Question

Q1:The data given for the following mechanisim are: Length of AB is 30 mm, length of BC is 40 mm, length of CD is 20 mm. All links are made of the same material and the mass per unit length of this material is 5 Kg/m. The angular velocity for link AB is 4 rad/sec CCW. Determine the following: • Mass moment of inertia for all links (4 points) • Instentanious center for all links (4 points) • Angular velocity for links BC and AB (10 points) • Angular momentum of link AB about it center of mass (G) (3 points) • Angular momuntum of link AB about point A (4 points) • Kinetic Energy of link CD (4 points)

Answer 1

1- mass moment Of Inertia Link AB mass = Length x mass per unit length 30x103 x - 0.15 kg, IG (IG) AB = this roof about axis through the center perpendiculao to length M1² T2 . (IG) AB 0-15* [ 30x1637 21.125x105 kgm2 12 Link BC muss 2 x length mags per unit length 40 x103 x 5 0a2ky, = (Ic) Все ML2 12 = 0·2x (40x103)2 12 = 2.6667x10-15 kgro2

Link CD mass = length x dineas density 20x163x5 Fool ky 2 (In) D = mi 2 12 0.1% (20x103)2 s 3-333xloʻlig mo 12 instan eous center's are 112, 123, I34, I41 Isu oni I g. ncht) 4x(44) = = 6 No of Incentera 2 scale 281 - (112 0 (1 23 112) 112123 = 3cm 300 1231245 3.2cm 124134 F 0.8cm I12I24 = 3.1 cm. (I2 2-4 cm. IzuI14= (134) 1 D I 41

# I28 is draw by be using a lines which are passing through C 116,712)&(143,132). inter secering point is I 24 is drawn by using 2 lines which are passing I 23) and (I 14, I43) Intersecting # I 13 through (I 121 is I24 point 8 WBC = ? WBCE Celine W2 = WAB= 4 radis W3 W2 I 23I12 123 I 13 I23112 = 3cm I 23 I 12 Iz3 Is = 5.8cm W3 = ҲО 2 I23 I 13 = 3x4 - 200689 gadis 5.8 (4 of link AB Ansgulay momentum IGUAB about center of mass - =IG WAB 2 Ilse ml2 T2 = 1.125x105 legs no

Of Dingular momentem about AB about center of mass 2 I GAWAB 5 = 1.125x4x10° Kgm'xvad 405X165 by 102 of link AB 5 Angular momentum about point A IAWAB CUBA 4 rudis IA - Igt M(5)? (foam parallel axis therren IA= ML² M22 ML 2 [ 4412 12 ML? 3 :) 12 t 4X12 4 MLL XWAB Onqulay momentum ofline AB about point A 3 3 M13 X WABE 0.15*[30x1õ)? - X4 3 . lo 8x10 + kome kg ma S

(6) CD = kinetic energy of link <Iw2 == // IGX xwell WCD= WA WAB = W22400dls I 24 I 12 W4 1 from above diacrum. WL I24 114 W4= I24 I 12 XW2 Izy I14 sad's = 3.1 2.4 -X4 5:1667 Link CD is rotating about point D, re Icinetic energy = 1/2 Ig west amuz so No lineau moment kinetic energy = ļı ID wano MIER =laxy 2 x WCD 3 - la x 0.1(20x103)2 -x501667) 3 =1147963 x1 Øt goule.

) w4 124 112 นงาน 124114 WHE Angular veloutey Of CD = ? WUD Iza Ilz XW2 I24 Ilse = 31 -5-1667.radis X 4 2.4 5.1667 gadis Angucav veloutey Of LO
i have done this question before and I used values of length of link is from questions not from graph.thank you