Question

A uniform rigid rod of mass $m$ and length $l$, is being rotated in a horizontal plane about a vertical axis, passing through one end $A$. If $T_B$ and $T_C$ are the tensions in the rod at points $B$ and $C$, then choose the correct option, if $B$ is the mid point of the rod.

• A. $T_B-T_C=\frac{3m{\omega }^{2}l}{8}$
• B. $T_C-T_B=\frac{3m{\omega }^{2}l}{8}$
• C. $T_C-T_B=\frac{3m{\omega }^{2}l}{4}$
• D. $T_B-T_C=\frac{3m{\omega }^{2}l}{4}$

Answer 1

The correct option is A $T_B-T_C=\frac{3m{\omega }^{2}l}{8}$


Let us choose a linear element of length $dx$ and mass $dm$.
$dm=\lambda dx\Rightarrow dm=\frac{m}{l}dx$
Applying Newton's 2nd law on element
$T-(T+dT)=\left(dm\right){\omega }^{2}x$
$\Rightarrow \underset{T_C}{\overset{T_B}{\int }}-dT=\underset{x=l}{\overset{x=\frac{l}{2}}{\int }}\frac{m}{l}{\omega }^{2}xdx$
$\Rightarrow -(T_B-T_C)=\frac{m{\omega }^2}{2l}[\frac{l^2}{4}-l^2]$
$\Rightarrow T_B-T_C=\frac{m{\omega }^2}{2\text{/}l}\frac{3l^{\text{/}2}}{4}$
$\Rightarrow T_B-T_C=\frac{3}{8}m{\omega }^{2}l$