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., $U_0=\frac{16}{9}{U}_B$