**Answer:**

(a) 94.25 m/min

(b)

(c) 1 min

(d) 10.996 kW and 15.027 kW

(e) 7 kN and 9.57 kN

**Explanation:**

The maximum cutting speed, is observed at the outer diameter hence

where N is the rate of rotation and is the outer diameter. Substituting 75mm for and 400 rpm for N we obtain

(b)

The material removal rate, MRR is given by

where is the average between the inner and outer diameters, d is the depth of cut, f is the feed which is given by where V is the axial velocity

In this case, the average diameter is

The feed f is

The depth of cut is

Therefore,

(c)

Time of cut is given by

where L is the length of rod. Substituting L for 200mm, f as seen in part b is 0.5 and 400 for N we obtain

(d)

The power is obtain by multiplying specific energy by material removal rate. Assuming specific energy range of to then

Power,

Power,

Therefore, power ranges between 10.996 kW and 15.027 kW

(e)

Cutting force is given by

where P is power and is already calculated in part a

First, is converted to m/s hence

From the power range in part d,

Therefore, the cutting force ranges from 9.57 kN and 7 kN