Lathe and Turning Operations Questions and Answers

Answer: D

The compound rest is swivelled to 10.62 degrees.

The compound rest is set to the half angle of the taper. Only the tapered portion counts, not the full 150 mm length of the work:

\[ \begin{aligned} \tan\alpha &= \frac{D - d}{2l} = \frac{80 - 50}{2 \times 80} \\ &= 0.1875, \quad \alpha = 10.62^\circ \end{aligned} \]

Setting the full included angle, about 21.2°, would cut a taper twice as steep, and 80.28°, 15° and 45.6° do not follow from these dimensions.

Answer: C

The machining time is 0.75 min.

The tool travels along the full 150 mm length once, at an axial speed of 200 mm/min:

\[ T = \frac{150}{200} = 0.75\ \text{min} \]

The spindle speed and the diameters are not needed, since the axial feed rate is given directly in mm per minute: 500 rpm and 200 mm/min only fix the feed per revolution, 0.4 mm. 2 min, 2.75 min and 3.75 min would need a much slower feed or a longer travel.

Answer: A

3 passes are required.

Turning from 60 mm to 50 mm diameter removes 10 mm from the diameter, which is 5 mm from the radius. Depth of cut is measured on the radius:

\[ \frac{5\ \text{mm}}{2\ \text{mm per pass}} = 2.5 \]

Part of a pass still needs a pass, so the answer rounds up to 3, for example cuts of 2, 2 and 1 mm. 2 passes would remove only 4 mm from the radius, and 4 or more is unnecessary.

Answer: A

The spindle speed is 159.2 rpm.

\[ \begin{aligned} N &= \frac{1000\,v}{\pi D} = \frac{1000 \times 100}{\pi \times 200} \\ &= \frac{100000}{628.3} = 159.2 \end{aligned} \]

The factor 1000 converts the cutting speed from metres to millimetres to match the diameter. 15.9 rpm and 1592 rpm are the same digits with the decimal point moved, the result of a slip in that conversion.

Answer: D

The answer is none of these: every method listed is used for taper turning.

A form tool turns short tapers by plunging a tool ground to the taper angle. Swivelling the compound rest turns short, steep tapers by hand feed. Combining longitudinal and cross feed moves the tool along a sloping line, which is how the taper turning attachment and CNC lathes produce tapers.

Since each of the first three is a recognised taper-turning method, none of them is the one 'not used'.

Answer: B

In facing, the length of tool travel is the radius of the workpiece.

Facing machines the end of the work. The tool is fed across the face at right angles to the axis, from the outside diameter to the centre (or from the centre outward), so it travels a distance equal to the radius, plus a small approach and overrun.

The length of the workpiece is the travel in turning, not facing. The tool does not cross the full diameter, since once it reaches the centre the whole face has been cut, and the travel cannot be all three at once.

Answer: D

The tailstock offset is 20 mm.

When only part of the job is tapered, the offset is worked out over the whole length between centres, since the whole job is tilted:

\[ \begin{aligned} \text{offset} &= \frac{(D - d)\,L}{2\,l} \\ &= \frac{(60 - 40) \times 200}{2 \times 100} = 20\ \text{mm} \end{aligned} \]

Here \(L\) is the total length of the job and \(l\) the length of the taper. Using the taper length for both gives \((60 - 40)/2 = 10\) mm, which would produce a gentler taper than required; 5 mm and 2.5 mm halve that again.