Question

The driver link AB of a four-bar mechanism is rotated at 5.0 rad/s in counter clockwise direction as shown in the figure below. The angular velocity (rad/s) of the coupler at instant when $\theta ={180}^{\circ }$ is ____rad/sec

• A. 2.5

Answer 1

The correct option is A 2.5
Given that ${\omega }_2=5rad/s$

Need to find out ${\omega }_3$ when $\angle BAD={180}^{\circ }$


According to Kennedy's theorem

$I_{24}$ is at the intersection of link 1 and 3 and $I_{13}$ at the intersection of links '2' and '4'.

$\therefore AtA-I_{12}$

$B-I_{23},I_{24}$

$C-I_{34}$

$D-I_{13},I_{14}$

From angular velocity-ratio theorem

$\frac{{\omega }_4}{{\omega }_2}=\frac{I_{24}-I_{12}}{I_{24}-I_{14}}=\frac{\ell }{2\ell }=0.5$

${\omega }_4=2.5rad/s$(counter clockwise)

$V_2=(I_{12},I_{23})\times {\omega }_2=(I_{13},I_{23})\times {\omega }_3$

$\Rightarrow l\times 5=2l\times {\omega }_3$

$\Rightarrow {\omega }_3=2.5rad/sec$