Question

Two uniform cylinders, each of weight W= 10 lb and radius r = 5 in., are connected by a belt as shown. Knowing that at the instant shown the angular velocity of cylinder B is 28 rad/s clockwise, determine the distance through which cylinder A will rise before the angular velocity of cylinder B is reduced to 5 rad/s and the tension in the portion of the belt connecting the two cylinders. (Round the final answers to two decimal places.) B The distance through which cylinder A will rise before the angular velocity of cylinder B is reduced to 5 radis is ft The tension in the portion of the belt connecting the two cylinders is Ib

Answer 1

Free Body Diagram are E vo ve woor Instantaneous c (V.:o) D VA? rwa Velocities Wis at point D&E o Assuming Center of cylinder A is point c. then, WA 2 Ja zor (we VE kru VWB vo 2 N CD 27 WB. Wor 28 + 2 (Vo) is at point o & velocity Vparwo r/27A) ..ZJA 2 Moment of Inertia of the system (I) Iz mra Here m is mass of uniform cylinder Total Kinetic Energy To mu+1 I WA + 고 Toim(rw) + 1/ mx)/+wa)? + 1 / 2 (2 m²) was

mr²wgh 2 t mr was + Ź mrug? 8 2 7 / 1 mr²w8² Equation of work & energy as: (0) 2 Tz Ti+ Una KE Here ToTz is at initial position & find position Uiya z work done by external forces Given (WB), ² 28 rad's .,(WB) (KE) inins al 2 (KE) 2 final Uiyz (WB) e sradls. To me (wgi)? T2 = 7 - my? (WB2) -Wh -mgh to mr²(WB1)² = mgh's / 1 m ² my? (WB2) 콩 mref wer-wei mo? (wait - wazo from (1), wg?) ngh hz T6 (2) mg ha 16(32.2)(28?52) (3) hz 1.79 ft Therefore rise of cylinder 1.79 ft FBO of uniform cylinders with Tension &

Tension in cord DE is of an We got, Vozava h Therefore point D moves 2 times distance that the cylinder A moves ie., twice of distance of a a(zh) i Tz2h Ž (WB) D c Tiz ķī (WB),? -20h Uixa & (2h) 2 ¿Î WB į i wbi - 20h. z Ļ I.( wgi -wg3) z 22h (wgł-we?), it I Cuish -wez) 25 4 From ② , I (woł-WB2 2 ( Igra (wsi"-w82) 4 I 44. (1 mo?) N 782 4- 2mg 2N 2 2 7 2 (20) 3.714 lb 2 g Tension in portion of belt connecting 2 cylinders Q25.714 lb

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