Question

The input link $O_{2}P$ of a four bar linkage is rotated at 2 rad/s in a counter clockwise direction as shown below. The angular velocity of the coupler PQ in rad/s, at an instant when $\angle O_4{O}_{2}P={180}^o$, is

• A. 4
• B. $2\sqrt{2}$
• C. 1
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C 1

Method II :

When
$\angle O_4{O}_{2}P={180}^o$

Applying the angular velocity theorem at $I_{23}$.

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

$2\left(a\right)={\omega }_3\left(2a\right)$

${\omega }_3=2/2=1$ rad/s

Method II:
The given four bar linkage is converted into an isosceles triangle as shown above.
$V_P=O_{2}P\times {\omega }_{O_{2}P}$

$=a\times 2=2a$m/s

Velocity $V_P$ also has two components, in which one component will be perpendicular to the link,

$PQ=V_{p}cos{45}^o$

$=2a\times \frac{1}{\sqrt{2}}=\sqrt{2}a$ m/s

$\because$ Angular velocity of link PQ,

${\omega }_{PQ}=\frac{V_{p}cos{45}^o}{PQ}=\frac{\sqrt{2}a}{\sqrt{2}a}$

$=1$ rad/s