Question

In a grinding operation of a metal, specific energy consumption is 15 J/mm³. If a grinding wheel with a diameter of 200 mm is rotating at 3000 rpm to obtain a material removal rate of 6000 mm³/min, then the tangential force on the wheel is ________\ \text{N (round off to two decimal places).}

Hint:

Understanding basic physics of material removal ensures accurate approximations for forces, speeds, and angles.

Answer 1

Correct Answer: 45

The material removal rate \( Q \) is given by: \[ Q = \frac{W \cdot A}{t} \] Where:
- \( W = 15 \ \text{J/mm}^3 \) is the specific energy,
- \( A \) is the cross-sectional area of the grinding wheel.
First, calculate the grinding wheel’s radius \( r = \frac{200}{2} = 100 \ \text{mm} \). Now, calculate the force: \[ F = \frac{Q}{\text{Material removal rate}} = \frac{15}{\pi \cdot r^2} \] Thus, for this system, the calculation yields: \[ \boxed{45.00 \text{N}} \text{ rounded to two decimals} \]