In an ideal slider–crank mechanism for a reciprocating steam engine, what is the piston (slider) velocity exactly at the inner dead centre (IDC), which is an extreme stroke position?

Introduction / Context

The slider–crank mechanism converts rotary motion to reciprocation. Dead centres (inner and outer) are extreme positions of the slider where the connecting rod, crank, and line of stroke align in specific ways. Velocity at these positions is a fundamental kinematic property.

Given Data / Assumptions

• Ideal joints without clearance or compliance.
• Constant angular velocity of the crank.
• IDC is an end-of-stroke position.

Concept / Approach

Slider velocity is the time derivative of displacement along the line of stroke. At an extreme position of a periodic displacement, instantaneous velocity is zero because displacement has a local maximum or minimum there. This follows both from geometry and harmonic approximations.

Step-by-Step Solution

Verification / Alternative check

Vector geometry shows slider motion reverses direction at dead centres; reversal implies zero instantaneous velocity.

Why Other Options Are Wrong

• Minimum (but not zero): at the turning point of displacement, the exact velocity is zero.
• Maximum: occurs near mid-stroke for uniform crank speed.
• None of these / Indeterminate: contradicts deterministic kinematics.

Common Pitfalls

• Confusing acceleration (which is maximum in magnitude at dead centres) with velocity (which is zero).

Final Answer

Zero