Question

A uniform meter stick of mass m is pivoted about a horizontal axis through its lower end O. Initially it is held vertical and is allowed to fall freely. Its angular velocity at the instant when it makes an angle of ${60^ \circ }$ with the vertical is
A. $\sqrt {\dfrac{g}{3}} $
B. $\sqrt {\dfrac{{3g}}{2}} $
C. $\sqrt {\dfrac{g}{4}} $
D. $\sqrt {\dfrac{{2g}}{3}} $

Answer 1

Hint: The sum of total initial energies is equal to the sum of total final energies because energy can neither be created nor be destroyed, it can only be transferred from one form to another i.e., initial potential energy is equal to the sum of rotational and potential energies.

Complete step by step answer:
A uniform meter stick of mass m is fixed about a horizontal axis through its lower end O. Initially when the stick is held vertically, it only possesses potential energy due to its height from the axis. Let the initial potential energy be ${P_1}$.
${P_1} = mg\left( {\dfrac{L}{2}} \right)$ [ $P = mgh$ and $h = \dfrac{L}{2}$ as the weight of the stick acts at its centre]
Now, when the stick is allowed to fall freely, it forms an angle of ${60^ \circ }$ with the vertical i.e., with the initial position of the stick. Thus, at this height, the stick has both potential and rotational kinetic energy (due to rotation).
Let the final potential energy be ${P_2}$ and rotational kinetic energy be R.
${P_2} = mg{h_1}$ [${h_1}$ is the final height that the stick has when it makes an angle of 60 degree]
$\implies {P_2} = mg\left\{ {\left( {\dfrac{L}{2}} \right)\cos {{60}^ \circ }} \right\}$
$\implies {P_2} = mg\left( {\dfrac{L}{2}} \right) \times \dfrac{1}{2}\left[ {\cos {{60}^ \circ } = \dfrac{1}{2}} \right]$
$\implies {P_2} = \dfrac{{mgL}}{4}$
According to conservation of energy, sum of total initial energies is equal to the sum of final energies.
${P_1} = {P_2} + R$
$\implies \dfrac{{mgL}}{2} = \dfrac{{mgL}}{4} + \dfrac{1}{2}I{\omega ^2}$
$\implies \dfrac{{mgL}}{2} - \dfrac{{mgL}}{4} = \dfrac{1}{2}\left( {\dfrac{{m{L^2}}}{3}} \right){\omega ^2}$
$\implies \dfrac{{2 mgL - mgL}}{4} = \dfrac{{m{L^2}{\omega ^2}}}{6}$
$\implies \dfrac{{mgL}}{4} = \dfrac{{m{L^2}{\omega ^2}}}{6}$
$\implies {\omega ^2} = \dfrac{{6mgL}}{{4m{L^2}}}$
$\implies {\omega ^2} = \dfrac{{3g}}{{2L}}$
$\implies {\omega ^2} = \dfrac{{3g}}{2}$ [ Since it is a metre scale i.e., $L = 1m$]
$\therefore \omega = \sqrt {\dfrac{{3g}}{2}} $

So, the correct answer is “Option B”.

Note:
When the metre stick falls freely and the point at which it makes an angle of ${60^ \circ }$ with the vertical, the stick has both rotational kinetic and potential energy because the stick is at certain height from the ground and rotational kinetic energy as it makes a turn of ${60^ \circ }$ with the initial direction.