Question

A copper rod of diameter 100 mm is turned on lathe machine at a feed of 0.50 mm/rev with depth of cut of 2 mm. The rotational speed of the rod is 320 RPM. MRR (material removal rate) in mm³/s is:
1. 16.755
2. 167.55
3. 1675.5
4. 16755

Answer 1

Correct Answer - Option 3 : 1675.5

Concept:

Material removal rate, MRR = V × f × d

Where V = cutting speed (mm/s), f = feed (mm/rev), d = depth of cut (mm)

Cutting speed V = \(\frac{{\pi DN}}{{60}}\)

Where D = diameter of shaft (mm), N = RPM

Calculation:

Given:

D = 100 mm, f = 0.50 mm/rev, d = 2 mm, N = 320 RPM

∴ V = \(\frac{{\pi DN}}{{60}}\)= \(\frac{{\pi \ \times\ 100\ \times \ 320}}{{60}}\)

= 1675.516 mm/s

MRR = V × f × d = 1675.516 × 0.50 × 2

= 1675.516 mm³/s