good one

sorry
i used torque a Nl/2 but it should be multiplied by sinθ
aneway that wont affect result as in rhs i used α instead αsinθ

acclereration of the end point will not be ay/2 it will be 2ay

and how are you getting
W-N=may/2 (shoudnt it be "may")

and how αl=ay?

should i solve it????????

and yes i observed centre of mass is moving along a circle
but how do i proceed

so as the com moves in a circle , its direction of velocity at any instant =tangent to the circle!!

so just move it from θ1 to θ2 and conserve energy (as normal rection doent "work" here)
then its up to you!!

wrting the soln for a general θ1 will be tedious (i feel lazy)

let the length of the rod be 2L

ok so let me write the full solution for he case when θ1=90 (i.e when the centre of mass is at a height L)

and let θ be the angle at which the rod loses contact from the wall

as it loses contact from the wall we can say that at that point of time the acceleration of the centre of mass of the rod towards the horizontal direction is zero (as there are no forces in that directions)

first the work done by normal reactions on both the sides is zero( as the point of application of normal reaction moves perpendicular to the normal reaction)

so we can conserve energy

intially kinetic energy = 0 and potential energy = mgL

finally potential energy =mgLsinθ

and let final kinetic energy = (1/2)mv²

then by conservation of energy

mgL=mgLsinθ+1/2mv²

we get v²=2gL(1-sinθ)

but as in the final state the rod loses contact with the wall ax=0

and further we observe that the centre of mass is moving in a circular path

hence the velocity of the centre of mass at any instant will be at the tangent to the path

so in the final state the velocity of the centre of mass makes an angle of 90-θ with the horizontal

so acceleration in that direction =
d(vsinθ)/dt=(d(vsinθ)/dθ)*dθ/dt

as the above quantity is zero in the final state

we get (d(vsinθ)/dθ)=0 or dθ/dt=0 --- (i)

but dθ/dt cannot be zero (af the centre of mass has velocity = Ldθ/dt as it is moving in a circle)

so d(vsinθ)/dθ=0

but v²=2gL(1-sinθ)

so v²sin²Î¸=2gLsin²Î¸(1-sinθ)

diffrerentiating both sides w.r.t θ

2vsinθd(vsinθ)/dθ=4gLsinθcosθ-6gLsin²Î¸cosθ

from (i)

0=4gLsinθcosθ-6gLsin²Î¸cosθ

we get sinθ = 2/3

that means it will leave the wall when sinθ=2/3

hence height of centre of mass at that instant = Lsinθ=2L/3

height of tip=2*(2L/3)=4L/3

intially the height of the tip was = 2L

so the tip now leaves the wall when its height is 2/3 of the original height

(just realized now , in the question it should have been 2/3rd)

oh! i apologize. its 2/3rd. anyway i had calculated it would fall 1/3rd by thge time :)).

if its 2/3 rd then my previous solution is perfectly right

but u didnt get that answer

u got y=8h/11

not my first post i realised my mistake of considering it as pure rotational

see my second post i got 2/3h easily

what is the base length

im getting a solution involving base length

no then its wrong

u will not require anything

you have to prove it!

normal reaction from other wall is 0 as it leaves the vertical wall

yes ive used torque as Nl/2 only

but l will vary !