Question

A rod of length $1m$ pivoted at one end is released from rest when it makes $30°$ from the horizontal as shown in the figure below.If $\omega$ of rod is $\sqrt{n}$ at the moment it hits the ground, then find $n$.

Answer 1

Step 1. Given data and drawing the diagram

Angle between the ground and the rod, $\theta$ $=30°$

Acceleration due to gravity, $g=10m{s}^{-2}$

Length of the rod $L$$=1m$

Step 2. To find $n$

Let $W$ be the work done by gravity from the initial to the final position. so the equation of work $W$ will be

$W=mg\frac{l}{2}\sin 30°$

on substituting $\sin 30°=\frac{1}{2}$ in the above equation we get

$\Rightarrow W=\frac{mgl}{4}$

according to the work-energy theorem,

$W=\frac{1}{2}\iota {\omega }^2$

on substituting moment of inertia $I=\frac{m{l}^3}{3}{\omega }^2$ in the above equation

we get $W=\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2$

so by comparing both the equations of work we get,

$\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2=\frac{mgl}{4}$

from this $\omega =\sqrt{3g/2l}$

on substituting the values we get $$\begin{gathered} \omega =\sqrt{15} \\ \omega =\sqrt{n} \end{gathered}$$

which indicates $n=15$

Hence, the value of $n$ is $15$.