Question

A rigid rod of mass $m$ and lengths $l$, is being rotated in horizontal plane about a vertical axis. passing through one end $A$. If $T_A,T_B$ and$T_C$ are the tensions in rod at point $A$, mid point $B$ and point $C$ of rod respectively, then:

• A. $T_C=0$
• B. $T_B=\frac{3}{4}{T}_A$
• C. $T_B=\frac{T_A}{2}$
• D. $T_A=m{\omega }^{2}l$

Answer 1

The correct option is B $T_B=\frac{3}{4}{T}_A$

We have torsion,

$T=\frac{M(l^2-x^2)w^2}{2l}$

where x is the distance from A,

$\therefore T_A=\frac{M{l}^2{w}^2}{2l}=\frac{Ml{w}^2}{2}{T}_B=\frac{M(l^2-{\left(\frac{l}{2}\right)}^2)w^2}{2l}=\frac{3}{4}\frac{Ml{w}^2}{2}=\frac{3}{4}{T}_A$