A uniform rod `AB` of mass `m = 2 kg` and length `l = 1.0 m` is placed on a sharp support `P` such that `a = 0.4 m` and `b = 0.6 m`. A. spring of force constant `k= 600 N//m` is attached to end `B` as shown in Fig. To keep the rod horizontal, its end `A` is tied with a thread such that the spring is `B` elongated by `1 CM`. Calculate reaction of support `P` when the thread is burnt.

Explanation

Just after burning of a thread, torque about point `P`.

`tau_(P)=mg(l/2-a)+kxb`
`I_(p)alpha=mg(l/2-a)+kxb`
`I_(p)=(ml^(2))/12+m(l/2-a)^(2)=56/300kgm^(2)`
`56/300alpha=28/5implies alpha=30 rad//s`
Equation of motion at this instant
`mg+kx-R=ma_(6)=m(l/20-a)`
`20+6-R=2x30x1/10`
`implies R=20 N`