**62**
Lokesh Verma
**·**2008-11-09 18:33:11
(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...

**24**
eureka123
**·**2008-11-16 10:39:39
i dont have answer.........will check and confirm

**1**
Adnan Kagzi
**·**2016-01-11 02:53:51
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/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