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

2.93

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5.86

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6.66

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13.33

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Solution

The correct option is A $2.93$
$t_{m} = \frac{π (d_{0}^{2} - d_{i}^{2})}{4 f_{r} . v} = \frac{π {{0.2}^{2} - {0.08}^{2}}}{4 \times 0.1 \times 10^{- 3} \times 90}$
$t_{m} = 2.93 m i n$

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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 $200 N$

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

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

Linked Question
A low carbon steel bar of $147 m m$ diameter with a length of $630 m m$ is being turned with an uncoated carbide insert. The observed tool lives are $24 m i n$ and $12 m i n$ for cutting velocities of $90 m / m i n$ and $120 m / m i n$ respectively. The feed and depth of cut are $0.2 m m / r e v$ and $2 m m$ respectively. Use the unmachined diameter to calculate the cutting velocity.

Neglect over-travel or approach of the tool. When too life is $20 m i n$ , the machining time in min for a single pass is

Q. The main cutting force acting on a tool during the turning (orthogonal cutting) operation of metal is

400 N

. The turning was performed using

2 m m

depth of cut and

0.1 m m / r e v

feed rate. The specific cutting pressure (in

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) is

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The correct option is A 2.93tm=π(d20−d2i)4fr.v=π{0.22−0.082}4×0.1×10−3×90 tm=2.93min

The correct option is A $2.93$
$t_{m} = \frac{π (d_{0}^{2} - d_{i}^{2})}{4 f_{r} . v} = \frac{π {{0.2}^{2} - {0.08}^{2}}}{4 \times 0.1 \times 10^{- 3} \times 90}$
$t_{m} = 2.93 m i n$