Question

A dentist using a dental drill brings it from rest to maximum operating speed of 382,000 rpm in 3.0 s. Assume that the drill accelerates at a constant rate during this time.
A) What is the angular acceleration of the drill in rev/s2?
B) Find the number of revolutions the drill bit makes during the 2.6 s time interval. rev

Answer 1

Answer:

a. $\alpha =2122.22\: rev/s^{2}$

b. $\Delta \theta =9,550.02\: rev$

Explanation:

The computation is shown below:

data provided in the question

The initial angular velocity  $\small \omega o$ = 0 rev/s,

f = Final angular velocity = $\small \omega$ = 382000 rpm i.e = $\frac{382,000}{60}$ = 6,366.67

And time = t = 3.0s

Based on the above information

a. For angular acceleration of drill

$\small \omega =\omega _{o}+\alpha t \\\\\ 6,366.67 = 0 + \alpha (3.0) \\\\ \alpha =2122.22\: rev/s^{2}$

b. For the number of revolutions

$\small \omega ^{2}-\omega _{o}^{2}=2\alpha \Delta \theta \\\\(6,366.67) ^{2}-(0)^{2}=2(2122.22) \Delta \theta \\\\ \Delta \theta =9,550.02\: rev$

We simply applied the above formulas for determining each parts