitude and direction. N, (a) v6=_.l .’M/i—?(/5ﬂ.g (a) V095 ropiﬂw w" 5’ lb) ws=m
; ﬂaring}? u (C) VB: M n WM/j) I 4730m¢r9
:3 (1.96?” 455%»? b V =0 a. 0’ =3}mhﬁv=é‘77;~,/» +13
() 915' lk 5' " :J?I;}v.)é)z/: g) l/ » .
Cd ; iii > 477;»: : X9
r atﬂalf ? 857m) “’7”? I (6) V5: 0‘53 My; 36mm.) .21, w; : may.”
“W77 urn/4 : wait/w 2. (15 Points) Determine the locations of all of the instant centers in the mechanisms below. Clearly identify the
location and identity of each one. Location accuracy is part of the grade. +46 MAE 364  Kinematics Exam 1  Summer, 2001 Page 2 3. (20 Points) The mechanism is sho one half of ful ' V ‘ elow. Link 6 is rotating counterclockwise at a rate of
1.10 radians per second. Determine the angular veloCities of links 2 and 4. A; 'ﬂ/m. , 1"=1 in/s or
1 mm=l mm/s C (04: ,7rM/5 A,
Work Here:
@ l/ : MM): Winn/,3. {Ml/Z) V , 4N
)f/ l// >(l/ 04;: A; t : ’é]f% A
Q; V9: (/5 + l/c/g 0:75 WM 0»
a we M (W Wm V
// ){J’ x/ U/qr ‘14: 7ﬂm’"/5: ZRMV' C: b’
@ VM V». Way/A. 56 76*" ‘ ‘
Q C
4. (10 Points) Determine the degrees of freedom in the following kinematic chains (show your work for credit).
0 DF: I
q I
F
3
I
0F: 3(g1) &(1/+3)~/[0)/[0] 0;: 3(7/)~ m +91) —/(0/~/{0) = haw—wow = mam 40:; MAE 364  Kinematics Exam 1  Summer, 2001 Page 3 5. (10 Points) The plate shown is made from aluminum ( p: 2700 kg/m3) and has ouside dimensions of 300 mm by 150 mm with a thickness of 20 mm. Four rectangular holes (120 mm by 60 mm each) were cut symmetrically
in the plate leaving the member shown. Determine the percent reduction in mass and mass moment of inertia
(about an axis going through the mass center perpendicular to the plane of the paper) of the member . % mass reduction = 6 % inertia reduction = 53 20 Edd A’x/t
150mm IQ = (,Il/(Jqu: , ﬂﬂﬂ/(lé/ ﬁagtyggi/V 5 t
_, M t
Ii =zi+mf 5 71%" WW” L mmtypicai : gléiéy'aﬁlw") + ,3XIY605’4 ‘2) 5%: (3)6076 w): (0007“? = ,ommu 4— mom» = ,0&3/M9
,4! 3/714 ,« 2710/0/07): MM; PW : 4’4; , 4; (39567 7 .2337ny J
1"; [114(4517: 3+?6351h/fy 1:1, — 71* = .mnm— ﬂaw/m) awmﬂfn ' I _ J " = la/Wy’
Z Iﬂirﬁﬁ Ila/"(#07  t0e17? — oscillation be? r= m2: = mm = WW 1
I” = I + ﬂr‘: ,p/o3t/+ my; (JWO’)
: was“ , 017/” : , 03751” kﬂ‘ : 331.. /7—6
T A77 My [er /) : a77 W
Lf7Qf/3 290%; “97% : . 777566 MAE 364  Kinematics Exam 1  Summer, 2001 Page 4 7. (15 Points) Determine the corrections that must be placed in balancing planes A and B to dynamically balance the
system shown. (mr>a=___/ 97" ” A Bdancinglees B
l \E I p
I 4 (3'3: 4 ea: ,/Q
A. (mufﬂe
0
9b: —°}
l #1 (1m) =o.2s kgm
Himen/r IVA/47‘ 8 7L”
/
r 6 M M W Mini/45 ,./1 — V
jit 342» To A75 /3.75" aim/527 (mm/A (43,30, 9,07) 0
D. .57 HO" 50 A? ~ £57 R '27: : ‘2??? EM;
3 "W “W Liar “r M 5 6m), a277 2’
v ’ ' ’l'" = I k
93.3 46477 (Mr/I; a _ fa 2; j”,
Mama/r401” & f5
9 ,6; w Mir/£9 grafl’l ,( 4 r “W
&! fig—:— 30 ‘f 2.570 «Iggy (Mrﬂjg:(l‘f13¢/) ' jg
3: M M w .20 «MW t r 9%, GM;
— ’ / 23 0 ~ I
3 “.1 ﬂ I5 :27ng ,1»? (5%: (:4) r (’5’! z I [275/90 8. (10 Points) Determine the velocity (in inches per second) of the follower in the cam mechanism shown. The mechanism is shown full
2 , ’ Vf 11 = I in/sec
[OW/w”, 7g 3;; , Mx/my, 0 ower I. We \— M j" . .
Véd/IW/ : Va“: M and» :WK/hwféééﬂ/ = 5,3117% 7 D. lief/91 Z 7 1/9/9472 Py/g' 0,, ...
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