Mechanisms — two links OA and OB are connected by a pin at O. If OA turns at ω1 rad/s (clockwise) and OB turns at ω2 rad/s (clockwise), the rubbing velocity at pin O (pin radius r) equals:

Difficulty: Easy

Correct Answer: (ω1 + ω2)r

Explanation:

Introduction / Context: Rubbing velocity at a pin joint arises from the relative surface speed between the two bush/eye surfaces rotating about the same pin. Each link contributes a peripheral speed at the pin’s contact surface.

Given Data / Assumptions:

  • Link OA rotates at angular speed ω1 (clockwise).
  • Link OB rotates at angular speed ω2 (clockwise).
  • Pin of radius r at O; both rotate relative to the pin surface.

Concept / Approach: The surface speed contributed by a rotating member at radius r is v = ω·r. When two members rotate relative to each other about the same pin in the same sense, their rubbing components add.

Step-by-Step Solution:

1) Peripheral speed due to OA at the rubbing surface = ω1·r.2) Peripheral speed due to OB at the rubbing surface = ω2·r.3) Net rubbing velocity = (ω1 + ω2)r.

Verification / Alternative Check: If the members rotate in opposite senses, the relative speed would be |ω1 − ω2|r; here they are both clockwise, so speeds add.

Why Other Options Are Wrong:
ω1·ω2·r — Dimensionally incorrect (rad²/s²).
(ω1 − ω2)r — Applies for opposite senses of rotation.
(ω1 − ω2)2r — Incorrect factor and opposite-sense assumption.

Common Pitfalls: Forgetting to multiply by pin radius r; confusing algebraic sum with difference based on rotation direction.

Final Answer: (ω1 + ω2)r.

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