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For the link and slider mechanism shown in figure, locate the instantaneous centre of rotation of link AB. Also find the angular velocity of link 'OA'. Take velocity of slider at B=2500 mm/sec.
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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=9030=60

 ABI=9060=30

  • Hence,AB=3IA

IA=13×AB

=13×0.4=0.2309 m;

AB=32×IB

IB=23×AB

IB=23×0.4=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
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