Question

The 40-lb flywheel A has a radius of gyration about its center of 4 in. Disk B weighs 55 lb and is coupled to the flywheel by means of a belt which does not slip at its contacting surfaces. A motor supplies a counterclockwise torque to the flywheel of M = (50t) lb. ft, where t is in seconds. (Figure 1) Figure < 1 of 1 M = (501) lb. ft 9 in. 6 in. А B

Answer 1

solution. Given, weight of flywheel WA = 40lb. Radius of gydation of flywheel , KA = 4 in. Weight of disk B WB = 55lb. Tosque on the flywheel = M = (50+) lb.ft of disk B, B = 50 gad/s. body Angulars veloity Free diagram (Feo) Flywheel Disk .B. 55lb 40lb. Ti T, 4 Coc. 0.57 0.75ft Doc M T2 Te Dy 2 Cy. moment of inestia of flywheel Mass flywheel A about point a 2 Ic = 32.2 0.1380 slug. ft² Angulas velocity of flywheed A, CA= OB. We DA 0.75 - x 50 0.5 = 75 sadls. 2 ta M dt = Ic We c Apply principle of impulse and momentum Is al, + ES (tve, sont de [Stedt] xo.5 - [ST, dt] xos st 50t dt t ot 0.1380 x 75

t + 0.5 dt - 10.35 S(1-11) dt = 10:39 254 + 05 S2-T.) tt dist about point D, ID = = 0.4805 Shug. ft² By poinciple of impulse and momentum from FBO , Mass moment of inestia of 2 55 :) xo.75 2 32.2 tz Ipo, + Esi Modt Ipula. ti Gtve , 2 Ot Todt|40.75 = 0.4803 x 50 -0.4803 x 50 0.75 [ST, dé] xo: 25 - [5 Tett] x0.75 S(T2-T.) dt S(T2-T1) dt Apply equation ® in equation ④ 0 95+* + 0.5*(-32-02) E - 32.02 2 / - 10.35 25+2= = 26. 36. 2 26.36 t = = 1.0544 25 time t=√ 1.0544 = 1.026 s. (Answer) Hence, time required for the disk to attain angular relocity of 50 sadlo t=1.026 S