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 $U_o$ represents elastic strain energy stored in the rod $OA$, $U_B$ represents elastic strain energy stored in the rod $\left(OB\right)$ then choose correct alternative (s).

• A. $T_o=\frac{4}{3}{T}_B$
• B. $T_o=2T_B$
• C. $U_o=\frac{16}{11}{U}_B$
• D. $U_o=4U_B$

Answer 1

Answer: (A), (C)

The correct options are
A $T_o=\frac{4}{3}{T}_B$
C $U_o=\frac{16}{11}{U}_B$

At fixed end,
$T_o=\frac{1}{2}m{\omega }^{2}L$
At $x,$ from fixed end
$T=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-x^2)$


At $B$,
$\therefore T_B=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-\frac{L^2}{4})\Rightarrow T_B=\frac{3}{8}m{\omega }^{2}L$

i.e., $T_o=\frac{4}{3}{T}_B$
Again, $T=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-x^2)$
Elastic strain energy
$\therefore dU=\frac{1}{2}\frac{m{\omega }^2}{L}(L^2-x^2)dx$
$\therefore U=\frac{1}{2}\frac{m{\omega }^2}{L}\underset{0}{\overset{L}{\int }}{L}^{2}dx-\underset{0}{\overset{L}{\int }}{x}^{2}dx=\frac{1}{2}\frac{m{\omega }^2}{L}(x{L}^2-\frac{x^3}{3})$
For $OA,x=L$
$\therefore U_o=\frac{1}{3}m{\omega }^2{L}^2$
and for $OB,x=\frac{L}{2}$
$U_B=\frac{1}{2}\frac{m{\omega }^2}{L}(\frac{L^3}{2}-\frac{L^3}{24})=\frac{m{\omega }^2}{L}\frac{11L^3}{48}$
i.e., $U_o=\frac{16}{11}{U}_B$