Question

A uniform rod of length $l$ is released from rest such that it rotates about a smooth pivot. Find the angular speed of the rod when it becomes vertical.

Answer 1

R.E.F image

According to the question the mass of whole

rod is m and length is l

$\ast$ we see the mass per unit length is m/l,

mass of length $l/4=\frac{l}{4}\times \frac{m}{l}=\frac{m}{4}$

mass of length $3l/4=\frac{3l}{4}\times \frac{m}{l}=\frac{3m}{4}$

$\ast$ we know, the moment of inertia of a rod

about its end print is

$I=\frac{1}{3}m{l}^2$ (where, m = mass of rod, l = length of rod)

Them,

$I_1=\frac{1}{3}\left(\frac{m}{4}\right)(\frac{l}{4}{)}^2=\frac{m{l}^2}{192}$

$I_2=\frac{1}{3}\left(\frac{3m}{4}\right)(\frac{3l}{4}{)}^2=\frac{9m{l}^2}{64}$

[$I_1$ is MOI of l/4 part & $I_2$ is MOI of 3l/4 part]

* Also, we know

Change in potential $=$ Change in Kinetic

energy energy

$-\left(\frac{m}{4}\right)g\left(\frac{l}{8}\right)+\left(\frac{3m}{4}\right)g\left(\frac{3l}{8}\right)=\frac{1}{2}{I}_1{w}^2+\frac{1}{2}{I}_2{w}^2$

$\frac{-mgl}{32}+\frac{9mgl}{32}=\left[\frac{1}{2}\right(\frac{ml2}{192})+\frac{1}{2}(\frac{9m{l}^2}{64}\left)\right]w^2$

$\frac{+8mgl}{32}=\frac{w^2}{2}[\frac{m{l}^2}{192}+\frac{9m{l}^2}{64}]$

$\frac{8mgl}{32}=\frac{w^2}{2}\left(\frac{m{l}^2+27m{l}^2}{192}\right)$

$\frac{8mgl}{32}=\frac{w^2}{2}\times \frac{28m{l}^2}{192}$

$w=\sqrt{\frac{24}{7l}}$

These frond the angular speed of the rod

to be $\sqrt{\frac{24g}{7l}}$ when it become vertical.