1
Manmay kumar Mohanty
·2010-03-25 01:27:09
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
1
bhaskar Chakraborty
·2010-03-25 02:06:55
I have seen the referred post. The problem is little difficult since the string is inclined. For vertical string, it is simpler.
1
akshay sharma
·2010-03-25 03:43:10

T cos30° = max
ax = √3T/2m ......(1)
mg - T sin30° = may
ay = g - T/2m ....................(2)
nw taking torque abt C.M. of the rod,
Tsin30°(l/2)=ml2α / 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.
=> ax cos37° + (αl/2)cos60° - aycos60° = 0
frm eqns (1),(2) nd (3) , we get
7T=2mg
T=(2mg) / 7
1
akshay sharma
·2010-03-25 03:47:26
is this the correct solution ??????
1
akshay sharma
·2010-03-25 04:48:26
so whts the correct answr ??????