solve this

no cena u r rong.

the rod will tend to topple as the com is not in line with normal force .......

we can also say that the part of rod above the edge is at rest just after release .....

applying NLM on the rod we get ....

mg(L/2-L/3) = [ml²/12 + m(l/6)²]α ....................(1)

mg - N = ma = mαL/6 ..........................................................(2)

solve for N ...........

i was confusd abt the direction of normal rxn whether it is perpendicular to the rod or not.

Can u give any explanations for takin the direction of normal rxn .

at the point it rotates how can u find the tangent.

yup i know dat.