rotation

is ans mg/2√3????

for concept in IAOR, look at this....
http://targetiit.com/iit-jee-forum/posts/rotation-instantaneous-centre-of-rotation-12434.html

tho doesnt give full concept but a basic part is given there...

@=60°

we know, Γ=Iα
so, mg.(^L/₂)=¹/₃mL₂.α
putting α=^a/(_^L/2)
the exp boils down to,
a=³/₄g

now, balancing forces in vertical direction,
mg-Tsin@=ma
thus, Tsin@=^mg/₄
thus, T=^mg/(_4sin@)=^mg/_2√3

hi ashis!!

was this a mcq question??
if yes then can u give the options given???

i doubt it coz we hav to draw perp on velocity vector....thus we get the point of tieing of string n rod...isnt it???

now i get the full explanation to the problem....

as we know, the IAOR lies on the point where the perpendicular of the velocity vector of the centre of mass meets the line on which pure rolling occurs(here the string)

thus the point where the string is attached to thee rod forms the IAOR..

now my solution in #4 follows..

@ashis...hope it has answered ur doubt in #5
also verify if it is a mcq...please mention the options given in case it is...i have a masssst 2 second technique to solve this in case we get the option...bt that needs the option...[1][1][1]

indeed then i was horribly wrong and the pt of rotation will be where the string is attached to the roof....right....i was damn fool....:P

so i was right???[5]hah ha.....he he.....yessssssssssssssssssssssssssssss!!!!![10][9][8][6]