rotational mechanics

Consider an element of width dy and having mass dm.

Centripetal force acting on this element ,
d\mathcal{F}_c=\omega^2 y\,dm=\frac{M}{L}\omega^2y\,dy

Net centripetal force acting on the rod till length x ,
\mathcal{F}_{c,x}=\int\limits_0^x \frac{M}{L}\omega^2y\,dy=\frac{M\omega^2 x^2}{2L}

Net centripetal force acting on the whole rod ,
\mathcal{F}_{c,l}=\int\limits_0^l \frac{M}{L}\omega^2y\,dy=\frac{M\omega^2 L}{2}

\mathcal{F}_{c,x}+T=\mathcal{F}_{c,l}
T=\frac{1}{2}M\omega^2 L\left(1-\frac{x^2}{L^2}\right)

Is the answer Mw^2L/2

Sry dude the answer must be M*w^2*x^2/2L....See assume a rod of lenght L which is hinged at 1 end.....its rotating with an angular velocity w......now take an element dy of mass dm at a distance y.......Now,
dm/dy=M/L -> dm=M*dy/L
Now...at d distance y,Tension=centrifugal force
So,dT=dm*w^2*y -> dT=M*w^2*y*dy/L
Integrating,we get.......T=M*w^2*x^2/2L.....