question_answer

A threaded rod with 12 turns/cm and diameter 1.18 cm is mounted horizontally. A bar with a threaded hole to match the rod is screwed onto the rod. The bar spins at 216 rev/min. How long will it take for the bar to move 1.50 cm along the rod?

Answer:

                Here, distance between two consecutive threads = pitch = \[\frac{1}{12}cm\]. Total distance to be moved = 1.5 cm      \[\therefore \] No. of rotations = \[\frac{1.5}{1/12}=18\] Total angle of turning, \[\text{ }\!\!\theta\!\!\text{ }=18\times 2\pi =36\pi \,radian\] angular speed,  \[\omega =2\pi n=\,2\pi \times \frac{216}{60}=7.2\pi \,rad\,/s\] \[\therefore \] time taken,  \[t=\frac{\theta }{\omega }=\frac{36\pi }{7.2\pi }=5s\]