Question

Check my work View previous attempt A system consists of three particles A, B, and C. We know that ma= 3 kg, mg= 2 kg, and mc=8 kg and that the velocities of the particles expressed in m/s are, respectively, VA= 4i + 2] + 2k, vg= 4i + 3j, and vc=-2i + 4j + 2k. Determine the angular momentum Ho of the system about O. 714 points eBook Hint Print . 3 m References 3 m 2.4 m 3.6 m 1.2 m The angular momentum Ho=-15 12 kg.m2/s)j +( 24 kg-m2/s) k.

Answer 1

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Solution write the coordinate of Point A, B and c as follow, Al0,3 m,o) BL 1.2m, 2.4m, 3m) c(3.6m, 0,0) > wite the position vectors for particles A, B ando VA = 3m 8B = (1.2m) i + (2.4m 9 + (3m) Ř Ve = 13.6m) i here, VA - Position vector of Pantole A 1 YA:- 1 B 1 C Determine the Angular Momentum of the system about o by Wing following relation - 16 = (VAX MAVA) + ( P2 X MBW) + We + Mc XV) Here, mA, ma and me:- mans af particles A, B and c. VA VA and Ve :- Velocity of particles A, B and C. putting mA = .3kg, MB= 2Kg, m, = 8 kg

8A = 3 B = 112 i +214) +32 re- 3.6 VA = 48 +2372k VB = 49 +39 Ve=-2 i +48 +22 Ho =T3 X 3 (48+05+23) 7 + (12i+243 +=R)x2641-3) +[(3-61)x 8 +23++2R)7 = (-368 + 18 17+[7.28 – 19.38 + 249–181] + [115.2 k _ 5766]] - oî - 33.6 7 + 67.2 Ho = -3367 +67, 2Ř /Kg m/s Therefore, the Angular momentom of the system about o is (-33.67 +67.2 ½ kgm/s.