Question

For the same quick return mechanism (crank length 250mm, pivot distance 500mm, crank speed 2 rev/s), the angular speed of the slotted lever in rev/s when the ram/block attains maximum speed during the forward stroke (the stroke with slower speed) is

Accepted Answer

The correct answer is (B) 2/3. Differentiating the lever angle as a function of crank angle θ (with O2 at origin, lever pivot at (0,500), crank pin at (r cosθ, r sinθ), r=250) shows the lever's angular speed has critical points only at θ=90° and θ=270° (where cosθ=0). At θ=90° (which falls within the FAST/return stroke, between the dead centers at 30° and 150°), |ω_lever|=ω_crank exactly (the global maximum over the full cycle). At θ=270° (which falls within the SLOW/forward stroke, between dead centers at 150° and 30° the long way around), |ω_lever|=ω_crank/3 — this is the relevant extremum for the question, giving (2 rev/s)/3 = 2/3 rev/s.

Explanation