Question

A uniform rod of mass $m$ and length $l$ rotates in a horizontal plane with an angular velocity $\omega$ about a vertical axis passing through one end. The tension in the rod at a distance $x$ from the axis is

• A. $\frac{1}{2}m{\omega }^{2}x$
• B. $\frac{1}{2}m{\omega }^2\frac{x^2}{l}$
• C. $\frac{1}{2}m{\omega }^{2}l(1-\frac{x}{1})$
• D. $\frac{1}{2}.\frac{m{\omega }^2}{l}[l^2-x^2]$

Answer 1

Answer: (D)

$\frac{1}{2}.\frac{m{\omega }^2}{l}[l^2-x^2]$

The mass of the element $=dm=\frac{m}{l}dx$
The force on the element towards the axis $=T-(T+dT)=-dT$
$\therefore -dT=\left(dm\right){\omega }^{2}x=\left(\frac{m}{l}dx\right){\omega }^{2}x$
$T=-\frac{m{\omega }^2}{l}.\frac{x^2}{2}+constant$
For $x=1$,$T=0$. $\therefore theconstant=\frac{1}{2}.\frac{m{\omega }^2{l}^2}{l}$
$\therefore T=\frac{1}{2}.\frac{m{\omega }^2}{l}(l^2-x^2)$