Question

Abar of mass m5kg is held as shown between four disks each of mass m2 kg and radius re75 mm. Knowing that the forces cortod on the disks are sufficient to prevent slipping and that the bar is released from rest, for each of the cases shown, determine the velocity of the bar after it has moved through the distance. Round the final answers to three decimal places.) Forcase (al, the velocity of the bar is Forcase (b), the velocity of the baris Forcase (c) the velocity of the baris ve ve VR

Answer 1

soin Tawing velocity ve the verocity and i' be the center G of the Upper left of Mass disk. Moment of inertia of one disn + 1 = m'r? 2 Kinetic energy. To 2. → Ta 512 +4 ( 2012 ar²a? t 2 4 u T = 2.5v2+ 45 +27°02. Here, are two position considened.. In position D + Inetial reeft position Ban has moved at a distance b. All . the dish moves down a distance h. ditangh of B 4migh' = 5ght 8gn - For Case (a): not The mass centen of each disy is moving % =0 &n=0 ř, rew=v. Taa=2.54² +0+2v²= 4.502 cos case (b): W = V vorwa ar =rw =internet Tab= 2.502 +41)+7()? 4 v2 Tab CS Scanned with CamScanner

Care -c : v=v, hah tv) Verw = v. ñ ar Tac = 2.50² +4 0²+21² 8.5v2 worn and energy; For case @ By principle of The velocity of - Ta ban is > Ti + ,-2 of 5gh t0 = 4.512 V= 1.054 vign An, Jone case the velocity of ban is 0+5gh tugh = 402 v=1.5 ign . An For Care © the velocity of ban is of5gh +8gh =8.50² va = 1.237 rgh AM (Thank you! CS Scanned with CamScanner