Question

A uniform thin rod of mass m and length l is standing vertically along the y axis

Answer 1

your question is incomplete . A complete question is ------> a uniform thin rod of mass m and length l is standing vertically along y- axis on a smooth horizontal surface, with its lower end at origin (0,) a slight disturbance at t = p cause the lower end to slope on smooth surface along positive x - axis and the rod starts falling the acceleration vector of vector centre of mass of rod the rod during its fall is ?

solve :- see free body diagram,
mg - N = ma { here a is acceleration of centre of mass } .........(1)

now, $\tau=I\alpha$

or, $N\frac{l}{2}cos\theta=\frac{1}{12}ml^2\alpha$

or, $\alpha=\frac{6Ncos\theta}{ml}$

at contact point , constraint equation is
$a=\frac{l}{2}cos\theta.\alpha$

or, $a=\frac{l}{2}cos\theta\frac{6Ncos\theta}{ml}$

or, $a=\frac{3Ncos^2\theta}{m}$

or, $N=\frac{am}{3cos^2\theta}$

now from equation (1),

$a=\frac{g}{\left(1+\frac{1}{3cos^2\theta}\right)}$