Question

A uniform rod of mass m, length L, area of cross-section A is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity $\omega$ in a horizontal plane. If Y is the Young’s modulus of the material of rod, the increase in its length due to rotation of rod is

• A. $\frac{m\omega^{2}L^{2}}{AY}$
• B. $\frac{m\omega^{2}L^{2}}{2AY}$
• C. $\frac{m\omega^{2}L^{2}}{3AY}$
• D. $\frac{2m\omega^{2}L^{2}}{AY}$

Answer 1

Answer: (C)

$\frac{m\omega^{2}L^{2}}{3AY}$

Answer 2

:

Consider a small element of length dx at a distance from the axis of rotation.

Mass of the element,

$dm = \frac{m}{L}dx = \mu dx$

Where $\mu = \frac{m}{L}$

The centripetal force acting on the element is

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

As this force is provided by tension in the rod (due to elasticity), so the tension in the rod at a distance x from the axis of rotation will be due to the centripetal force due to all elements between x = x to x = L.

$\therefore T = \int_{x}^{L}{\mu\omega^{2}xdx}$

$= \frac{\mu\omega^{2}}{2}\lbrack L^{2} - x^{2}\rbrack$

Let dl be increase in length of the element. Then

$Y = \frac{T/A}{dl/dx}$

$dl = \frac{Tdx}{YA} = \frac{\mu\omega^{2}}{2YA}\lbrack L^{2} - x^{2}\rbrack dx$ [Using (i)]

So the total elongation of the whole rod is

$l = \int_{0}^{L}{\frac{\mu\omega^{2}}{2YA}\lbrack L^{2} - x^{2}\rbrack dx}$

$= \frac{\mu\omega^{2}}{2YA}\left\lbrack L^{2}x - \frac{x^{3}}{3} \right\rbrack_{0}^{L}$

$= \frac{1}{3}\frac{\mu\omega^{2}L^{3}}{YA} = \frac{1}{3}\frac{m\omega^{2}L^{2}}{YA}$