Question

MECE 3336 Dynanics Exern 2- Protected View - 5aved to this PC Autasove CH Caprlavile, Cena C Share Comments Ногги Sh File Inser Daи Design Layut Relecnces Mailings Revicw View Hrlp nable ditng PROTECTED VIEW B cancf. ies from the Intemat can contain virusas. Unless vou nacd to adit, it's safer to stay in Protond vicw 3. (35) Knowing that at the instant shown bar AB is vertical and bar DE is horizontal. Given that bar AB has a clockwise angular velocity of 10 rad/s and it is slowing down at a rate of 2 rad/s2 (a) (5) Locate the instantaneous center of zero velocity for bar BD (b) (10) Determine angular velocity of bar BD and DE (c) (20) Determine angular acceleration of bar BD and DE A 0.2 m В 0.25 m E D 0.2 m 0.6 m Pee 415 s 15% 757 PM OH w- Type here to search ENG 14נו1/2ר/1ו

Answer 1

Sol: Given that, velocity. Analysis , WAB = 10 rad.si V8 = (ABW AB = (0,200) 10) = 2 m/s TE > VB = VB. : Vo = Vol. -> Locating the instantaneous center of bon BD by noting that velocity directions at Bld. are known. = BC ton to vp & Dctor tovo. => 680= yol 0.95 -8 rad/sec.gr VO = CCB) WBD = 20.6)x8 = 408 misi WDE = you 4.9 - 24 rad/s. 0.21 Now, Acceleration Analysis: LAB = 2 radis24 WAB = lorad Isv. ) unit vectors are, 21, =1 , K=19

* A8 = QA+ LAB X VB/A - W2AB*B/A. = 0+(2k) x (h0.2) 5 -C102-(-0.2); - 6.4m/838 +(20 m/s2); - QD = a8+ BD x YOUR-WPBD 'D/B = 0.41 +905 + XBOX XC-0.61 -0.963 -82 (-0.61-0-25). » Qp = (38.8 +0.95 Lede+ (36 -0.62BD) And =) ao= aetaDex role - CODE DIE 50+LDEKX60-21)-(24) M(-0.21) = 115-91 - 0.9 1069 → ® From eans @ and 6. eluating the components of an » i : 38.8+0.25L80 = 115.9. >>8D = 305 6 rad/s2. j: 36-06 LBD = -0.9.XDE. - XDE = 30BD -180 = 3(305.6) -180. = 73618 rad 152, → Angular acceleration XBD²306 radis KDEEP 137 rad/s29