Question

A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity ωabout 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. $\frac{1}{2}m\omega^{2}l\left( 1 - \frac{x}{l} \right)$
• D. $\frac{1}{2}.\frac{m\omega^{2}}{l}\left\lbrack l^{2} - x^{2} \right\rbrack$

Answer 1

Answer: (D)

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

Answer 2

The mass of the element = dm = (m/l)dx.

The force on the element towards the axis = T – (T + dT)

= - dT.

∴ - dT = (dm)ω² x = $\left( \frac{m}{l}dx \right)\omega^{2}x.$

T = - $\frac{m\omega^{2}}{l}.\frac{x^{2}}{2} + cons ⥂ \tan ⥂ t$

For x = 1, T = 0.

∴ the constant = $\frac{1}{2}.\frac{m\omega^{2}l^{2}}{l}$

∴ T = $\frac{1}{2}.\frac{m\omega^{2}}{l}\left( l^{2} - x^{2} \right)$