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

Given, frequency n = 216 rpm `=(216)/(60)`rps
Length of one turn `=(1)/(2)cm`
`therefore` Number of rotations required to move a distance of `1.5cm`,
`N=("Distance")/("Length of one turn")=(1.5)/(1//12)=18`
Therefore, angular displacement `theta=2piN=2pixx18=36pi` rad
`therefore" Angular velocity of the bar",omega=2pin=2pixx(216)/(60)`
`=72pi"rad s"^(-1)`
`therefore" Required time, t"=("Angular displacement"(theta))/("Angular velocity"(omega))=(36pi)/(7.2pi)=5s`