What is the velocity of the crosshead - crank:

In the slider crank mechanism shown in Figure, the crank is rotating clockwise at 120 revolution/minute. What is the velocity of the crosshead when the crank is in the 60^o phase?

Solution

We have, crank AB = 200 mm = 0.2 metres.

ω _AB  = 120 revolutions / minute

= 120 × 2 π /60

= 4 π rad / sec

As B rotates about A, v_B  = AB . ω _AB  = 0.2 × 4 π = 0.8 π = 2.51 m / sec , and v_B is perpendicular to AB.

Crosshead C moves along the horizontal line as shown.

 We know v_B fully, its magnitude and direction.

Consider Δ ABC,

AB /sin α   =    BC /sin 60^o  =  AC / sin (120^o  - α)

∴ sin α = (AB / BC ). sin 60^o  = (200/900) × 0.866 = 0.1924

 ∴ α = 11.09^o

 (i)        By Instantaneous Centre Method

Draw perpendiculars to v_B at B and to v_C at C. They meet at O.

Consider Δ BOC.

BC /sin 30^o =   CO /sin 71.09 ^o = BO / sin 78.9^o

∴ OC = 0.9 ×( sin 71.09 /sin 30^o) = 1.702 m

OB = 0.9 × (sin 78.9 /sin 30^o) = 1.76 m

∴ ω_BC =  vB /OB =( 2.51/1.76) = 1.426 rad / sec

v_C  = ω_BC  × OC = 2.42 m / sec

 (ii)       By Velocity Diagram Method

From any point O draw a line perpendicular to AB to represent v_B. From end of v_B draw a line perpendicular to BC to represent v_BC. Next, draw from O a line parallel to AC to represent v_C.

∴          v_B /sin 78.91^o  =           v_BC /sin 30^o    = v_C /sin 71.11^o

∴ v_C  = v_B . (sin 71.11^o /sin 78.91^o)

= 2.42 m / sec

v_C = v_B.   (sin 30 ^o / sin 78.91^o)

= (2.51 × 0.5 )/0.98

= 1.2806 m / sec

∴ ω_BC    = v_BC / BC

 = 1.2806 /0.9 = 1.423 rad/sec