A unifrom rod `AB` of length `4m` and mass `12 kg` is thrown such that just after the projection the centre of mass of the rod moves vertically upwards with a velocity `10 m//s` and at the same time its is rotating with an angular velocity `(pi)/(2) rad//sec` about a horizontal axis passing through its mid point. Just after the rod is thrown it is horizontal and is as shown in the figure. Find the acceleration (in `m//sec^(2)`) of the point `A` in `m//s^(2)` when the centre of mass is at the highest point. (Take ` = 10 m//s^(2)` and `pi^(2) = 10`)

Answer

5

Explanation

Time in which `C.M.` reaches its highest point `= 1 sec`. (from `v = u + at`, putting `v = 0, u = +10, a = 10 m//sec^(2)`) after projection angular velocity will not change as the torque of external forces is zero.
In `1 sec`., the rod will rotate by an angle `= omegat = (pit)/(2) xx t` rad.
The rod will be vertical with point `A` at the lowest point.
`:. a_(A) = g - omega^(2)L//2 = 10 - (pi^(2)4)/(4 xx 2) = 5 m//sec^(2)`.