A Challenging One Indeed

For the semi - infinite rod to remain in equillibrium , the torques due to the two pieces of it separated by the support must be balanced .

Let the mass density of the rod be " ρ ( x ) " .

Let the rod be cut off at a distance " y " from the leftmost end .

Then , the support is at a distance of " y + L " from the leftmost end .

Balancing the torques due to the two pieces , we obtain , Γ = _y ∫ ^∞ ρ ( x ) [ x - ( y + L ) ] dx = 0

Now , we want " Γ = 0 " identically for any " y " , so , " Γ " must be a constant .

For that , we must have : -

dΓdy = 0

Or , L ρ ( y ) = _y ∫ ^∞ ρ ( x ) dx

Or , L ρ ' ( y ) = ρ ( y )

Or , ρ ( y ) = C e ^{- y / L} .....................( Solving the differential equation )

I think this is the required answer .

Not a olympiad level problem though , according to me of course , so no hard feelings : ) .

Balancing the torques due to the two pieces

Yeh wala step thoda elaborate karna na, plz!

I just differentiated " Γ " with respect to " y " . Since the limits of the integration also involve " y " , hence you have to use " Newton - Liebniz " rule here .

As for the doubt of Vivek : - Just write down the torques due to the two masses considering that the rod lies on the positive " x - axis " or something . Then you ' ll see it yourself .

I am too lazy right now Vivek , so if you can't " see " , then please drop a note in my chatbox : ) Sorry :( :D

d/dy of y ∫ ∞p[x]*[x-l]dx = p [y]*[y-l]...

rhs --same as above-- y ∫ ∞ ρ ( x ) dx

i split it into 2 integals--- x-l vala alag, y vala alag...

then applied d formulae of newton-leibn..

ddy ( _y ∫ ^∞ ρ ( x ) y dx ) = ρ ( y ) y + _y ∫ ^∞ ρ ( x ) dx

Not what you have written : )

oh,,,i was applying constant limits vala formulla,,,,

thanks a TON for d help...:]