A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg

Step 1 
${\displaystyle \mathrm{△}U=m_{2}g{y}_2+m_{3}g{y}_3=g\times (0.05\times (-0.4)+0.02\times 0.4)=-0.118N.m}$ 
Determine the system's overall potential change. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential. 
Step 2 
${\displaystyle I_1=2\rho {\int }_0^r{r}^{2}dr=2\times \frac{0.12}{0.8}{\int }_0^{0.4}{r}^{2}dr=0.0064kg.m^2}$ 
Determine the rod's moment of inertia.
Step 3 
${\displaystyle I_{2+3}=\sum m{r}^2=(0.05+0.02)\times {0.4}^2=0.0112kg.m^2}$ 
determine the end masses' moment of inertia
Step 4 
${\displaystyle \mathrm{△}U=K=\frac{1}{2}(I_1+I_{2+3}){\omega }^2}$ 
compare the system's kinetic energy to the change in potential energy.
Step 5 
${\displaystyle =0.118=\frac{1}{2}\times 0.0176{\omega }^2}$ 
Calculate velocity at lowest point:
Step 6 
${\displaystyle \omega =3.66ra\frac{d}{s};v=r\omega =0.4\times 3.66=1.46\frac{m}{s}}$

Answer:
Explanation:
Given that,
Slender rod
Length of rod ${\displaystyle =80cm=0.8m}$
Mass of slender rod ${\displaystyle =0.12kg}$
Sphere Bob at one end
Mass ${\displaystyle M1=0.02kg}$
Sphere Bod at the other end
Mass ${\displaystyle M2=0.05kg}$
Linear speed of mass 2 at the lowest point
We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.
So, at the lowest point,
${\displaystyle \mathrm{△}U=M2\cdot g\cdot y2+M1\cdot g\cdot y1}$
Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is ${\displaystyle -40cm(-04m)}$ from the midpoint
${\displaystyle \mathrm{△}U=M2\cdot g\cdot y2+M1\cdot g\cdot y1}$
${\displaystyle \mathrm{△}U=0.05\cdot 9.81\cdot (-0.4)+0.02\cdot 9.82\cdot 0.4}$
${\displaystyle \mathrm{△}U=-0.1962+0.07848}$
${\displaystyle \mathrm{△}U=-0.11772Nm}$
Now, the moment of inertia of the rod is given as
${\displaystyle I=\sum r^{2}dm}$
${\displaystyle dm=2pdr}$
${\displaystyle I=2p\sum r^{2}dr}$
${\displaystyle I=2\times \frac{0.12}{0.8}\sum r^{2}dr}$; from ${\displaystyle r=0\to 0.4}$
${\displaystyle I=0.3\left[\frac{r^3}{3}\right]\mathfrak{o}mr=0\to 0.4}$
${\displaystyle I=0.3[\frac{{0.4}^3}{3}-0]}$,from ${\displaystyle r=0\to 0.4}$
${\displaystyle I=0.3\times 0.02133}$
${\displaystyle I=0.0064k\frac{g}{m^2}}$.
calculating of inertia of the end masses.
${\displaystyle I(1+2)=\sum m{r}^2=(m1+m2)r^2}$
${\displaystyle I(1+2)=(0.02+0.05){0.4}^2}$
${\displaystyle I(1+2)=0.07\times {0.4}^2}$
${\displaystyle I(1+2)=0.0112k\frac{g}{m^2}}$
Now, the Energy of the masses due to angular velocity is given as
${\displaystyle K.E=\frac{1}{2}(I+I(1+2))w^2}$
${\displaystyle K.E=\frac{1}{2}(0.0064+0.0112)w^2}$
${\displaystyle K.E=0.0088w^2}$
Using conservation of energy
The potential energy is equal to the kinetic energy of the system
${\displaystyle K.E=P.E}$
${\displaystyle 0.0088w^2=0.11772}$
Then, ${\displaystyle w^2}$