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 $T_B-T_C=\frac{3m{\omega }^{2}l}{8}$

Taking any position at a distance $x$ from $A$.


From F.B.D. of strip $PC$ :


Mass of element $PC=m^{\prime }=\frac{m}{l}(l-x)$
Radius of rotation of strip $PC$ :
$r=x+\frac{(l-x)}{2}=\frac{(l+x)}{2}$

Here,
$T=m^{\prime }{\omega }^{2}r$
$T=\frac{m}{l}(l-x)\times {\omega }^2\times \frac{(l+x)}{2}$
$T=\frac{m{\omega }^2}{2l}(l^2-x^2)$
$T=\frac{m{\omega }^{2}l}{2}(1-\frac{x^2}{l^2})$

Now,
$T_B=\frac{m{\omega }^{2}l}{2}(1-\frac{l^2}{4l^2})=\frac{3m{\omega }^{2}l}{8}$
($\because x=\frac{l}{2}\text{for}B$)

Similarly, $T_c=0$ ($\because x=l$ for C)

$\Rightarrow T_B-T_C=\frac{3m{\omega }^{2}l}{8}$

Hence, option $\left(a\right)$ is the correct answer.