length of the rod =2l mass=M
ant hit the rod with V=3m/s at a distance 2l/3 from the string T2 , which string will broken????
1concider that the ant colliding with the rod and coming to rest happens in an interval ∂t...
now we have, mV=(T1 +T2)∂t.....1)
and also for angular impulse about the centre of mass of the rod,
mVl/6 = {T1(l/2)-T2(l/2)}∂t.....2)
u cud also concider angular impulse about one of the tension points say
T2 (here u include the wt of the rod Mg as well)
so thus u can work out which one of T1 and T2 is greater thats the one that breaks first...this is the technical way of going about it...
u cud do this one roughly as well, going by common sense,(which is so uncommon, hence no gaurentee), it shud be T1 that breaks first..
9T2 > T1 but that doesnt mean any string needs to break
but if at all somethin breaks strg 2 breaks first
9hmm gordo uve said exactly opp of mine
wait checkin wats wrong with me
9k change the part following and urs will be correct
mVl/6 = {T1(l/2)-T2(l/2)}∂t.....2)
its actually mVl/6 = {T2(l/2)-T1(l/2)}∂t.....2)
also l =2l in q
9Dharun wats the ans
as u ve thanked gorodo alone i assume ans is T1 but are u sure ???
1If we consider the string to be inextensible .. then notice that if the rod rotates then one of the string will shorten and the another will lengthen .. so there should alwayts remain rotational equillibrium
hence the angular impulse given by the ant will have to balance the one given be the tensions
f T2>T1 then we have an opposite angular impulse which will help it remain in rotational equillibrium..
1sorry guyz...the answer shud be T2, i messed it up, actually used the length of the rod as L in think....u see thats the difference between me and someone else, i mess it up b'coz of my bloody silly mistakes, although the concept is rite, any ways, my sincere apologies to every one, including darun and especially pritam...
