Question

N 3321-006: Dy X s Connect х C The 30-kg Uniform Disk A And T X (25,697 unread) - jasonbeckford X + ezto.mheducation.com/ext/map/index.html?_con=con&external_browser=0&launchUrl=https%253A%252F%252Flms.mnheducation.com%252Fmghmiddleware%252Fmheproducts%252Fims Saved Help Save The 38-kg uniform disk A and the bar BC are at rest and the 9 kg uniform disk D has an initial angular velocity wh of magnitude 440 rpm when the compressed spring is released and disk D contacts disk A. The system rotates freely about the vertical spindle BE. After a period of slippage, disk Drolls without slipping. Knowing that the magnitude of the final angular velocity of disk Dis 176 rpm. determine the final angular velocities of bar BC and disk A. Neglect the mass of bar BC. ences 300 mm 115 mm C 0 The angular velocity of bar BC is WBC The angular velocity of disk Ais wa rpm 2 rpm 7 of 9 HE < Prev Next >

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solution mass of Disk ArmA = 38 kg Thitius Angulur velocity WA=0 muss of Disk 3 mg = g kg Initius Angular velocity = Wi= 400 rpm = 46-0769 ruals Finus Angulur velocity = W2= 186 8pm= 18 43067 ruols be the finly Angular velocity Let wb of diska, Vo va p and oh are contact points on disks A and disk B respectively, from left figure, (1) I WO Igwa Up = -80 WAI from right figure, vq=riorg woi va = (v2-8 Wy); TAWA IA WA Disk I rolls without slipping, so VQ = = VP (3) Vo = r. Wg ~TAWA V = n W₂ -NAWA From principle of Impulse and momentum, moments about D, IgW,+ o= Is W2 t moro (VATND) - IAWA Ig (wp-W2) = movo (²A +80) - FAWA -(4)

here, Ig: myria x 4x (0-1153 2 2 2 x 9 x (0.115² = 0.059 legms² 2 IA = MATA? 1/2 x 38+ (0-300² = 1.71 kgms² 2 here, Vo = 0.115 x 18.43067 - 0-3 WA = 2.119 -0.3 WA ALSO, 0.069 (46.0969 - 18.43067) = 9 * VO * (0.3 + 0.115) -1.7) WA 0.059 (46.0969-18:43067) = g *(2.119-0-3 WA) - 1.41 WA (0.3+0.115) 1.6311 = 3.735 (2.119 - 0-3 WA) – 1 9I WA 1.631 = 7.152 (1) - 1,0125 WA- 1.71 WA WA = 7.152-16311 1.0125 + 1:3) WA = 19.37 opm or WA = 2.027 rad/s Therefore, Vo = 2.119 -0.3* 2.027 = 1.5108 mis

V 1.5109 WBC = 11 S 3.6385 radis YATINO 0.3+ 0.115 WBC = 3.6385 ×60 = 34.76 rpm 2x 3.14 WB C = 34.76 rpm