Question

$\text{Two rotors are mounted on a shaft. Both the rotors are moving in same direction. Inertia of both rotors is same say}{I}_R\text{. The natural frequency of vibration is}$

• A. ${\omega }_n=\sqrt{\left(\frac{GJ}{l{I}_R}\right)}$
• B. ${\omega }_n=\sqrt{\left(\frac{GJ}{l{I}_R}\right)}$
• C. ${\omega }_n=\sqrt{\left(\frac{2GJ}{l{I}_R}\right)}$
• D. ${\omega }_n=0$

Answer 1

The correct option is D ${\omega }_n=0$


$I_1{\ddot{\theta }}_1+\frac{G{I}_P}{l}){\theta }_1-{\theta }_2)=0$

$I_1{\ddot{\theta }}_1+k_t{\theta }_1=k_t{\theta }_2$

$\text{Similarly}{I}_2{\ddot{\theta }}_2+k_t{\theta }_2=k_t{\theta }_1$

$\text{Let}{\theta }_1={\beta }_1\sin \omega t$

${\theta }_2={\beta }_2\sin \omega t$

On substituting

$k_t-I_1{\omega }^2\left){\beta }_1\sin \right(\omega t)=k_t{\beta }_2\sin (\omega t)$

$k_t-I_2{\omega }^2\left){\beta }_2\sin \right(\omega t)=k_t{\beta }_1\sin \omega t$

$\therefore \frac{{\beta }_1}{{\beta }_2}=\frac{k_t}{k_t-I_1{\omega }^2}------\left(1\right)$

$\frac{{\beta }_1}{{\beta }_2}=\frac{k_t-I_2{\omega }^2}{k_t}------\left(2\right)$

on solving (1) & (2)

$\text{(a)}\omega =0\Rightarrow \frac{{\beta }_1}{{\beta }_2}=+1$

$\text{(a)}\omega =\sqrt{\frac{k_t(I_1+I_2)}{I_1{I}_2}}\Rightarrow \frac{{\beta }_1}{{\beta }_2}=\frac{-I_2}{I_1}$

$\therefore \text{for the same direction of rotation of rotors,}\frac{{\beta }_1}{{\beta }_2}=+1$
$\therefore {\omega }_n=0$