A shuttle loom having 1.75 m reed width is running at 180 rpm. The shuttle enters and leaves the shed at $120^\circ$ and $240^\circ$ angular positions of crankshaft, respectively. If length of the shuttle is 0.25 m, then the mean velocity (in m/s) of the shuttle within the shed is________.
This solution calculates the mean velocity of a shuttle within the shed of a loom.
First, determine the time it takes for the crankshaft to complete one revolution and then calculate the specific time the shuttle is active within the shed.
$ \text{Speed} = \frac{180 \text{ rev}}{60 \text{ s}} = 3 \text{ rps} $
$ T = \frac{1}{\text{Speed}} = \frac{1}{3} \text{ s} $
$ \text{Angle traversed} = \text{Exit Angle} - \text{Entry Angle} = 240^\circ - 120^\circ = 120^\circ $
$ \Delta t = \frac{\text{Angle traversed}}{360^\circ} \times T = \frac{120^\circ}{360^\circ} \times \frac{1}{3} \text{ s} = \frac{1}{3} \times \frac{1}{3} \text{ s} = \frac{1}{9} \text{ s} $
The total distance the shuttle effectively travels during this active period is the sum of the reed width and the shuttle's own length.
$ D = W + L = 1.75 \text{ m} + 0.25 \text{ m} = 2.0 \text{ m} $
Finally, calculate the mean velocity using the effective distance and the time interval.
$ V_{\text{mean}} = \frac{D}{\Delta t} = \frac{2.0 \text{ m}}{1/9 \text{ s}} = 2.0 \times 9 \text{ m/s} = 18.0 \text{ m/s} $
The calculated mean velocity is 18.0 m/s, which falls within the expected range.
| Group I | Group II |
| P. Multiphase | 1. Matched cam |
| Q. Projectile | 2. Profile reed |
| R. Air-jet | 3. Crank shaft |
| S. Shuttle | 4. Weaving rotor |
The diameter and length of torsion rod used in a projectile loom are increased by 5% and 20%, respectively. If the torque required to twist the rod increases by X%, then the value of X, accurate to two decimal places, is_______.