Question

If a light rod is rotated by angle $\theta$ and torque equation is given by $\iota =k\theta$. If rod is rotated by angle ${\theta }_0$ then the maximum tension in the rod is

• A. $\frac{3k({\theta }_0{)}^2}{\ell }$
• B. $\frac{2k({\theta }_0{)}^2}{\ell }$
• C. $\frac{k({\theta }_0{)}^2}{\ell }$
• D. $\frac{k({\theta }_0{)}^2}{2\ell }$

Answer 1

The correct option is C $\frac{k({\theta }_0{)}^2}{\ell }$

$\begin{array}{c}\text{Given, restoring torque}\left(\tau \right)=-k\theta \\ \Rightarrow I\cdot \alpha =-k\theta \\ \Rightarrow \begin{array}{cc}\Rightarrow -\left(\frac{k}{I}\right)\theta & \text{(here, I is moment of inertia of}\\ & \text{system about chord axis, and}\end{array}\\ \alpha \text{is angular acceleration.)}\end{array}$

$\begin{array}{c}\text{So,}\text{D (angular velocity)}=\sqrt{\frac{k}{I}}\text{.}\\ \text{Note that, I}=m{\left(\frac{L}{3}\right)}^2+\frac{m}{2}{\left(\frac{2L}{3}\right)}^2=\frac{m{L}^2}{3}\end{array}$

$\text{Here,}{\left(V_{max}\right)}_A=r\times \omega =\left(\frac{2l}{3}\right){\theta }_0\cdot \sqrt{\frac{k}{I}}$

$\text{When rod rotates by}\theta \text{o, max" tension is when velocity}$

$\text{is maximum and, tension equals the centripetal force.}$

$\text{Therfore,}{T}_{max}=\frac{\left(\frac{m}{2}\right)\cdot {\left(V_{max}\right)}_A^2}{\left(\frac{2l}{3}\right)}$

$\begin{array}{c}=\frac{ml}{3}\times {\theta }_0^2\times \frac{k}{I}(I=\frac{m{l}^2}{3})\\ =\frac{k{\left({\theta }_c\right)}^2}{l}\end{array}$