• $(Given)...Velocity\ at\ B=2500 mm/sec\ i.e.\ V_B= 2500 mm/sec\ $

$= 2.5\ m/s$

$AB = 400\ mm = 0.4 \ m,\ OA= 200\ mm = 0.2m.$

• $Let\ \omega_{AB}\ and\ \omega_{OA}\ be\ the\ angular\ velocities\ of\ links\ AB\ and\ OA\ respectively.$
• $Let\ 'I'\ be\ the\ instantaneous\ center\ of\ rotation\ of\ link\ AB. $

$\angle\ CBV_A=30^\circ$

$\angle\ CBI=90^\circ - 30^\circ = 60^\circ$

$\angle\ ABI=90^\circ - 60^\circ = 30^\circ$

• $Hence, AB= \sqrt3 IA$

$IA= \dfrac {1} {\sqrt3} \times AB $

$=\dfrac {1} {\sqrt3} \times0.4 = 0.2309\ m; $

$AB= \dfrac {\sqrt3}{2} \times IB $

$IB= \dfrac {2} {\sqrt3} \times AB $

$IB= \dfrac {2} {\sqrt3} \times 0.4 = 0.4619\ m$

• $Hence, Angular\ velocity\ of\ the\ link\ AB= \omega_{AB} = \dfrac {V_B} r$

$=\dfrac {V_B} {IB} = \dfrac {2.5} {0.4619}= 5.4124\ rad/s$

• $Hence, Instantaneous\ velocity\ of\ A= r.\omega_{AB} = IA \times \omega_{AB} $

$= 0.2309 \times 5.4124 = 1.25 m\ /s$

• $Hence, Angular\ velocity\ of\ the\ link\ OA= \omega_{OA} = \dfrac {V_A} r = \dfrac {V_A} {OA}$

$ =\dfrac {1.25} {0.2} = 6.25\ rad/s$

• $Angular\ velocity\ of\ the\ link\ OA= 6.25\ rad/s$