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

Hint: 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
\[\theta = \dfrac{1}{2}\left( {\dfrac{{3F}}{{mL}}} \right){t^2} \\
\Rightarrow \theta = \dfrac{{3F{t^2}}}{{2mL}} \]
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.