Question

A dental drill accelerates from rest to $900\,rpm$ in $2\sec $ what is angular acceleration? How many revolutions does it make in coming to full speed?

Answer 1

Hint: In order to solve this question we need to understand angular acceleration. Angular acceleration is defined as the magnitude of torque exerted on the body per unit moment of inertia of an object which is revolving or it is defined as acceleration with which it rotates in a plane. Also like the translation motion we can use here three equations of motion provided that angular acceleration remains constant throughout the motion just like translation where we use the equation of motion only when the acceleration is constant.

Complete step by step answer:
Let the initial angular velocity be ${\omega _i} = 0$ (given in question).
Also the final angular velocity be, ${\omega _f} = 900\,rpm$ (given in question).
${\omega _f} = \dfrac{{900}}{{60}}\,rps$
$\Rightarrow {\omega _f} = 15rps$
where rpm stands for rotations per minute while rps stands for rotations per second.
Let the angular acceleration be, $\alpha $. And time taken to revolve is given as, $t = 2\sec $.So using equation of motion in rotation we get,
${\omega _f} = {\omega _i} + \alpha t$
Angular acceleration is,
$\alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t}$

Putting values we get,
$\alpha = \dfrac{{15 - 0}}{2}$
$\Rightarrow \alpha = 7.5\,rp{s^{ - 2}}$
So the angular acceleration is given as, $\alpha = 7.5\,rp{s^{ - 2}}$.
To find the number of rotations, let the number of rotations be by $\theta $. So using equation of motion we get,
\[\theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2}\]
Putting values we get,
\[\theta = (0)(2) + \dfrac{1}{2}(7.5){(2)^2}\]
\[\Rightarrow \theta = \dfrac{1}{2}(7.5)(4)\]
\[\therefore \theta = 15\,revolution\]

So it would take a $15$ revolution to come in full speed.

Note: It should be remembered that we can use equations of motion in rotation only when the angular acceleration remains constant through the revolution. If the angular acceleration varies then we would not use an equation of motion rather than we use a differential definition of angular acceleration to find the equation of motion.