Question

A thin uniform rod of length l and mass m rotates uniformly with angular velocity w in the horizontal plane passing through one of its ends. what is the tension in the rod as a function of x as rotational motion?

Answer 1

A thin uniform rod of length l and mass m rotates uniformly with angular velocity w in the horizontal plane passing through one of its ends. The tension in the rod as a function of x as rotational motion is

$\boxed{T=\frac{m\omega^2(l^2-x^2)}{2l}}$

Explanation:

Taking a small mass of dm at the distance x

The centrifugal force acting on this mass

$dF=dm \times x\times \omega^2$

Again

Mass per unit length $=m/l$

Thus mass of length dx $dm=\frac{mdx}{l}$

Thus

$dF=\frac{m\omega^2xdx}{l}$

Tension in the mass dm

$dT=dF$

$\implies dT=\frac{m\omega^2xdx}{l}$

Tension in the whole rod

$\int dT=\int\Bigr|_x^l\frac{m\omega^2xdx}{l}$

$\implies T=\frac{m\omega^2}{l}\int\Bigr_x^l xdx$

$\implies T=\frac{m\omega^2}{l}\frac{x^2}{2}\Bigr|_x^l$

$\implies T=\frac{m\omega^2}{2l}(l^2-x^2)$

$\implies T=\frac{m\omega^2(l^2-x^2)}{2l}$

Hope this helps.

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