Common Data:
A disc of $200 m m$ outer and $80 m m$ inner diameter is faced at a feed of $0.1 m m / r e v$ with a depth of cut of $1 m m$ . The facing operation is undertaken at a constant cutting speed of $90 m / m i n$ in a CNC lathe The main (tangential) cutting force is $200 N$

Neglecting the contribution of the feed force towards cutting power, the specific cutting energy in $J / m m^{3}$ is

0.2

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2

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200

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2000

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Solution

The correct option is B $2$
$S_{e} = \frac{F_{c}}{d \times f} = \frac{200}{0.1 \times 1 \times 1000} = 2 J / m m^{3}$

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Similar questions

Common Data:
A disc of $200 m m$ outer and $80 m m$ inner diameter is faced at a feed of $0.1 m m / r e v$ with a depth of cut of $1 m m$ . The facing operation is undertaken at a constant cutting speed of $90 m / m i n$ in a CNC lathe The main (tangential) cutting force is $200 N$

Assuming approach and over-travel of the cutting tool to be zero, the machining time in min is

Q. Common Data:
In orthogonal turning of a bar of

100 m m

diameter with a feed of

0.25 m m / r e v

, depth of cut of

4 m m

and cutting velocity of

90 m / m i n

, it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting force is

1500 N

.

The orthogonal rake angle of the cutting toll in degree is

Common Data:
In orthogonal turning of a bar of $100 m m$ diameter with a feed of $0.25 m m / r e v$ , depth of cut of $4 m m$ and cutting velocity of $90 m / m i n$ , it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting force is $1500 N$ .

The normal force acting the chip-tool interface in N is