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)
Sushovan Halder thanks a lot
Upvote·0· Reply ·2013-11-07 06:43:34