A drinking straw of length 3a/2 and mass 2m is placed on a square table of side 'a' parallel to one of its sides such that one third of its length extends beyond the table.An insect of mass m/2 lands on the inner end of the straw(i.e the end which lies on the table)and walks along the straw until it reaches the outer end.It doesn't topple even when another insect lands on top of the first one.Find the largest mass of the second insect that can have without toppling the straw (neglect friction)

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3 Answers
(Our origin is the edge of the table)
This is a very good question indeed!
The CM is 1/4a from the edge of the table.
The first insects is of mass m/2 on the side of the table so at position a
The CM of the whole set up is
2m(a/4)+(m/2)(a) = C(2m+m/2)
C=2/5.a
If this insect moves to the other end of the straw...
Assume that the CM of the straw moves x towards the table
The insect moves 2ax away from the current position .. so its new position will be
x.2m+(2ax)m/2=0
5x/2+a=0
x=2/5a
so the insect will move to 8a/5 from its old position which was a
so its new position will be 3a/5
But the CM will not shift..
The old CM was 2/5a
The old mass was 2m+m/2 =5/2m
extra mass that can sit = w let..
so CM should become the edge of the table
2/5a.(5m/2) + w(3a/5) = 0
am+3/5aw=0
w=5/3m
check this whole thing for errors if any :)
But i think this is the whole method...
My ans is 5m
I am considering length as 'a'. It should not mater at the end.
when insect moves from inner end to outer let COM of pipe moves x distance inwards.( com of pipe is initially a/6 m inside of edge)
then COM of system does not move> 2m(l/6)+m/2(2a/3) = 2mx  m(a/2x)
x= 11a/30 a/2x=2a/15( distance of insect from edge later)
Let w be the max mass of another insect and now after it lands COM of System should lie on the edge for rotational equilibrium. Second insect is also at distance 2a/15 from edge.
w(2a/15) + m/2(2a/15) = 2m(11a/10)
w(2a/15)=10ma/15
w = 5m