System of Particles and Rotational Motion

The position vectors of two 1 kg particles, (A) and (B), are given by rA= ( 1t2 i+ 2t j+ 3t k )m and rB= ( 1t i+ 2t2 j+ 3t k )m, respectively; ( 1=1 m/s2, 2=3n m/s, 3=2 m/s, 1=2 m/s, 2=-1 m/s2, 3=4p m/s), where t is time, n and p are constants. At t=1 s, | VA |= | VB | and velocities VA and VB of the particles are orthogonal to each other. At t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is L kg m2 s-1. The value of L is .

Answer

90

Solution

At t=1 s: VA=(2,3n,2), VB=(2,-2,4p).

VA VB: 4-6n+8p=0; equal magnitudes: 9n2=16p2 p=- 14, n= 13

VA=(2,1,2), rA- rB=(1,1,2)-(2,-1,-1)=(-1,2,3)

L=( rA- rB) VA=(1,8,-5)

| L|= 90 L=90

Answer: 90