Question

In a single point turning operation of steel with a tool, Taylor’s tool life exponent is 0.2. What is the increase in the tool life if the cutting speed is halved?

• A. 8 times
• B. 16 times
• C. 32 times
• D. 64 times

Hint:

Taylor’s tool life equation \( V T^n = C \) shows that reducing cutting speed increases tool life exponentially: \( \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{1/n} \).

Answer 1

The Correct Option is C

Step 1: Understand Taylor’s tool life equation.
Taylor’s tool life equation relates the cutting speed (\( V \)) to the tool life (\( T \)) in machining: \[ V T^n = C, \] where:
\( V \): Cutting speed,
\( T \): Tool life (time until the tool fails),
\( n \): Taylor’s tool life exponent (given as 0.2),
\( C \): A constant depending on the material and tool.
Step 2: Set up the equation for the two conditions.
Initial condition: Cutting speed \( V_1 \), tool life \( T_1 \).
New condition: Cutting speed \( V_2 = \frac{V_1}{2} \) (halved), tool life \( T_2 \).
Using Taylor’s equation:
For the initial condition: \( V_1 T_1^n = C \),
For the new condition: \( V_2 T_2^n = C \).
Since \( C \) is the same, equate the two: \[ V_1 T_1^n = V_2 T_2^n. \] Substitute \( V_2 = \frac{V_1}{2} \): \[ V_1 T_1^n = \left( \frac{V_1}{2} \right) T_2^n. \] Step 3: Solve for the ratio of tool life.
Divide both sides by \( V_1 \): \[ T_1^n = \left( \frac{1}{2} \right) T_2^n, \] \[ \left( \frac{1}{2} \right) T_2^n = T_1^n, \] \[ T_2^n = 2 T_1^n, \] \[ \frac{T_2^n}{T_1^n} = 2, \] \[ \left( \frac{T_2}{T_1} \right)^n = 2, \] \[ \frac{T_2}{T_1} = 2^{1/n}. \] Given \( n = 0.2 \): \[ \frac{1}{n} = \frac{1}{0.2} = 5, \] \[ \frac{T_2}{T_1} = 2^5 = 32. \] Thus, the tool life increases by a factor of 32, meaning \( T_2 = 32 T_1 \). Step 4: Interpret the increase in tool life.
The question asks for the "increase in tool life," which typically means the factor by which the tool life increases:
If \( T_1 \) is the original tool life, \( T_2 = 32 T_1 \),
The tool life increases to 32 times the original, which matches the phrasing of the options (e.g., "32 times").
Step 5: Evaluate the options.
(1) 8 times: Incorrect, as \( 2^3 = 8 \). Incorrect.
(2) 16 times: Incorrect, as \( 2^4 = 16 \). Incorrect.
(3) 32 times: Matches the calculated factor \( 2^5 = 32 \). Correct.
(4) 64 times: Incorrect, as \( 2^6 = 64 \). Incorrect.
Step 6: Select the correct answer.
The tool life increases by a factor of 32 when the cutting speed is halved, matching option (3).