Question

The small cylinder C has a mass of 10 kg and is attached to the end of a rod whose mass may be neglected. If the frame is subjected to a couple M = (8t² + 5) Nm where t is in seconds, and the cylinder is subjected to a force of 60 N, which is always directed as shown, determine the speed of the cylinder (in m/s) when t=2 seconds. The cylinder has a speed of vo 2m when t = 0. 60 N 0.75 m у M = (81? + 5) N·m

Answer 1

ti F 3 we have to calculate the speed of the cylinder out t = 2 see using the principle of Angular impube, and momentum. (Hg), * Et Me dt = (her), muin & Feat t ++ J Made Madt = mvar Here, m, mans a lokg Vr = intial velocity = amis, rodius a 0.75 Mz is the couple V2 is the final velocity, (Gox ŹN and tz 2 sec. (8t + 5) am from mz :, 10 x 2x0.75 + fret'r sjut * *0.75*2* 5)dt = POxV, X 0.75 → 15+54+ [813+s+] = 7.5v9 1,5 va = 100,333 7 V2 = 13-378 ~ 13.4 "Therefore the speed of the cylinder at t=2 V2 = 13.4 ms. mis, lis

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