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A thin uniform rod XY of length 2l is hinged at one end X to the floor and stands vertical. When allowed to fall, the angular speed with which the rod strikes the floor is
A. $\sqrt{\dfrac{3g}{4l}}$
B. $\sqrt{\dfrac{3g}{2l}}$
C. $\sqrt{\dfrac{3g}{l}}$
D. $\sqrt{\dfrac{g}{2l}}$

Last updated date: 23rd Feb 2024
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IVSAT 2024
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Hint:Calculate the work done on the rod by the gravitational force when it comes down from its initial position. Then by using the work energy theorem equate the work done to change in kinetic energy of the rod.

Formula use:
$W=mgh$, where W is the work done by gravity on the rod of mass m when its centre of mass comes down by a height h. g is acceleration due to gravity.
$K=I{{\omega }^{2}}$, where K is the rotational kinetic energy of a body rotating with an angular velocity $\omega $ and its moment of inertia about the rotational axis is I.

Complete step by step answer:
It is said that the rod is released from rest when it is vertical or perpendicular to the floor. The rod will fall down under the influence of the gravitational force acting downwards. And since the end X of the rod is hinged, it will rotate about the point X. Here, we can use the work energy theorem. According to the work-energy theorem, the work done on a body is equal to the change in its kinetic energy.
i.e. $W=\Delta K$.

Let us assume that the mass of the rod is uniformly distributed and its centre of mass is at its centre. Therefore, in this case, the centre of mass of the rod comes down by a height of $h=\dfrac{l}{2}$.
Therefore, the work done on the rod is equal to $W=mgh=mg\dfrac{l}{2}$ ….. (i)
Since the rod is under rotational motion, it gains rotational kinetic energy. Let the kinetic energy of the rod when it comes down be $K=I{{\omega }^{2}}$
We know that the rod was at rest initially. Then this means that the change in its kinetic energy is $\Delta K=I{{\omega }^{2}}$ ….. (ii)
The moment of inertia of a rod about its end point is equal to $I=\dfrac{m{{l}^{2}}}{3}$.

Substitute the value of I in (ii).
$\Rightarrow \Delta K=\dfrac{m{{l}^{2}}}{3}{{\omega }^{2}}$ ….. (iii).
Now, equate (i) and (iii).
$\Rightarrow mg\dfrac{l}{2}=\dfrac{m{{l}^{2}}}{3}{{\omega }^{2}}$
$\Rightarrow {{\omega }^{2}}=\dfrac{3g}{2l}$
$\therefore\omega =\sqrt{\dfrac{3g}{2l}}$.
This means that the angular speed of the rod just before it strikes the floor is equal to $\sqrt{\dfrac{3g}{2l}}$.

Hence, the correct option is B.

Note:We know that gravitational force is conservative force. Therefore, we can also use the law of conservation of mechanical energy of the system, where mechanical energy is the sum of kinetic energy and potential energy. The potential energy is equal to the negative of work done by the gravitational force.
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