Question

A four bar mechanism is shown in the figure. The link numbers are mentioned near the links. Input link 2 is rotating anticlockwise with a constant angular speed ${\omega }_2$. Length of different links are:
$O_2{O}_4=O_{2}A=L,AB=O_{4}B=\sqrt{2}L$
The magnitude of the angular speed of the output link 4 is ${\omega }_4$at the instant when link 2 makes an angle of ${90}^o$ with $O_2{O}_4$ as shown. The ratio $\frac{{\omega }_4}{{\omega }_2}$is (round off to two decimal places).

Answer 1

According to Arnold's Kennedy theorem,
$\Rightarrow tan{75}^o=\frac{\left(I_{24}{I}_{12}\right)}{\left(L\right)}$
$\left(I_{24}{I}_{12}\right)=\left(Ltan{75}^o\right)=3.732L$
$\left(I_{24}{I}_{14}\right)=4.732L$
${\omega }_2\times \left(I_{24}{I}_{12}\right)={\omega }_4\left(I_{24}{I}_{14}\right)$
$\frac{{\omega }_4}{{\omega }_2}=\frac{I_{24}{I}_{12}}{I_{24}{I}_{14}}$
$\frac{{\omega }_4}{{\omega }_2}=\left(\frac{3.732}{4.732}\right)=0.7886$