Question

The configuration of a mechanism is shown in the figure, where the link AB is collinear with link AD and both are horizontal. The link DC is vertical. The link AB has an instantaneous angular velocity of 3 rad/sec in CCW direction. AB = l 0 cm, BC = 50 cm, CD = 30 cm, AD = 50 cm. The angular velocity of link DC is

• A. 3/4 rad/sec CCW
• B. 4/3 rad/sec CW
• C. 3/4 rad/sec CW
• D. 4/3 rad/sec CCW

Answer 1

The correct option is C 3/4 rad/sec CW


Given date:

$AB=10cm$

$BC=50cm$

$CD=30cm$

${\omega }_2=3rad/sec\left(CCW\right)$

From Kennedy theorem, we know that

${\omega }_2\left(I_{23}{I}_{12}\right)={\omega }_3\left(I_{23}{I}_{13}\right)$

$I_{13}$ isnot knownso we locate itusing Kennedy theorem as shown in figure

It lies at $I_{41}$

$I_{23}{I}_{12}=AB=10cm,$

$I_{23}{I}_{13}=BD=40cm,$

$3\times 10={\omega }_3\times 40$

${\omega }_3=\frac{3}{4}rad\left(CW\right)$

Direction is CW because $I_{12}$ & $I_{23}$ lie opposite to $I_{23}$