InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A uniform rod of mass `m` and length `l` is placed over a smooth horizontal surface along the `y`-axis and is at rest as shown in Fig. An impulsive force `F` is applied for a small time `/_ ` along `x`-direction at point `A`. The `x`-coordinate of end `A` of the rod when the rod becomes parallel to `x`-axis for the first time is [initially, the coordinate of centre of mass of the rod is `(0, 0)`] A. `(pil)/12`B. `l/2(1+pi/12)`C. `1/2(1-pi/6)`D. `l/2(1+pi/6)` |
|
Answer» Correct Answer - D As torque `=`change in momentum `F/_ =mv`(linear momentum)………i and `(Fl/2)/_ =(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]` |
|