A uniform rod of mass m, length l is placed over a smooth horizontal surface along y -axis and is at rest as shown in figure. An impulsive force F is applied for a small time `Deltat` along x-direction at point A. The x-coordinte of end A of the rod when the rod becomes parallel to x-axis for the first is (in itially the coordinates opf centre of mass of the rod is (0,0).

Options

A. `(pil)/12`
B. `l/2(1+pi/12)`
C. `1/2(1-pi/6)`
D. `l/2(1+pi/6)`

Answer

D

Explanation

As torque `=`change in momentum
`F/_\t=mv`(linear momentum)………i
and `(Fl/2)/_\t=(ml^(2))/12 omega` (angular momentum)..ii
Dividing eqn i and ii we get
`2=(12v)/(omegal)implies omega=(6v)/l`
using `S=ut`
Displacement of `CM` is `pi/2=omegat=((6v)/l)t`
and `=vt`
Dividing we get
`(2x)/pi=l/6impliesx=(pil)/12`
coordinates of A will be `[(pil)/12+l/2, 0]`