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Machining Processes
GATE 2017
Expert
Two cutting tools with tool life equations given below are being compared: Tool 1: VT^0.1 = 150 Tool 2: VT^0.3 = 300 where V is cutting speed in m/minute and T is tool life in minutes. The breakeven cutting speed beyond which Tool 2 will have a higher tool life is __________ m/minute.
106 m/minute(± 1 accepted)
Explanation
Equating tool lives: (150/V)^10 = (300/V)^(10/3). Taking logs and simplifying: 3 ln(150/V) = ln(300/V) → (150/V)³ = 300/V → 150³ = 300V² → V² = 150³/300 = 11250 → V ≈ 106.07 m/min.
Source: GATE ME 2017 Session 1 (ME-1) (2017)
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