Question

A rod OA of mass m and length L is rotating in a horizontal circle about point O as shown in figure. B is the mid-point of rod. If T represents tension in the rod at any point and U represents elastic potential energy per unit volume at any point of rod, then choose correct alternative (s).

• A. $T_0=\frac{4}{3}{T}_B$
• B. $T_0=2T_B$
• C. $U_0=\frac{16}{9}{U}_B$
• D. $U_0=4U_B$

Answer 1

Answer: (A), (C)

The correct options are
A $T_0=\frac{4}{3}{T}_B$
C $U_0=\frac{16}{9}{U}_B$

$T_0=\frac{1}{2}m{\omega }^{2}L$ (at fixed end)
and as $T=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-x^2)$

$\therefore T_B=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-\frac{L^2}{4})\Rightarrow T_B=\frac{3}{8}m{\omega }^{2}L$
i.e., $T_0=\frac{4}{3}{T}_B$
again, $T=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-x^2)$
$\therefore dW=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-x^2)dx\therefore W=\frac{1}{2}\frac{m{\omega }^2}{L}{\int }_0^L{L}^{2}dx={\int }_0^L{x}^{2}dx$
$\frac{1}{2}\frac{m{\omega }^2}{L}(L^3-\frac{L^3}{3})\therefore U_0=\frac{1}{3}m{\omega }^2{L}^{2}and{U}_B=\frac{1}{2}\frac{m{\omega }^2}{L}(\frac{L^3}{2}-\frac{L^3}{8})=\frac{3m{\omega }^2{L}^2}{16}$
$i.e.,U_0=\frac{16}{9}{U}_B$