solve this

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 .

the normal rxn is perpendicular to the rod .... its always perpendicular to the common tangent na ....

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

3mg/4??

what iitcoming says...,

eqn 1) translatory motion of the rod...

eqn 2) rotational motion of the rod...

eqn 3) connection of the rot. and trans. motion of the rod...

yup i know dat.

third axis