A thin uniform rod `AB` of mass `m` and length `L` is placed on a smooth horizontal table. A constant horizontal force of magnitude `F` starts acting on the rod at one of the ends `AB`, Initially, the force is perpendicular to the length of the rod. Taking the moment at which force starts acting as `t=0`, find.
`(a)` Distance moved by centre of mass of the rod in time `t_(0)`.
`(b)` Magnitude of initial acceleration of the end `A` of the rod.

Explanation

Let the rod be placed in `x-y` plane (horizontal) with initial orientation along `y`-axis. Further, assuming that the force acting is along `x`-axis , the situation is represented by the following figure.

`(a)` the acceleration of centre of mass is given by
`veca_(cm)=(vecF)/(m)=(F)/(m)hati`
The displacement is given by `x_(cm)=(1)/(@)a_(cm)t_(0)^(2)=(1)/(2)(F)/(m)t_(0)^(2)` `(:'a_(cm)` is constant)
`(b)` Initial acceleration of end `A` of the rod is
`veca_(A)=veca_(cm)+vecalphaxxvecr_(AO)` `(:'omega=0)`
`veca_(A)=(F)/(m)hati+((-6F)/(mL)hatk)xx((L)/(2)hatj)=(4F)/(m)hati`