Question

A thin uniform rod of mass $m$, length $L$, area of cross section $A$ and young's modulus $Y$ rotates at angular velocity '$\omega$' in a horizontal plane about a vertical axis passing through one of its ends, then

• A. tension in rod at distance 'r' from the axis of rotation is $\frac{m{w}^2(L^2-r^2)}{2L}$.
• B. elongation of rod is $\frac{2}{3}\frac{m{w}^2{L}^2}{AY}$.
• C. elongation of rod is $\frac{m{w}^2{L}^2}{3AY}$.
• D. elongation of small element of rod decreases linearly as we move away from axis.

Answer 1

Answer: (A), (C)

elongation of rod is $\frac{m{w}^2{L}^2}{3AY}$.



$T=m^{\prime }{w}^2[r+\frac{(L-r)}{2}]=\frac{m}{L}(L-r)w^2\left(\frac{L+r}{2}\right)$

$T=\frac{m{w}^2(L^2-r^2)}{2L}$


$Y=\frac{T}{A\frac{dl}{dr}}\Rightarrow dl=\frac{T}{AY}dr$

$\Delta l=\int dl={\int }_0^L\frac{m{w}^2}{2L}\frac{(L^2-r^2)dr}{AY}$

$\Delta l=\frac{m{w}^2}{2LAY}.\frac{2}{3}{L}^3=\frac{m{w}^2{L}^2}{3AY}$

Since tension does not decrease linearly with radius, elongation of small element also does not decrease linearly with radius.