Question

A threaded rod with 12 turns per cm and a diameter of 1.18cm is mounted horizontally. A bar with a threaded hole to match the rod is screwed onto the rod. The bar spins at a rate of 216rev/min. How long will it take to move 1.5cm along the rod?

Answer 1

Hint: First, we need to calculate the number of turns in $1.5\,cm$ . Then calculate the revolutions of the bar in one second. If we divide the total number of turns by the number of turns travelled by the bar in one second, we will get the total time taken.

Complete step by step answer:
Given that a threaded road has 12 turns per cm.
The diameter of the rod is given as $1.18\,cm$.
A bar with a threaded hole to match the rod is screwed on the road. This is similar to the case of nut and bolt.
The spin rate of the bar is given as $216\,\,{\text{rev}}/\min $.
We need to find the time taken to move a distance of $1.5\, cm$ along the rod.
Since $1\,cm$ contains 12 turns,$1.5\,cm$ will contain
$1.5\, \times 12 = 18\,{\text{turns}}$
So, we need to find the time taken for 18 turns.
Now let us calculate the number of revolutions per second of the bar.
Since $216\,{\text{rev}}$ happened in $1\min $, revolution in $1\,s$ can be calculated by dividing this value by 60.
$ \Rightarrow \dfrac{{216\,{\text{rev}}}}{{60\,s}} = 3.6\,rev/s$
This means in $1\,s$ the bar will undergo $3.6\,{\text{rev}}$.
We know one revolution corresponds to one turn. So, we can say that in $1\,s$ the bar will cover $3.6\,{\text{turns}}$
So, the time taken to cover $18\,{\text{turns}}$ will be the total number of turns divided by the number of turns in one second.
$ \Rightarrow \dfrac{{18}}{{3.6}} = 5\,s$
$t = 5\,s$

The time required to move along the rod is 5sec.

Note:
Alternative solution:
We can also calculate this by converting the number of rotations into the total angle of turning $\theta $.
$ \Rightarrow \theta = 18 \times 2\pi \,rad$
$ \Rightarrow \theta = 36\pi \,rad$
If n is the number of revolutions per second then the angular speed $\omega $ can be calculated as
$\omega = 2\pi n$
$ \Rightarrow \omega = 2\pi \dfrac{{216}}{{60}}$
$ \Rightarrow \omega = 7.2\pi \,rad/s$
We know that angular speed is the ratio of angular displacement by time.
$ \Rightarrow \omega = \dfrac{\theta }{t}$
Thus, time is given as
$ \Rightarrow t = \dfrac{\theta }{\omega }$
On substituting the values, we get
$t = \dfrac{{36\pi }}{{7.2\pi }}$
$\therefore t = 5\,s$