Question

A batch of $10$ cutting tools could produce $500$ components while working at $50rpm$ with a tool feed of $0.25mm/rev$ and depth of cut of $1mm$. A similar batch of $10$ tools of the same specification could produce $122$ components while working at $80rpm$ with a feed of $0.25mm/rev$ and $1mm$ depth of cut. How many components can be produced with one cutting tool at $60rpm$?

• A. $29$
• B. $31$
• C. $37$
• D. $42$

Answer 1

The correct option is A $29$

Given data:
Velocity of first cutting tool
$V_1=50\times 0.25=12.5mm/min$

Velocity of second cutting tool
$V_2=0.25\times 80=20mm/min$

since tool life = no.of components produced $\times$ tool constant

$\therefore$ Tool life for first tool,
$T_1=500K$
Tool life for second tool,
$T_2=122K$
Taylor's tool life equation:
$V_1{T}_1^n=V_2{T}_2^n$
$\left(12.5\right)\times (500K{)}^n=20\times (122K{)}^n$
$\therefore {\left(\frac{500}{122}\right)}^n=1.6$
$\therefore n=0.333$

Hence number of components produced by one cutting tool at $60rpm$
$12.5\times {\left(\frac{500K}{10}\right)}^{0.333}=60\times 0.25\times (xK{)}^{0.333}$
${\left(\frac{x}{50}\right)}^{0.333}=\frac{12.5}{60\times 0.25}$
$\therefore x=29$