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 ?

Explanation

As the distance between two consecutive threads on the rod, i.e., pitch is =(1/12) cm, so the number of rotations to be made in moving a distance of 1.5cm.

`n=("distance")/("pitch")=(1.5)/((1//12)=18`
So, angu lar displacement of bar
`theta=2pin=2pixx18=36pi` rad
And as angu lar speed of bar
`omega=(2pin)/(t)=(2pixx216)/(60)`
=7.2 `pi` rad/s So, time taken by bar to travel a distance of 1.5 cm on the rod (as `theta=omegat` with `alpha=0`)
`t=(theta)/(omega)=(36pi)/(7.2pi)=5s`