Question

In the composite filament winding process as shown in the figure, the mandrel diameter is 700 mm and is rotating at a speed, \( N = 6 \) rev/min.
If a \( 45^\circ \) helical winding angle is needed, the axial velocity, \( v_c \), of the slider should be ______ m/s (rounded off to two decimal places).
Assume \( \pi = 22/7 \). Ignore the thickness of the composite layers already wound on the mandrel.

Note: Figure is not to scale.

Hint:

Surface velocity of the mandrel equals \( \pi D N \); relate it to slider speed using tan of the winding angle.

Answer 1

Correct Answer: 0.22

Step 1: Understanding the Concept:
In filament winding, the fibre wraps around a rotating mandrel while a slider carrying the fibre moves along the mandrel's axis.
The winding angle is set by the ratio of two speeds: how fast a point on the mandrel surface is moving around, versus how fast the slider is moving along the axis.

Step 2: Key Formula or Approach:
First convert the rotational speed into an angular velocity, then use it with the mandrel radius to get the surface (tangential) speed:
\[ \omega = 2\pi N, \qquad v_{surface} = \omega \times r \]
Then relate the surface speed to the axial slider speed through the winding angle:
\[ \tan(\alpha) = \frac{v_{surface}}{v_c} \]

Step 3: Detailed Explanation:
Given data: mandrel diameter \( D = 700 \) mm \( = 0.7 \) m, so radius \( r = 0.35 \) m. Rotational speed \( N = 6 \) rev/min. Winding angle \( \alpha = 45^\circ \). \( \pi = \frac{22}{7} \).
Convert rotational speed to angular velocity, keeping minutes as the time unit for now:
\[ \omega = 2\pi N = 2 \times \frac{22}{7} \times 6 = \frac{264}{7} = 37.71 \text{ rad/min} \]
Multiply by the radius to get the tangential surface speed of the mandrel:
\[ v_{surface} = \omega \times r = 37.71 \times 0.35 = 13.2 \text{ m/min} \]
Convert this speed into m/s by dividing by 60:
\[ v_{surface} = \frac{13.2}{60} = 0.22 \text{ m/s} \]
Now use the winding angle relation. Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{v_{surface}}{v_c} \implies v_c = v_{surface} = 0.22 \text{ m/s} \]

Final Answer:
The axial velocity of the slider required to produce a 45 degree helical winding angle is: \[ \boxed{v_c = 0.22 \text{ m/s}} \]