Question

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The flywheel shown has a radius of 20 in., a weight of 290 lbs, and a radius of gyration of 15 in. A 70-lb block Ais attached to a wire that is wrapped around the flywheel, and the system is released from rest. The effect of friction is neglected. Determine the acceleration of block A. (You must provide an answer before moving on to the next part.) The acceleration of block A is tus21.

Answer 1

A . 3in B bin. P agt = 20ft/s² - Arcolection at B (tergent Coponent) *Solutionia) PAGE the conditions podemeter from the question. sin → Hese coloceleting few peremeters seguind. LB n = 8AB WAB? (276) (12)? Lon. = 30ft/s Aceleration at sa proognel component) (3x = Lor _ TAB WAR? - b) / 80 3o)? of B locity y Calvidlake ve Vo = SABI TAB ) (2) = 3ft/s (3 4 y Anguld velocity of AB. VA= TBC WBD. =(6+1/4) WBD

PAGE-2 le 3 = 16x162) UBD UBnF band /s. of red By using Pythage sous the son I 3 I IBC 2 + I? - Calve date the length 2 0282=100 F13102 TIBD Froin * Eevirlent diaglen for Rod Bo. 6 Normal Component of Accodention XD1B, n = IBD web? - Ello x 1/2) 304412 ) (BD) 2D/31+ of BD TiBa =110 xV2 21/12&B) → Accedektion diaggen for system :-)

y component of G- G.y' PAGE 3 OB 3 color 3 abr obit Apply equilibaium condition in x-firection -dod -145) (0/3+) +1 %-) (%D182) = By god/s2 Accelestion of a 180- OG/B + % =% Q3 +12 (114 +80/134) Xx corporary of Acceleration to GB = on the loo Bn TOP124) =-20-%2("/s xis the 10 (88) E-loff Ish = 34.4's ( }vis **s * loro -7-3152 I= 2 2 I (123) ( % ) 0.0053/5-64

PAGE 4 Applying moment of equilibrium EMO = 4MB (%) - له " درا f - (%) و ره (3) و (*) co 3 *(79) (%) 0:50 - 3 ) = 0.0053 (8) + 32.2 ( ا ر ) ( (2202 دل 71 -2 1 PLEASE UPNOTE * Its very much important for met