Question

1.67 m OD = 1 m Problem 1: The angular velocity and angular acceleration of bar AB are WAB 2 rad/s and AAB 10 rad/sec2. The dimensions of the rectangular plate are 1 mx 2 m. What are the angular velocity and angular accelerations of the rectangular plate? ΟΙ Β AB "АВ 45° 45°

Answer 1

solt The instantaneous center for bar AB is point B. by definition the instantaneous center for bar CD velplities is point o. by definition the inter at points A and a normal to the bars AB and co are these bars respectively. However, by inspection these are parallel at the instant shown. So that lies perpendicules to to the velocities at A and a will never interseet the instantaneous center for of the plate AC is at infinity, hence the plate only translates the instant shown at and Waco. L If the plate it not rotating the velocity at en every point on the plan must be the same and in particular the velocity at A and must be identical. The vector AB is с r - i cous- j sinus = l (i+j) (m) ALB = The velocity at point a is - IC - WAB VAZWAo ar 0 0 1 - AIB :-(J2) Ci-j) (m/s)

2 The vector clo js Iclo-(1.67) (-icosus°-;sinux) = -1.179 (i +;) m) The velocity at point c is Ne=-11179 WCB [: : Jarmaci o i 21.179 Wed li-;] (mis) Equate the velocities veava, separate components and solve uco 21-2 rad I see To determine the accelerations, the acceleration of point is dA=LABAMAIB WAB TALB -10 i jk 1) t 2 Cena (i+j) 10 = 9.91 -4.24j (m187) The acceleration of point a relative to point A is a c = a Ata SILIA jk O LAC Ac a At [ 0 2 -9.97 +622Ac-4.24 )j (m): The acceleration of point a relative to point o is a c=apt&co & cofrelo مد . reld CD Noting apa 0

a c=-1.179&co ol/+1.179w0 Ci+j) =(1.179 dc8+1697); +6-1-1794c0+1-697); (mb22) Equate the two expressions for the acceleration at point c and separate components (-9.9+1.1794c9+1.697) i 20 (2 <AC -4-24+1:1794c0-1-697);=0 solue above two equations <co -1.697 +9.9 -6.957 19 lco 26.96 rad 152 | (counter clockwise) 1.697-61.179x6.96)+4.24 LACE LAC -1.13 radis 2 (clockwise)