Question

In a turning operation, doubling the cutting speed \(V\) reduces the tool life \(T\) to \( \frac{1}{8} \) of the original tool life. The exponent \(n\) in the Taylor’s tool life equation, \( V T^n = C \), is

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{8} \)

Hint:

The Taylor tool life equation shows the inverse relationship between cutting speed and tool life, with the exponent \(n\) describing how sensitive the tool life is to changes in speed.

Answer 1

The Correct Option is B

In machining, the Taylor tool life equation relates the cutting speed (\(V\)) and tool life (\(T\)) with an empirical constant \(C\) and the exponent \(n\): \[ V T^n = C \] We are told that when the cutting speed \(V\) is doubled, the tool life \(T\) reduces to \( \frac{1}{8} \) of its original value. Let’s substitute this information into the equation. Let the initial values be \( V_0 \) and \( T_0 \). Then, \[ V_0 T_0^n = C \] After doubling the cutting speed: \[ 2V_0 \left( \frac{T_0}{8} \right)^n = C \] Since \(V_0 T_0^n = C\), we can equate: \[ 2V_0 \left( \frac{T_0}{8} \right)^n = V_0 T_0^n \] This simplifies to: \[ 2 \times 8^{-n} = 1 \] Solving for \(n\), we get: \[ 8^{-n} = \frac{1}{2} \] Taking the logarithm: \[ -n \log(8) = \log\left(\frac{1}{2}\right) \] \[ -n \log(8) = -\log(2) \] \[ n = \frac{1}{3} \] Thus, the correct answer is (B) \( \frac{1}{3} \).