ANGULAR ACCELERATION

this thread might help, check it http://www.targetiit.com/iit-jee-forum/posts/rotation-instantaneous-centre-of-rotation-12434.html

the same question as urs is there at abt post #11 or something

I have seen the referred post. The problem is little difficult since the string is inclined. For vertical string, it is simpler.

T cos30° = ma_x
a_x = √3T/2m ......(1)

mg - T sin30° = ma_y
a_{y = g - T/2m ....................(2)}

nw taking torque abt C.M. of the rod,
Tsin30°(l/2)=ml²Î± / 12
αl=3T/m ....(3)

nw since the string neither slacks nor breaks , thus the acceleration of tied end must be zero along the string.

=> a_{x cos37° + (αl/2)cos60° - aycos60° = 0}

frm eqns (1),(2) nd (3) , we get

7T=2mg

T=(2mg) / 7

THE ANSWERS ARE NOT CORRECT