Cutting speed distribution on a twist drill Considering a rotating twist drill, is the cutting speed zero at the periphery and maximum at the centre?

Introduction / Context:
Cutting speed in drilling is the tangential velocity of a point on the cutting edge and is proportional to radius times angular speed. Understanding this radial variation explains why the chisel edge at the center behaves differently than the lips at the periphery.

Given Data / Assumptions:

• Circular motion with angular speed omega.
• Local cutting speed v(r) = omega * r.
• Standard right-hand twist drill.

Concept / Approach:
Because v(r) increases linearly with radius, the cutting speed is zero at the center (r = 0) and maximum at the periphery (r = R). The statement in the question reverses this reality and is therefore false.

Step-by-Step Solution:
Define v(r) = omega * r.At the center (r = 0): v = 0.At the periphery (r = R): v = omega * R (maximum for the tool).

Verification / Alternative check:
High thrust and rubbing at the chisel edge occur because the cutting speed there is near zero, explaining why web thinning reduces thrust.

Why Other Options Are Wrong:
Hand of the drill or feed mode does not change the basic kinematics; speed depends on radius, not on these factors.

Common Pitfalls:
Using cutting speed values based on diameter without accounting for the center’s low speed; excessive feed causes chisel-edge extrusion rather than cutting.

Final Answer:
No