Question

2. In the mechanism shown, each of the two wheels has a mass of 30 kg and a centroidal radius of gyration of 100 mm. Each link OB has a mass of 10 kg and may be treated as a slender bar. The 7 - kg collar at B slides on the fixed vertical shaft with negligible friction. The spring has a stiffness k = 30 kN/m and is contacted by the bottom of the collar when the links reach the horizontal position. If the collar is released from rest at the position 0=45° and if friction is sufficient to prevent the wheels from slipping, determine (a) the velocity vg of the collar as it first strikes the spring and (b) the maximum deformation x of the spring. of 150 150

Answer 1

a) UB=? Forum wore wore Energy Energy Principle (KE), + (PF), = (KF), f(PE)2 kibetic Enersy 177 7 171 potential Eneray. (kel = 0 -> (PE), = 2x mg [ ]+ Mgh Cosus こ h- 0.375 0.37581 2 0.265m Here H+ 7x9-81 x 0.265 (PE), . 2x10x9:8\x{0.20534 CPEN) = = 44.28 ] KÖE = (KE), = Fonal orrorath (k#polling + (KF haunslation w2 ) x2] x2 link [ům mu) cellar = [(CA/ X Io + Thanks

where То 11 Yz mi? Đ Link (KE), - [ \g m[>w<) + + [ { mu?] cullar on the other + %*7* Vg? f - 1/3* 10 x (0.375² 4 (KE), = 6183 Va li o tial Energy PE), - potential then from above equanin (KE), + (PE), e (kF), +CPE 2) 44.28 = 6.83 Us mis VB 2.546 =X. maximum de formatics b) To culculate (KE) ₂ t (PE); (PF), + (KF), (KE), + (PE), = 44028

넉 की 22 f. 2x10x9181 (342) - #xq181 +}30%20% -goal oc - 68 670 +1500022 (ICE), + CPE) 15000 22429C13 -44.2 (KF) g so PE), 2x mg (342). - 7xg ath k x² Then 'CkE), 4(6), 44.2 PET *2 = - qoulu_68067971500022 15000 72 f 29.439-44.220 15000x²_66 67 07 (90.12-44*2) So 15000 22?–68.672745-9 By soiving 0-05331 (2=-6.055273 SO Answer is 450.05331m = 53.03 m