Determine the angular velocity of connecting rod BC and velocity of Piston C using ICR method. AB = 0.3m and CD = 0.8m.

Angular velocity of the rod $OA= \omega_{AB}=2rad/s\\ AB=0.3 m$

Instantaneous centre of rotation of rod AB.

In $\triangle $ABC, By Sine Rule.

$\dfrac {AB}{\sin ACB}=\dfrac {BC}{\sin A} \\ \therefore \dfrac {0.3}{\sin ACB}=\dfrac {0.8}{\sin 30 } \\ \therefore \sin ACB =\dfrac {0.3\sin 30}{0.8}=0.1875 \\ \therefore \angle ACB=\sin^{-1}0.1875=10.8069^\circ \\ \therefore \angle ABC=180^\circ -(30^\circ +10.8069^\circ ) =139.1931^\circ \\ \therefore \angle IBC =180^\circ -130.1931^\circ =79.1931^\circ \space \text { and } \\ in \space \triangle IAC , \angle AIC =180^\circ -30^\circ -90^\circ =60^\circ \\ in \triangle IBC , \text { By sin erule } \\ \dfrac {BC}{\sin I}=\dfrac {IB}{\sin ICB}=\dfrac {IC}{\sin IBC} \\ \therefore \dfrac {0.8}{\sin 60 } =\dfrac {IB}{\sin ICB} =\dfrac {IC}{\sin IBC} \\ \therefore \dfrac {0.8}{\sin 60} =\dfrac {IB}{\sin 79.1931} =\dfrac {IC} {\sin 40.8069}$

Consider

$\therefore \dfrac {0.8}{\sin 60}=\dfrac {IB}{\sin 79.1931} \text {and } \dfrac {0.8}{\sin 60} =\dfrac {IC}{\sin 40.8069} \\ \therefore IB=\dfrac {0.8\sin 79.1931}{\sin 60}=0.9074 \space \space \text {and } \space IC \\ =\dfrac {0.8\sin 40.8069}{\sin 60}=0.6037$

Angular velocity of the rod BC= $\omega_{BC}=\dfrac {v_s}r=\dfrac {v_s}{IB}=\dfrac {0.6}{0.9074}=0.6612 \space rad/s$

Instantaneous velocity of the piston

$C=rw_{BC}= IC \times \omega_{BC}=0.6037\times 0.6612 =0.3992 m/s \rightarrow$