1this 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
The length of rod is 'L' .What is acceleration of the rod and tension in the string just after the one of the string is snapped.
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6 Answers
11I have seen the referred post. The problem is little difficult since the string is inclined. For vertical string, it is simpler.
1
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