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The weight is used to turn the gear system shown below. Gear A has a mass of 15 kg and a radius of gyration of 100 mm about i

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page, povestiom solution briven that Two years are of different diameter in mesting with each other and carrying a weight witpage- we have to determine each gear il speed of the weight 5s-kg after it has beem Lowered Bom. (ii) Angular velocity of iepage. Free body diagram of bear B doumac and weight o Carbine QB B > със IT at starting IT assg 55 Kg D 99 Apply Togare. copage- → (5589.911* 2 X pins (o Fx200 ino les 29% 0.115 2B 558 volt (53.955) -(0.2)F (1.23 dB (1 = > Free body diagram of hearPage - F = List okol ofis (1) A and / from equation 6 and ② put the value of F = xal in eavation 2 153.955-(0.27 LA & (1023)page- IA TB VA = 2067 X5 XA z 4. 3 dA=.4 dB 3 So from equation ③ - 55.955 (0:2)x d8 +(1.239 dB V. LB 255.gsS [ 3. 0.8.+ 1:23]page-① so, LA 1 2 3 4 X 37:37 7 = (49.82] sad 15? Amgular acceleration of hear A. Now, li) for Velocity of D (55kg) do = dB (Pagento Apply Linear tho tiom equatran ffbarch* velo = (UE) @ + 2001 s = > (4) + 2X 3073723 Velo- 6 x 3.737 i Velo (4.735 m ipage-@ We WB Because Aniq of Jo Aation in same WB= (17.35) sadis Rowy Ve= fronte >> WA NB = 4 20 1877 =413 WB NA 3 → WA x WisPagino IVA VO at point of contact (WA) VA = VB NA (WB) OB . NA= (WA) NA = 163-138 0.15 [NA = VB = (3. 469) mis velocity of m

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