Let's analyze and solve the first question step-by-step: --- # Question 1: **Given Data:** - Diameter of the work-piece, \( d = 2 \text{ inches} \) - Rotations per minute, \( N = 360 \) - Carbide tool life, \( T = 1 \text{ minute} \) - Tool life exponent, \( n = .3 \) - Cutting speed of the reference tool, \( V_{ref} = 100 \text{ ft/min} \) **Objective:** Find the tool life \( T_2 \) for a carbide tool under the given conditions. --- # Step 1: Convert the cutting speed of the reference tool to surface speed The reference cutting speed is 100 ft/min. --- # Step 2: Calculate the cutting speed \( V \) for the operation The surface speed \( V \) is related to the rotational speed \( N \) and the diameter \( d \): \[ V = \frac{\pi \times d \times N}{12} \] *(convert inches to feet by dividing by 12)* \[ V = \frac{\pi \times 2 \text{ inches} \times 360 \text{ rpm}}{12} \] \[ V = \frac{\pi \times 2 \times 360}{12} \] \[ V = \pi \times 2 \times 30 = 60\pi \text{ ft/min} \] \[ V \approx 60 \times 3.1416 \approx 188.5 \text{ ft/min} \] --- # Step 3: Use the tool life equation (Taylor's Tool Life Equation) Taylor's equation relates the tool life \( T \), cutting speed \( V \), and a constant \( C \): \[ V \times T^n = C \] For the reference tool: \[ C = V_{ref} \times T_{ref}^n \] \[ C = 100 \times 1^n = 100 \] *(since \( T_{ref} = 1 \text{ min} \) and \( n = .3 \))* --- # Step 4: Find the tool life \( T_2 \) at the calculated cutting speed \( V \) Using the same relation: \[ V \times T_2^n = C \] \[ 188.5 \times T_2^{.3} = 100 \] \[ T_2^{.3} = \frac{100}{188.5} \approx 5.31 \] --- # Step 5: Solve for \( T_2 \) \[ T_2 = (5.31)^{1/.3} \] \[ T_2 = (5.31)^{3.33} \] Calculate: \[ \ln T_2 = 3.33 \times \ln 5.31 \] \[ \ln 5.31 \approx 1.67 \] \[ \ln T_2 \approx 3.33 \times 1.67 \approx 5.56 \] \[ T_2 = e^{5.56} \approx 260 \text{ minutes} \] --- # **Final Answer:** **The carbide tool will last approximately \(\boxed{260 \text{ minutes}}\) under the given conditions.**