Question

A rod of mass $m$ and length $L$, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude $F$ acts in the rod at a distance of $\dfrac{L}{4}$ from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time $t$ after the motion starts.
A. $\dfrac{{3F{t^2}}}{{5mL}}$
B. $\dfrac{{5F{t^2}}}{{2mL}}$
C. $\dfrac{{3F{t^2}}}{{2mL}}$
D. $\dfrac{{2F{t^2}}}{{3mL}}$

Answer 1

As, the rod is free to rotate about a vertical axis, find the moment of inertia of the rod. Recall how the motion of a rotating object is affected when a force acts on the object, apply those conditions. And use the equation of motion for rotational motion to find the angular displacement of the rod.

Complete step by step answer:
Given, the mass of the rod is $m$, Length of the rod is $L$. A horizontal force of constant magnitude $F$ acts in the rod at a distance of $\dfrac{L}{4}$ from the centre and the force is perpendicular. We are asked to find, with how much angle the rod will rotate at time $t$ after the rod starts rotating.

As, the rod is free to rotate about the vertical axis through its centre, there will be moment of inertia and moment of inertia for a rod rotating about an axis through its centre and perpendicular to it is given by,
$I = \dfrac{{m{L^2}}}{{12}}$ (i)
When the force acts on the rod, it will exert torque on the rod which can be written as,
$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = rF\sin \theta$
$r$ is the distance from the axis where the force acts, here $r$ is given as $\dfrac{L}{4}$
$\theta$ is the angle between the force and the rod, here it is given that force is acting always perpendicular so $\theta = {90^ \circ }$.
$\therefore \tau = \dfrac{L}{4}F$ (ii)

Torque can also be written in terms of moment of inertia as,
$\tau = I\alpha$
$\alpha$ is the angular acceleration of the rod.
Putting the value of $I$, we get
$\tau = \dfrac{{m{L^2}}}{{12}}\alpha$ (iii)
Now, equating equation (ii) and (iii), we get
$\dfrac{{m{L^2}}}{{12}}\alpha = \dfrac{L}{4}F$
$\alpha = \dfrac{{12F}}{{4mL}}$
$\Rightarrow \alpha = \dfrac{{3F}}{{mL}}$ (iv)
Now, from equation of motion for rotational motion we have,
$\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$
$\theta$ is the angular displacement,
${\omega _0}$ is the initial angular velocity and
$t$ is the time taken.
Initially the rod was at rest so, initial angular velocity will be zero, that is
$\theta = \dfrac{1}{2}\alpha {t^2}$
Now, substituting the value of angular acceleration $\alpha$ in the above equation we get

\Rightarrow \theta = \dfrac{{3F{t^2}}}{{2mL}} $$ Therefore, the angle rotated by the rod after a time $$t$$ is $$\dfrac{{3F{t^2}}}{{2mL}}$$ **Hence, the correct answer is option C.** **Note:** For such questions, first check what the possibilities are when force acts on the body. Here it was a rotating body so when force acted upon it, a torque was produced. But in case force acts upon a body moving in a straight line, then there will be no torque produced and we have to apply different conditions as per the question.