The ends `A` and `B` of a eod of length `l` have velocities of magnitudes `|vecv_(A)|=v` and `|vecv_(B)|=2v` respectively. If the inclination of `vecv_(A)` relationn to the rod is `alpha` find the
a. Inclination `beta` of `vecc_(B)` relative of the rod.
b. angular velocity of the rod.

Explanation

We know in a rigid body the velocities of two points along the line joining two should remain same.
i.e., `(v_(A))_(||)=(v_(B))_(||)`
Hence `v costheta=2vcosbeta`
`cosbeta=1/2cosalpha` or `beta=cos^(-2)(1/2cosalpha)`
b. for angular velocity of the rod we can write
`omega=(|(vecv_(AB))_(_|_)|)/l=(|(vecv_(A)-vecv_(B))_(_|_)|)/l`
`|vecv_(AB)|=|vecv_(A)-vec_(B)|=vsinalpha(-2vsinbeta)=v(sinalpha+2sinbeta)`
`implies omega=((sinalpha+2sinbeta)v)/l` (in clockwise direction)
we have calculated `cosbeta=(cosalpha)/2.` Thus,
`sinbeta=sqrt(1-cos^(2)beta)implies sinbeta=sqrt((3+sin^(2)alpha)/2)`
Hence `omega=(sinalpha+sqrt(3+sinalpha))/l` ( in clockwise direction)