Question

Fundamental Problem 16.23 < 3 of 9 > I Review At the instant shown, the wheel rotates with an angular velocity of w = 15.2 rad's and an angular acceleration of a = 7.6 rad's . (Figure 1) Part A Determine the angular acceleration of link BC at the instant shawn measured counterclockwise. Express your answer using three significant figures. Enter positive value if the angular acceleration is counterclockwise and negative value if the angular acceleration is clockwise. Figure < 1 of 1 > VG AZ 1 vec O 2 ? CERC = - rad/a² De 0.3 m 45°C 0.3 m Submit Request Answer -1.2 m Provide Feedback Next >

Answer 1

I calculate the velocity at B Ve = whxre = -15.2  x 10:37) → V8 = -4.56 i 2 0 Calculate the velocity of į wort B as follow V = y + cock x Veb 1 = (-hisb) + abe ( (123) → V = -4156 +1.2001 - Calculate the velocity of į wrt D as follows co = chockaro = Woc  x (0-3 (as 45° ; – 0.3 Sin 450;) → Vcp = 0.212 Woc j + 0.212. Cope Equate terms → 0.212 Wbc =0 → Wbc=0 Equate the j terms from I and 2 egns as follows → -40% + 2C8C =0 7 WBC = 3.8 radls Acceleration of the B g = &RX () -/40) = 76 6x(0-31) - 6*(0:37) de = -2.28 – 69.312

Acceleration of C' wort c' qe = ag + dag hx reo - color* reb (2.28j –69312;) + 12 dbac j - 384(1-21) şq = -2.28; –69312; + 1-2 O AC j - 17-328 i Acceleration of wist Q = (ak x Yeo-wöc rcps → (-2:28j –69.312 i +12 dBj-173281) = 0.212 Xoc j +0.212 doc! Equate my terme -69.312 -17.328 = 0.212 doc - » *pc = -408-67 Equate ☺ terms - 2:28 + 1.2 dec = 0.212 doc 1.2 dec = 0.2126-408-67) + 2i28 dec = = 70.3 rad|s?