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. $\frac{1}{2}\mspace{6mu} m\omega^{2}x$
• B. $\frac{1}{2}\mspace{6mu} m\omega^{2}\frac{x^{2}}{l}$
• C. $\frac{1}{2}m\omega^{2}l\left( 1 - \frac{x}{l} \right)$
• D. $\frac{1}{2}\frac{m\omega^{2}}{l}\lbrack l^{2} - x^{2}\rbrack$

Answer 1

Answer: (D)

$\frac{1}{2}\frac{m\omega^{2}}{l}\lbrack l^{2} - x^{2}\rbrack$

Answer 2

Let rod AB­ performs uniform circular motion about point A. We have to calculate the tension in the rod at a distance x from the axis of rotation. Let mass of the small segment at a distance x is dm

So

$dT = dm\omega^{2}x$

$= \left( \frac{m}{l} \right)dx.\omega^{2}x = \frac{m ⥂ \omega^{2}}{l}$

[x d x]

Integrating both sides $\int_{x}^{l}{dT} = \frac{m\omega^{2}}{l}\int_{x}^{l}{xdx} \Rightarrow$

$T = \frac{m\omega^{2}}{l}\left\lbrack \frac{x^{2}}{2} \right\rbrack_{x}^{l}$

∴ $T = \frac{m\omega^{2}}{2l}\left\lbrack l^{2} - x^{2} \right\rbrack$