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) m*l*ω^{2}

(b) m*l*ω^{2}/4

(c) 3m*l*ω^{2}/8

(d) m*l*ω^{2}/8

Ans: (c)

Solution: Consider a elemental mass at a distance x from one end is

dm = (m/l)dx

dF = (dm)ω^{2} 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 $