Question

a uniform rod of mass m, lengh L, arwa of cs A is rotated about an axus passing through one of its ends and perpendicular to its length with constant angular velocity w in a horizontal plane, If Y is younds modulus of the material of the road, the increase in its lengtg due to rotation of rod is...​

Answer 1

First of all let's consider a small element of length dx at a distance x from the axis of rotation (which passes through the centre).

Mass of the element, dm = m/L dx

Therefore dm = μ dx

• μ = m/L

Where μ is the linear mass density.

Let's calculate centripetal force acting on the small element! :D

$\sf:\implies\:dT=\dfrac{dm\:v^2}{x}$

$\sf:\implies\:dT=\dfrac{dm(x\omega)^2}{x}$

$\sf:\implies\:dT=dm\:\omega^2x$

$\sf:\implies\:dT=\mu\omega^2x\:dx$

Here centripetal force works as the tension force.

In order to find net tension force in the string, we have to integrate the above equation from x = x to x = L.

$\displaystyle\sf:\implies\:T=\int\limits_x^{L}\:\mu \omega^2x\:dx$

$\sf:\implies\:T=\dfrac{\mu\omega^2}{2}(L^2-x^2)$

Let dl be increase in length of the element. It can be calculated by using formula of Young's modulus.

$\sf:\implies\:Y=\dfrac{T}{A}\times\dfrac{dl}{dx}$

$\sf:\implies\:dl=\dfrac{T\:dx}{YA}$

$\sf:\implies\:dl=\dfrac{\mu\omega^2}{2YA}[L^2-x^2]\:dx$

Total elongation of the rod;

$\displaystyle\sf:\implies\:l=\int\limits_0^L\:\dfrac{\mu\omega^2}{2YA}[L^2-x^2]\:dx$

$\sf:\implies\:l=\dfrac{\mu\omega^2}{2YA}\left[L^2x-\dfrac{x^3}{3}\right]_0^L$

$\sf:\implies\:l=\dfrac{1}{3}\cdot\dfrac{\mu\omega^2L^3}{YA}$

$:\implies\:\underline{\boxed{\bf{\orange{l=\dfrac{m\omega^2L^2}{3YA}}}}}$