A uniform rod of mass m and length l rotates in a horizontal plan… - Vidyadip

MCQ

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
$\frac{1}{2}\,m{\omega ^2}l\,\left( {1 - \frac{x}{l}} \right)$
✓
$\frac{1}{2}\,\frac{{m{\omega ^2}}}{l}\,\left( {{l^2} - {x^2}} \right)$

✓

Answer

Correct option: D.

$\frac{1}{2}\,\frac{{m{\omega ^2}}}{l}\,\left( {{l^2} - {x^2}} \right)$

d
The mass of the element $=d m=\frac{m}{l} d x$

The force on the element towards the axis $=T-(T+d T)=-d T$

$\therefore-d T=(d m) \omega^{2} x=\left(\frac{m}{l} d x\right) \omega^{2} x$

$T=-\frac{m \omega^{2}}{l} \cdot \frac{x^{2}}{2}+$ constant

For $x=1, T=0 . \quad \therefore$ the constant $=\frac{1}{2} \cdot \frac{m \omega^{2} l^{2}}{l}$

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

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