A thin uniform rod of mass `M` and length `L` is lying on a smooth horizontal floor. It is struck by a force `F` normally at one end. The contact lasts for a very small time `t_(0)`. Find the distance traveled by the rod in the time in which it is turned by `pi//2`.

Explanation

Linear impulse `Ft_(0)=Deltap=M(v-0)=Mv` (`i`)
Angular impulse `F(L)/(2)t_(0)=DeltaL=I_(0)(omega-0)=(ML^(2))/(12)omega` (`ii`)
Dividing (`ii`) by (`i`), we get
`(L)/(2)=(L^(2)omega)/(12v)impliesomega=(6v)/(L)`
Angular motion, `theta=omegat` (`iii`)
Linear motion, `d=vt` (`iv`)
Dividing (`iv`) by (`iii`), we have
`(d)/(theta)=(v)/(omega)=(L)/(6)impliesd=(L)/(6)theta=(L)/(6)xx(pi)/(2)=(piL)/(12)`