Question

farm AB has constant C.W. angular veloc- ity of 0.9 rad/s, arm BC has constant an- gular velocity of 0.6 rad/s, and arm CD re- mains vertical, what is the acceleration of part D? х Figure 1: Distance AB-0.25, BC=0.22, CD=0.11 01=45°, 02=3

Answer 1

CA 0.41 10,- 45° We have to The BLA Given Dala: IF arm AB has Constant CN angular velocity 0.9 radite Arm BC has Constant angular Velocity 0.6 rad/seu Arm со remains Vertical then Calculate the acceleration of pasto. vectors Component at B with respect to A is given by “BA = Length of ABCs Coso, +j sino) = 0.25 (i cos 45° tj sin 45) -10.1767 +0.1781 The vectors Component at U with respect to B & 0.22 (1 Cos 3 - 3 sim2) a [0.2199 -0. 1848 j. Acceleration of Doint B &p= Kasa TBA ang=0 an WAX x YBA =-0.91* (0.1761+0.1763) (ap =-6.1421 +0.1423) ac = abt ABC NNC/B W CBX rc/B aq-wic B X rc/=(-0.142; -0.142J) - 0.8 (0.2152-0.164) ac = -0.1421 -0.1424 -0.070841 -0,06624 ) -0.22061 – 0.2005 acceleration Finally we mequired at point D. can see the Fisure to obtain a ac 120 -0.2108 1 – 0.2009 Similarly mclo = *BLA then acs aco ONC we