1I have seen the referred post. The problem is little difficult since the string is inclined. For vertical string, it is simpler.
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
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
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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