Question

A rod of length $L$ is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. The value of $x$ for point $P$, where tension is ${\left(\frac{1}{4}\right)}^{th}$ of the tension at $O$

• A. $\frac{L}{2}$
• B. $\frac{L}{4}$
• C. $\frac{3L}{4}$
• D. $\frac{\sqrt{3}L}{2}$

Answer 1

The correct option is D $\frac{\sqrt{3}L}{2}$


The net inward force should provide the centripetal force.
$-(T+dT)+T=dm{\omega }^{2}x$
$-dT=\frac{m}{L}{\omega }^{2}xdx[\because dm=\frac{m}{L}dx]$
Let $T_0$ be tension at O, tension at $L$ is $0$
$-{\int }_{T_0}^{0}dT=\frac{m{\omega }^2}{L}{\int }_0^{L}xdx$
$\Rightarrow T_0-0=\frac{m{\omega }^{2}L}{2}$
$\Rightarrow \frac{m{\omega }^2}{2}=\frac{T_0}{L}$
Now integrating from $x$ to $L$,
$-{\int }_T^{0}dT=\frac{m{\omega }^2}{L}{\int }_x^{L}xdx$
Let $T$ be tension at $P$
$T=\frac{m{\omega }^2}{2L}[L^2-x^2]$
Since $T=\frac{T_0}{4},$
$\frac{T_0}{4}=\frac{T_0}{L^2}[L^2-x^2]\Rightarrow x^2=\frac{3L^2}{4}$
$\Rightarrow x=\frac{\sqrt{3}L}{2}$