Determine the dynamical equivalent two mass system of the connecting rod. If one of the mass is at the small end centre.

Solution: $L_1 + L_2 = 0.3$

$m k_2^2 = IG = 7000$

$k^2 = \frac{7000}{15} = 466.67 \ mm$

$k = 21.6 \ m^2$

$l_1 = 0.2 \ m$

$k^2 = l_1 . l_2 = l_2 = 233.33 \ mm$

$m_1 \frac{ml_2}{l_1 + l_2}$

$=\frac{15 \times 0.233}{0.2 + 0.233}$

$= \frac{15 \times 0.233}{0.433}$

$m_1 = 13.81 \ kg$

$m_2 = 15 – 13.81$

$m_2 = 1.19 \ kg$