Question

In the four bar mechanism shown in figure, BC is parallel with AD. AB is vertical. DC makes an angle of ${135}^{\circ }$ with AD and has an instantaneous angular velocity of one rad/sec. The angular velocity of the couple BC is

• A. 1 rad/sec
• B. zero
• C. 1/3 rad/sec
• D. $\frac{1}{\sqrt{18}}rad/sec$

Answer 1

The correct option is D $\frac{1}{\sqrt{18}}rad/sec$

The respective instantaneous centers are shown in the figure.

${\omega }_4=1rad/sec$

$I_{34}{I}_{14}=20cm$

$BC=60cm$

$\therefore I_{34}{I}_{13}=60\sqrt{2}$

From Kennedy theorem
${\omega }_3\left(I_{34}{I}_{13}\right)={\omega }_4\left(I_{34}{I}_{14}\right)$

${\omega }_3\times 60\sqrt{2}=1\times \left(20\right)$

${\omega }_3=\frac{20}{60\sqrt{2}}=\frac{1}{\sqrt{18}}rad/sec$