The angular velocities and angular accelerations are calculated
using the given formulas above.
(a) The angular velocity of the rod BD (BD)
= 45.33 rad/sec
(b) The angular velocity of the rod DE (DE)
= 16 rad/sec
(c) The angular acceleration of the rod BD (BD)
= 1520.925 rad/sec2
(d) The angular acceleration of the rod DE (DE)
= 1982.268 rad/sec2
We were unable to transcribe this image
We were unable to transcribe this image
We were unable to transcribe this image
We were unable to transcribe this image
Givey WAB 34 Maafsee writing vector forms ß 굿 0.2? + 0.3 Men 30-3 î + o. 0.075 A de -0.425 DE +0.425 We how that, WART TBA = 34 h + (0.27 +0.39) - 10.2 î + 6.8 > ܟ Jo Cox Top - Jest was R + (0.31 +0.0758) (10.2-0.075 Woo) ↑ + (6.8+ 0.3 wao ĵ or 2 - Jole - woe trole Jo = (-0.425 wool -0.425 W pe ) karating both the equation ① 2@, we we get
-0.075 W BO - 0.425 WDR 10.2 0.3 WBO - 0.425 WOE - 6.8. 6 solving, we get BD- us.333 rad/ge DE -16 radoc BO 45.33 rad/sec (clock wire) (dock wire) 16 rad/sec u. Agular velocity of line Aguler velocity of lit de - Ave Anti-cled win to win clockwire 45-33 B -vez Byrds > 16 Angular acceleration: we know that za wa Rach To & Fast -(34) (0-2 1 + 0.35 -23 1:27 – 346 8 9 wege - 0 (021 40-39 110
2 a = a + XBOX F/ 10/B Bp = (231.2 ? - 34687) + Root (0.21 +0.075) - (45.33) ² (0.37 +0.075 0.075 2 - (-231.2- 616-333-0.075 ano] ît (-346-8-154.133 + 0-3® Alo, &- ade Roke DE 2-40Couns î = o-u2734 x OE (-0.423 + 0.48 o 2- (108.8 – 0.423% of] 1 + (-1888 – 0.425 *DE] î Eauasting Equation ③ & A - -2312 - 616:533 -0.075 280 = 108.8 -0.495 X DE - 0.075 x + BD 0.425 X DE 956.5332 2 -346.8 – 154.133 + 0.30BD = – 108.8 -0.425 XE 0.30 BD to.425 XE 392 133
soling Ecation ③ & & , we get مرا 0.3 & BD +0.925 XDE = 392.133 0.07S XBO ours & DE 956.5332 "BD - 1520.925 red/sec2 DE- 1982- 268 rad/see? BD = 1520.925 (clockwire) Angular acceleattle of lint Angular acceleration of lint DE = 1982.268 (anti-clock wil)