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