Homework 2 Solution.pdf

# Homework 2 Solution.pdf - (cu 0-9 Wek/M(b 03 = V4k(M m...

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Unformatted text preview: @ (cu) 0-9": We‘k/M (b) 03,, = V4k/(M + m) await. .csr-A‘ffenii r——“ Velocity of falling mass m = V: 23 .2 ('3 9‘50 at A‘L‘ai'a'c azuftibrium P0951950!)- xo: 1(‘l:=o)=_ “Jaiét = .. 2': ConSErvafl‘qn of momentum: 04+") i0 = m .1, g 1-,. #231 . 92° = qc (19:9) Complete solwtion : .3; (t) = A0 sfn(CJ,,t .4. ’50) where A ___ 7(2. + 12.0.. 2. - m1 32' 0 can — z (6 #1 and ¢ = Ml<ﬁcown>= tam—d —- W a 71‘ J:24<(M+m) (.3: I'm: '='.___E'I'+ 43211; “ ""m 12,3 w (05) £05555 {MH5MA} 3K 5“ 55K 3% 5‘— 53 \$93 wbmﬁyéﬂ V€§§>mseﬂ _. 5 ,. n . h :45“ w 15505} ,5, « ; if 1mm Cow‘difiong, 55555555me A 5%,%W ~- 37.1584 5555;5255 C9554 550519" 5555535555555 511,595,534” "5552‘ Sjgéem 5: Wmey- ~55 amigg} if; sigma" Qimtiwm 9035-55555: ; xﬂiiﬁvw m N W n m AAV‘I'CE {Zia/23555555»: {95556564 ;L i w 0 Mmmwlww :fémgs 555mg” ,5 5; 32mm. Xoﬂ‘. Mi“! 3 3 83%? gaffﬁm \$55334 , M Z 2:: ‘5’)55” 5R 5 >3 555* 4-5» am; {,1 :51” 5:2 M05 5,359 3mm bezsgmaﬁm 6f msmémgr: 5’ -‘ «33: W3) :5 mg’52 5605305555; W55»? 2/52 ”"3 5555;,555‘535 ; W 5‘55 55a M xwﬁz ,5 f/Secw "“ 5. 53\$ 0 : r We, 5’ 5’55": {wt} 5.; &@g wmﬁ ~55 b 551560515 (365:5?) 2 05 1 3C»: "3‘535533 M w»: 005/5 ( ass‘wnt + 5583mm y£ﬁ§w 83%;“ 51:. 3,55%} m} bi; wkwux £{75ﬁgg 5" 55515555 \r Kw 555555 05555 5555375553555) 55173555 may? 555591») (5) mm; 5555953 "55559652. My: 1;. V; mg :2 3333?: , 3a; ZCS‘» 5x}; 25:5 W MH VH if 48:“: {39 554 f :3 2 {’3 {Sari/w “5/156 ‘“ MA "‘ “ {CG X 6 /§€ 7 mwazc ,occabnm ~22. :5 @5555’555 5‘55 (i) (a) Viscous damping, (b) Coulomb damping. w (iii) (a) rd— _ 0.2 sec, fd = 5 Hz, cud = 31. 416 rad/sec. (Mn Tn = 0.2 sec, 1', = 5 Hz, w,— = 31.416 rad/sec. Ur e511! 2x; + In 1 =ln2-06931- xi+1 §;1_§3 or 39.9590 ('2 - 0.4804 or g = 0.1090 Sincewd-wn V 1- ,weﬁnd 31.416 w“: 77: V0. 98798 k= mw‘ =[5—_°——° if] (31. 9095)2 = s. 0916 (104) N/m = 31.6065 rad/sec _°_ = one: 2 m «I: Hence 0 = 2 m (an §= 2 (—) (31.0005) (0.1096) =- 353.1194 N—s/m (b) From Eq. (2.135): 11 =- m w” = 95‘: (31.410)2 = 5.0304 (10‘) N/m Using N = w = 500 N, 0.002 I: .. (0.002) (5.0304 10‘) _00503 # p“ 4w 4(500) ...
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