Question

In a mechanism, a follower link is pivoted at a fixed point P, with its free end resting in sliding/tangential contact against the rim of an eccentric circular disc of radius 25 mm. The disc rotates about a separate fixed shaft axis S, with the disc's geometric centre offset (eccentric) from S by 5 mm. At the instant shown, P, S, and the disc's centre are collinear, with the distance from P to S equal to 45 mm. If the angular velocity of the eccentric circular disc is 1 rad/s, the angular velocity (rad/s) of the follower link for this instant is

Accepted Answer

The correct answer is (B) 0.1. By the three-centres-in-line theorem, the instant centre I23 between the disc and the follower link lies on the line joining their two fixed pivots (P and S), AND on the common normal to the contact (the line through the disc's current centre O, perpendicular to the tangent follower rod). Since P, S, and O are collinear at this instant, and the common normal necessarily passes through O (the radius to any tangent point is always perpendicular to the tangent line), the common normal intersects line PS exactly at O itself — so I23=O. Equating the velocity of this shared point computed via each link: ω_disc·(S to I23) = ω_follower·(P to I23), i.e. ω_disc·(SO) = ω_follower·(PO), where PO=PS+SO=45+5=50mm. So ω_follower = ω_disc·SO/PO = 1×5/50 = 0.1 rad/s. (Cross-checked via an independent published solution using the tangent-length/similar-triangles method — same numeric answer.)

Explanation