Answer:
The answer is below
Explanation:
v = velocity = 30 m/min = 30 * 10³ mm/min, D = diameter = 10 mm, f = feed = 0.25 mm/rev, point angle = 118, cutting time = Tm, d = depth = 60 mm
[tex]a)\\N=\frac{v}{\pi D}=\frac{30*10^3}{\pi * 10}=954.9\ rev/min\\\\f_r=Nf =954.9(0.25)=238\ mm/min\\\\A=0.5Dtan(90-\frac{point\ angle}{2} )=0.5*10*tan(90-\frac{118}{2} )=3\ mm\\\\T_m=\frac{(d+A)}{f_r} =\frac{60+3}{238}=0.265 \ s\\\\b)\\metal\ removal\ rate(R_{MR})=0.25\pi D^2f_r\\\\R_{MR}=0.25\pi (10)^2(238)=18692\ mm^3/min[/tex]
