Question

A thin rod of length $L$ and mass $M$ is held vertically with one end on the floor and is allowed to fall. Find the velocity of the other end when it hits the floor, assuming that the end on the floor does not slip:
(A) $\sqrt {\dfrac{{3g}}{L}}$
(B) $\sqrt {3gL}$
(C) $\sqrt {\dfrac{L}{{3g}}}$
(D) $\sqrt {\dfrac{g}{{3L}}}$

Answer 1

First of all, we will find the expression for potential energy and then the rotational kinetic energy. The potential energy is converted into rotational kinetic energy. We will equate the two expressions and then manipulate accordingly and obtain the result.

Complete step by step answer:
In the given problem, we are supplied the following data:
The thin rod is of length $L$ and its mass $M$ .
The rod falls vertically downward and it hits the floor.
We are asked to find the velocity of the other end when it hits the floor.
In this numerical, we will apply the concept of conservation of energy. This states that energy can neither be destroyed nor be created but only can be transformed from one form to the other.
When the rod is at a position above the ground, then it contains potential energy, due to its position and tendency to come down due to acceleration of gravity. As soon as the rod is dropped, it falls off and the potential energy starts to transform into rotational kinetic energy. In this case, the rod will possess a moment of inertia due to the rotational motion along with angular velocity.
So, we can say that the potential energy is equal to the rotational kinetic energy.
The expression of potential energy is given by:
$P.E = mgh$ …… (1)
Where,
$P.E$ indicates potential energy.
$m$ indicates mass of the body.
$g$ indicates acceleration due to gravity.
$h$ indicates position of the object from the ground.
Again, the rotational kinetic energy is given by the expression:
$R.K.E = \dfrac{1}{2}I{\omega ^2}$ …… (2)
Where,
$R.K.E$ indicates the rotational kinetic energy.
$I$ indicates the moment of inertia.
$\omega$ indicates the angular velocity.
Now, from equation (1) and (2), we can write:

$$P.E = R.K.E \\\ mgh = \dfrac{1}{2}I{\omega ^2} \\\ mg\left( {\dfrac{L}{2}} \right) = \dfrac{1}{2} \times \left( {\dfrac{{m{L^2}}}{3}} \right) \ {\left( {\dfrac{v}{L}} \right)^2} \\\ mgL = \dfrac{{m{v^2}}}{3} \\\$$

Again, further manipulation gives:
$v = \sqrt {\dfrac{{3g}}{L}}$
Hence, the velocity of the other end when it hits the floor is $\sqrt {\dfrac{{3g}}{L}}$ .
The correct option is A.

Note: In the given problem, remember that in the case of rotational kinetic energy, angular velocity comes into play but not the linear velocity. Many students seem to have such a confusion with it. Whenever a body changes its position, then the energy transformation takes place.