62(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 2a-x away from the current position .. so its new position will be
-x.2m+(2a-x)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...
24i dont have answer.........will check and confirm
1My 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/2-x)
x= 11a/30 a/2-x=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
