Question

A uniform rod of mass $m$ and length $l$ rotates in a horizontal plane with an angular velocity $\omega $ about a vertical axis passing through one end. The tension in the rod at a distance $x$ from the axis is
A. $\dfrac{1}{2}m{\omega ^2}x$
B. $\dfrac{1}{2}m{\omega ^2}\dfrac{{{x^2}}}{l}$
C. $\dfrac{1}{2}m{\omega ^2}l\left( {1 - \dfrac{x}{l}} \right)$
D. $\dfrac{1}{2}\dfrac{{m{\omega ^2}}}{l}\left[ {{l^2} - {x^2}} \right]$

Answer 1

Hint: Consider a small portion of $dx$ in the rod at a distance $x$ from the axis of the rod. So , the mass of this portion will be $dm = \dfrac{m}{l}dx$ (as the uniform rod is mentioned). Angular velocity can be defined as the rate of change of angular displacement with time. As the body follows a curved path centripetal force will be acting on it. Centripetal force is defined as the force required to keep a body in a circular path.
Formula Used: Tension on the part will be the centrifugal force acting on it , i.e.
 $dT = - dm{\omega ^2}x$

Complete step by step answer:
 Let a uniform rod perform circular motion about a point. We have to calculate the tension in the rod at a distance $x$ from the axis of rotation. Let mass of the small segment at a distance $x$ is $dm$
So,
$dT = dm{\omega ^2}x = \left( {\dfrac{m}{l}} \right)dx{\omega ^2}x = \dfrac{{m{\omega ^2}}}{l}\left[ {xdx} \right]$
Integrating both sides
$\int\limits_x^l {dT} = \dfrac{{m{\omega ^2}}}{l}\int\limits_x^l {xdx} $
$ \Rightarrow T = \dfrac{{m{\omega ^2}}}{l}\left[ {\dfrac{{{x^2}}}{2}} \right]_x^l$
$\therefore T = \dfrac{{m{\omega ^2}}}{{2l}}\left[ {{l^2} - {x^2}} \right]$

Hence, option D is correct.

Note: In this question it is said that a uniform rod of mass and length is given that is to be rotated on the horizontal plane with a velocity at a vertical axis passing through one end it is said to find the tension in the rod at a distance $x$from the axis.
  Here, we have assumed a small portion of rod at a fixed distance from the axis of the rod and a mass of the small portion of uniform rod as mentioned .So the tension on that part will be the centrifugal force acting on it because , tension is directed away from the centre whereas,$x$ is being counted towards the centre ,if you solve it considering centripetal force, then the force will be positive but limit will be counted from $x$to $l$.And hence we got the correct answer.