Q: A thin rod of mass m and length l is rotating in the horizontal plane about its one of the ends with constant angular velocity ω . The tension at the middle point of the rod is
(a) mlω2
(b) mlω2/4
(c) 3mlω2/8
(d) mlω2/8
Ans: (c)
Solution: Consider a elemental mass at a distance x from one end is
dm = (m/l)dx
dF = (dm)ω2 x
$ \displaystyle T = \int dF = \int_{l/2}^{l}\frac{m}{l}\omega^2 x dx $
$\displaystyle = \frac{m}{l}\omega^2 \int_{l/2}^{l} xdx $
$ \displaystyle = \frac{m}{l}\omega^2[\frac{x^2}{2}]_{l/2}^{l} $
$ \displaystyle = \frac{3}{8}m\omega^2 l $