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- (Given)...Velocity at B=2500mm/sec i.e. VB=2500mm/sec
=2.5 m/s
AB=400 mm=0.4 m, OA=200 mm=0.2m.
- Let ωAB and ωOA be the angular velocities of links AB and OA respectively.
- Let ′I′ be the instantaneous center of rotation of link AB.
∠ CBVA=30∘
∠ CBI=90∘−30∘=60∘
∠ ABI=90∘−60∘=30∘
IA=1√3×AB
=1√3×0.4=0.2309 m;
AB=√32×IB
IB=2√3×AB
IB=2√3×0.4=0.4619 m
- Hence,Angular velocity of the link AB=ωAB=VBr
=VBIB=2.50.4619=5.4124 rad/s
- Hence,Instantaneous velocity of A=r.ωAB=IA×ωAB
=0.2309×5.4124=1.25m /s
- Hence,Angular velocity of the link OA=ωOA=VAr=VAOA
=1.250.2=6.25 rad/s
- Angular velocity of the link OA=6.25 rad/s