Question

Saved of 2 Required information NOTE: This is a multi part question. Once an answer is submitted, you will be unable to return to this part Each of the double pulleys shown has a centroidal mass moment of inertia of 0.25 kg.m2 an inner radius of 100 mm, and an outer radius of 150 mm. The mass m= 17 kg. Neglect the bearing friction. ces с Determine the velocity of the cylinder 3 s after the system is released from rest. (You must provide an answer before moving to the next part.) The velocity of the cylinder 3 s after the system is released from rest is m/s 1 < Prev 5 6 of 9 BA Next > search BI

Answer 1

Sketch the free body diagram of the system as follows: WB Т. D 0.1 m B А. 0.15 m C Uc

The angular Velocity of Pulley it is written as Vc WA WA Vc 100mm WA Vc olm 0 WA i8 R X WA. = 10 vc the Velocity of the cond D is givers by the relation Vo Vo = RX love Vo = 150mmx love Vo = 0.15 m X love Vo = 1.5 Vc The angular Velocity of Pulley B is Writters WB Vs 9 WB 1.5Vc WB loomm 1.5 V Ooimy 15 Vc WB = (3) Apply the principle of momentum and impulse for Pulley B.

r = 0.1m Toi 1 = 0 1(wo), = 0 (1(wo) = 0 1 = 3s II B B B Syst momenta, + Syst Ext Impia Syst Momenta,

Apply moment about point B foos pulley B. EMB=0 0+ (TH) 9 = IWB), tot) (0.1) = 0.25 (15Vc) Tot = 37.5 Vc Apply the impulse and momentum Principle for Pulley A. 4

or 1 = 0 T (w.), =0 1(w.) = 0 1 = 3s с C C mc (rc),=0 me(rc), Syst momenta, Syst Ext Impia Syst Momenta,

Apply moment about point A foos Pulley A. EMA = 0 of (Wetes - (Fot) R I Wat me Veles 0+ (17x9.8143)(0.1) – (37.5Vc)(0.16) = (0.25 Xlove) +(17 Vc)(0.1) 50.031-5.625 Vc 2.5 Vc +1.7V 50.031 = 9.825 Vc 50.031 9.825 Vc = 5.0922 m/s Vc